cpp-codingbat-12-Recursion-1.ipynb


{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
	"#include <iostream>\n",
	"#include <iomanip>\n",
	"#include <vector>\n",
	"\n",
	"using namespace std;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
	"namespace testns { template<class A> void printTest( string str, A run, A expected )\n",
	"{\n",
	"    if( run == expected )\n",
	"        cout << boolalpha << \"\\033[1;32m\"\n",
	"            << \"Expected: \" << setw(60) << left << str << \" Run: \" << setw(20) << left << run << \"\\t\\tOK\"\n",
	"            << \"\\033[0m\" << endl;\n",
	"    else\n",
	"        cerr << boolalpha << \"\\033[1;31m\"\n",
	"            << \"Expected: \" << setw(60) << left << str << \" Run: \" << setw(20) << left << run << \"\\t\\tX\"\n",
	"            << \"\\033[0m\" << endl;\n",
	"} } using namespace testns;\n",
	"void printTest( string str, string run, const char* expected ){ printTest( str, \"\\\"\"+run+\"\\\"\", \"\\\"\"+string( expected )+\"\\\"\" ); }\n",
	"template<class A> void printTest( string str, vector<A> run, vector<A> expected )\n",
	"{\n",
	"    string runStr = \"[\"; for( A a : run ) runStr += to_string( a ) + \", \";\n",
	"    if( runStr.length() > 1 ) runStr = runStr.substr(0, runStr.length()-2); runStr += \"]\";\n",
	"    string expStr = \"[\"; for( A a : expected ) expStr += to_string( a ) + \", \";\n",
	"    if( expStr.length() > 1 ) expStr = expStr.substr(0, expStr.length()-2); expStr += \"]\";\n",
	"    printTest( str, runStr, expStr );\n",
	"}"
   ]
  },
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > factorial"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given n of 1 or more, return the factorial of n, which is n * (n-1) * (n-2) ...\n", "1. Compute the result recursively (without loops).\n", "\n", "factorial(1) \u2192 1\n", "factorial(2) \u2192 2\n", "factorial(3) \u2192 6\n", "*/\n", "\n", "int factorial(int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"factorial(1) \u2192 1\", factorial(1), 1 );\n", "printTest( \"factorial(2) \u2192 2\", factorial(2), 2 );\n", "printTest( \"factorial(3) \u2192 6\", factorial(3), 6 );\n", "printTest( \"factorial(4) \u2192 24\", factorial(4), 24 );\n", "printTest( \"factorial(5) \u2192 120\", factorial(5), 120 );\n", "printTest( \"factorial(6) \u2192 720\", factorial(6), 720 );\n", "printTest( \"factorial(7) \u2192 5040\", factorial(7), 5040 );\n", "printTest( \"factorial(8) \u2192 40320\", factorial(8), 40320 );\n", "printTest( \"factorial(12) \u2192 479001600\", factorial(12), 479001600 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > bunnyEars"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "We have a number of bunnies and each bunny has two big floppy ears. We want to\n", "compute the total number of ears across all the bunnies recursively (without\n", "loops or multiplication).\n", "\n", "bunnyEars(0) \u2192 0\n", "bunnyEars(1) \u2192 2\n", "bunnyEars(2) \u2192 4\n", "*/\n", "\n", "int bunnyEars(int bunnies) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"bunnyEars(0) \u2192 0\", bunnyEars(0), 0 );\n", "printTest( \"bunnyEars(1) \u2192 2\", bunnyEars(1), 2 );\n", "printTest( \"bunnyEars(2) \u2192 4\", bunnyEars(2), 4 );\n", "printTest( \"bunnyEars(3) \u2192 6\", bunnyEars(3), 6 );\n", "printTest( \"bunnyEars(4) \u2192 8\", bunnyEars(4), 8 );\n", "printTest( \"bunnyEars(5) \u2192 10\", bunnyEars(5), 10 );\n", "printTest( \"bunnyEars(12) \u2192 24\", bunnyEars(12), 24 );\n", "printTest( \"bunnyEars(50) \u2192 100\", bunnyEars(50), 100 );\n", "printTest( \"bunnyEars(234) \u2192 468\", bunnyEars(234), 468 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > fibonacci"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "The fibonacci sequence is a famous bit of mathematics, and it happens to have a\n", "recursive definition. The first two values in the sequence are 0 and 1\n", "(essentially 2 base cases). Each subsequent value is the sum of the previous two\n", "values, so the whole sequence is: 0, 1, 1, 2, 3, 5, 8, 13, 21 and so on. Define\n", "a recursive fibonacci(n) method that returns the nth fibonacci number, with n=0\n", "representing the start of the sequence.\n", "\n", "fibonacci(0) \u2192 0\n", "fibonacci(1) \u2192 1\n", "fibonacci(2) \u2192 1\n", "*/\n", "\n", "int fibonacci(int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"fibonacci(0) \u2192 0\", fibonacci(0), 0 );\n", "printTest( \"fibonacci(1) \u2192 1\", fibonacci(1), 1 );\n", "printTest( \"fibonacci(2) \u2192 1\", fibonacci(2), 1 );\n", "printTest( \"fibonacci(3) \u2192 2\", fibonacci(3), 2 );\n", "printTest( \"fibonacci(4) \u2192 3\", fibonacci(4), 3 );\n", "printTest( \"fibonacci(5) \u2192 5\", fibonacci(5), 5 );\n", "printTest( \"fibonacci(6) \u2192 8\", fibonacci(6), 8 );\n", "printTest( \"fibonacci(7) \u2192 13\", fibonacci(7), 13 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > bunnyEars2"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "We have bunnies standing in a line, numbered 1, 2, ... The odd bunnies (1, 3,\n", "..) have the normal 2 ears. The even bunnies (2, 4, ..) we\u02bcll say have 3 ears,\n", "because they each have a raised foot. Recursively return the number of \"ears\" in\n", "the bunny line 1, 2, ... n (without loops or multiplication).\n", "\n", "bunnyEars2(0) \u2192 0\n", "bunnyEars2(1) \u2192 2\n", "bunnyEars2(2) \u2192 5\n", "*/\n", "\n", "int bunnyEars2(int bunnies) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"bunnyEars2(0) \u2192 0\", bunnyEars2(0), 0 );\n", "printTest( \"bunnyEars2(1) \u2192 2\", bunnyEars2(1), 2 );\n", "printTest( \"bunnyEars2(2) \u2192 5\", bunnyEars2(2), 5 );\n", "printTest( \"bunnyEars2(3) \u2192 7\", bunnyEars2(3), 7 );\n", "printTest( \"bunnyEars2(4) \u2192 10\", bunnyEars2(4), 10 );\n", "printTest( \"bunnyEars2(5) \u2192 12\", bunnyEars2(5), 12 );\n", "printTest( \"bunnyEars2(6) \u2192 15\", bunnyEars2(6), 15 );\n", "printTest( \"bunnyEars2(10) \u2192 25\", bunnyEars2(10), 25 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > triangle"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "We have triangle made of blocks. The topmost row has 1 block, the next row down\n", "has 2 blocks, the next row has 3 blocks, and so on. Compute recursively (no\n", "loops or multiplication) the total number of blocks in such a triangle with the\n", "given number of rows.\n", "\n", "triangle(0) \u2192 0\n", "triangle(1) \u2192 1\n", "triangle(2) \u2192 3\n", "*/\n", "\n", "int triangle(int rows) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"triangle(0) \u2192 0\", triangle(0), 0 );\n", "printTest( \"triangle(1) \u2192 1\", triangle(1), 1 );\n", "printTest( \"triangle(2) \u2192 3\", triangle(2), 3 );\n", "printTest( \"triangle(3) \u2192 6\", triangle(3), 6 );\n", "printTest( \"triangle(4) \u2192 10\", triangle(4), 10 );\n", "printTest( \"triangle(5) \u2192 15\", triangle(5), 15 );\n", "printTest( \"triangle(6) \u2192 21\", triangle(6), 21 );\n", "printTest( \"triangle(7) \u2192 28\", triangle(7), 28 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > sumDigits"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a non-negative int n, return the sum of its digits recursively (no loops).\n", "Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide\n", "(/) by 10 removes the rightmost digit (126 / 10 is 12).\n", "\n", "sumDigits(126) \u2192 9\n", "sumDigits(49) \u2192 13\n", "sumDigits(12) \u2192 3\n", "*/\n", "\n", "int sumDigits(int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"sumDigits(126) \u2192 9\", sumDigits(126), 9 );\n", "printTest( \"sumDigits(49) \u2192 13\", sumDigits(49), 13 );\n", "printTest( \"sumDigits(12) \u2192 3\", sumDigits(12), 3 );\n", "printTest( \"sumDigits(10) \u2192 1\", sumDigits(10), 1 );\n", "printTest( \"sumDigits(1) \u2192 1\", sumDigits(1), 1 );\n", "printTest( \"sumDigits(0) \u2192 0\", sumDigits(0), 0 );\n", "printTest( \"sumDigits(730) \u2192 10\", sumDigits(730), 10 );\n", "printTest( \"sumDigits(1111) \u2192 4\", sumDigits(1111), 4 );\n", "printTest( \"sumDigits(11111) \u2192 5\", sumDigits(11111), 5 );\n", "printTest( \"sumDigits(10110) \u2192 3\", sumDigits(10110), 3 );\n", "printTest( \"sumDigits(235) \u2192 10\", sumDigits(235), 10 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > count7"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a non-negative int n, return the count of the occurrences of 7 as a digit,\n", "so for example 717 yields 2. (no loops). Note that mod (%) by 10 yields the\n", "rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost\n", "digit (126 / 10 is 12).\n", "\n", "count7(717) \u2192 2\n", "count7(7) \u2192 1\n", "count7(123) \u2192 0\n", "*/\n", "\n", "int count7(int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"count7(717) \u2192 2\", count7(717), 2 );\n", "printTest( \"count7(7) \u2192 1\", count7(7), 1 );\n", "printTest( \"count7(123) \u2192 0\", count7(123), 0 );\n", "printTest( \"count7(77) \u2192 2\", count7(77), 2 );\n", "printTest( \"count7(7123) \u2192 1\", count7(7123), 1 );\n", "printTest( \"count7(771237) \u2192 3\", count7(771237), 3 );\n", "printTest( \"count7(771737) \u2192 4\", count7(771737), 4 );\n", "printTest( \"count7(47571) \u2192 2\", count7(47571), 2 );\n", "printTest( \"count7(777777) \u2192 6\", count7(777777), 6 );\n", "printTest( \"count7(70701277) \u2192 4\", count7(70701277), 4 );\n", "printTest( \"count7(777576197) \u2192 5\", count7(777576197), 5 );\n", "printTest( \"count7(99999) \u2192 0\", count7(99999), 0 );\n", "printTest( \"count7(99799) \u2192 1\", count7(99799), 1 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > count8"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a non-negative int n, compute recursively (no loops) the count of the\n", "occurrences of 8 as a digit, except that an 8 with another 8 immediately to its\n", "left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the\n", "rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost\n", "digit (126 / 10 is 12).\n", "\n", "count8(8) \u2192 1\n", "count8(818) \u2192 2\n", "count8(8818) \u2192 4\n", "*/\n", "\n", "int count8(int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"count8(8) \u2192 1\", count8(8), 1 );\n", "printTest( \"count8(818) \u2192 2\", count8(818), 2 );\n", "printTest( \"count8(8818) \u2192 4\", count8(8818), 4 );\n", "printTest( \"count8(8088) \u2192 4\", count8(8088), 4 );\n", "printTest( \"count8(123) \u2192 0\", count8(123), 0 );\n", "printTest( \"count8(81238) \u2192 2\", count8(81238), 2 );\n", "printTest( \"count8(88788) \u2192 6\", count8(88788), 6 );\n", "printTest( \"count8(8234) \u2192 1\", count8(8234), 1 );\n", "printTest( \"count8(2348) \u2192 1\", count8(2348), 1 );\n", "printTest( \"count8(23884) \u2192 3\", count8(23884), 3 );\n", "printTest( \"count8(0) \u2192 0\", count8(0), 0 );\n", "printTest( \"count8(1818188) \u2192 5\", count8(1818188), 5 );\n", "printTest( \"count8(8818181) \u2192 5\", count8(8818181), 5 );\n", "printTest( \"count8(1080) \u2192 1\", count8(1080), 1 );\n", "printTest( \"count8(188) \u2192 3\", count8(188), 3 );\n", "printTest( \"count8(88888) \u2192 9\", count8(88888), 9 );\n", "printTest( \"count8(9898) \u2192 2\", count8(9898), 2 );\n", "printTest( \"count8(78) \u2192 1\", count8(78), 1 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > powerN"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given base and n that are both 1 or more, compute recursively (no loops) the\n", "value of base to the n power, so powerN(3, 2) is 9 (3 squared).\n", "\n", "powerN(3, 1) \u2192 3\n", "powerN(3, 2) \u2192 9\n", "powerN(3, 3) \u2192 27\n", "*/\n", "\n", "int powerN(int base, int n) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"powerN(3, 1) \u2192 3\", powerN(3, 1), 3 );\n", "printTest( \"powerN(3, 2) \u2192 9\", powerN(3, 2), 9 );\n", "printTest( \"powerN(3, 3) \u2192 27\", powerN(3, 3), 27 );\n", "printTest( \"powerN(2, 1) \u2192 2\", powerN(2, 1), 2 );\n", "printTest( \"powerN(2, 2) \u2192 4\", powerN(2, 2), 4 );\n", "printTest( \"powerN(2, 3) \u2192 8\", powerN(2, 3), 8 );\n", "printTest( \"powerN(2, 4) \u2192 16\", powerN(2, 4), 16 );\n", "printTest( \"powerN(2, 5) \u2192 32\", powerN(2, 5), 32 );\n", "printTest( \"powerN(10, 1) \u2192 10\", powerN(10, 1), 10 );\n", "printTest( \"powerN(10, 2) \u2192 100\", powerN(10, 2), 100 );\n", "printTest( \"powerN(10, 3) \u2192 1000\", powerN(10, 3), 1000 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > countX"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively (no loops) the number of lowercase 'x' chars\n", "in the string.\n", "\n", "countX(\"xxhixx\") \u2192 4\n", "countX(\"xhixhix\") \u2192 3\n", "countX(\"hi\") \u2192 0\n", "*/\n", "\n", "int countX(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"countX(\\\"xxhixx\\\") \u2192 4\", countX(\"xxhixx\"), 4 );\n", "printTest( \"countX(\\\"xhixhix\\\") \u2192 3\", countX(\"xhixhix\"), 3 );\n", "printTest( \"countX(\\\"hi\\\") \u2192 0\", countX(\"hi\"), 0 );\n", "printTest( \"countX(\\\"h\\\") \u2192 0\", countX(\"h\"), 0 );\n", "printTest( \"countX(\\\"x\\\") \u2192 1\", countX(\"x\"), 1 );\n", "printTest( \"countX(\\\"\\\") \u2192 0\", countX(\"\"), 0 );\n", "printTest( \"countX(\\\"hihi\\\") \u2192 0\", countX(\"hihi\"), 0 );\n", "printTest( \"countX(\\\"hiAAhi12hi\\\") \u2192 0\", countX(\"hiAAhi12hi\"), 0 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > countHi"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively (no loops) the number of times lowercase\n", "\"hi\" appears in the string.\n", "\n", "countHi(\"xxhixx\") \u2192 1\n", "countHi(\"xhixhix\") \u2192 2\n", "countHi(\"hi\") \u2192 1\n", "*/\n", "\n", "int countHi(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"countHi(\\\"xxhixx\\\") \u2192 1\", countHi(\"xxhixx\"), 1 );\n", "printTest( \"countHi(\\\"xhixhix\\\") \u2192 2\", countHi(\"xhixhix\"), 2 );\n", "printTest( \"countHi(\\\"hi\\\") \u2192 1\", countHi(\"hi\"), 1 );\n", "printTest( \"countHi(\\\"hihih\\\") \u2192 2\", countHi(\"hihih\"), 2 );\n", "printTest( \"countHi(\\\"h\\\") \u2192 0\", countHi(\"h\"), 0 );\n", "printTest( \"countHi(\\\"\\\") \u2192 0\", countHi(\"\"), 0 );\n", "printTest( \"countHi(\\\"ihihihihih\\\") \u2192 4\", countHi(\"ihihihihih\"), 4 );\n", "printTest( \"countHi(\\\"ihihihihihi\\\") \u2192 5\", countHi(\"ihihihihihi\"), 5 );\n", "printTest( \"countHi(\\\"hiAAhi12hi\\\") \u2192 3\", countHi(\"hiAAhi12hi\"), 3 );\n", "printTest( \"countHi(\\\"xhixhxihihhhih\\\") \u2192 3\", countHi(\"xhixhxihihhhih\"), 3 );\n", "printTest( \"countHi(\\\"ship\\\") \u2192 1\", countHi(\"ship\"), 1 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > changeXY"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively (no loops) a new string where all the\n", "lowercase 'x' chars have been changed to 'y' chars.\n", "\n", "changeXY(\"codex\") \u2192 \"codey\"\n", "changeXY(\"xxhixx\") \u2192 \"yyhiyy\"\n", "changeXY(\"xhixhix\") \u2192 \"yhiyhiy\"\n", "*/\n", "\n", "string changeXY(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"changeXY(\\\"codex\\\") \u2192 \\\"codey\\\"\", changeXY(\"codex\"), \"codey\" );\n", "printTest( \"changeXY(\\\"xxhixx\\\") \u2192 \\\"yyhiyy\\\"\", changeXY(\"xxhixx\"), \"yyhiyy\" );\n", "printTest( \"changeXY(\\\"xhixhix\\\") \u2192 \\\"yhiyhiy\\\"\", changeXY(\"xhixhix\"), \"yhiyhiy\" );\n", "printTest( \"changeXY(\\\"hiy\\\") \u2192 \\\"hiy\\\"\", changeXY(\"hiy\"), \"hiy\" );\n", "printTest( \"changeXY(\\\"h\\\") \u2192 \\\"h\\\"\", changeXY(\"h\"), \"h\" );\n", "printTest( \"changeXY(\\\"x\\\") \u2192 \\\"y\\\"\", changeXY(\"x\"), \"y\" );\n", "printTest( \"changeXY(\\\"\\\") \u2192 \\\"\\\"\", changeXY(\"\"), \"\" );\n", "printTest( \"changeXY(\\\"xxx\\\") \u2192 \\\"yyy\\\"\", changeXY(\"xxx\"), \"yyy\" );\n", "printTest( \"changeXY(\\\"yyhxyi\\\") \u2192 \\\"yyhyyi\\\"\", changeXY(\"yyhxyi\"), \"yyhyyi\" );\n", "printTest( \"changeXY(\\\"hihi\\\") \u2192 \\\"hihi\\\"\", changeXY(\"hihi\"), \"hihi\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > changePi"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively (no loops) a new string where all\n", "appearances of \"pi\" have been replaced by \"3.14\".\n", "\n", "changePi(\"xpix\") \u2192 \"x3.14x\"\n", "changePi(\"pipi\") \u2192 \"3.143.14\"\n", "changePi(\"pip\") \u2192 \"3.14p\"\n", "*/\n", "\n", "string changePi(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"changePi(\\\"xpix\\\") \u2192 \\\"x3.14x\\\"\", changePi(\"xpix\"), \"x3.14x\" );\n", "printTest( \"changePi(\\\"pipi\\\") \u2192 \\\"3.143.14\\\"\", changePi(\"pipi\"), \"3.143.14\" );\n", "printTest( \"changePi(\\\"pip\\\") \u2192 \\\"3.14p\\\"\", changePi(\"pip\"), \"3.14p\" );\n", "printTest( \"changePi(\\\"pi\\\") \u2192 \\\"3.14\\\"\", changePi(\"pi\"), \"3.14\" );\n", "printTest( \"changePi(\\\"hip\\\") \u2192 \\\"hip\\\"\", changePi(\"hip\"), \"hip\" );\n", "printTest( \"changePi(\\\"p\\\") \u2192 \\\"p\\\"\", changePi(\"p\"), \"p\" );\n", "printTest( \"changePi(\\\"x\\\") \u2192 \\\"x\\\"\", changePi(\"x\"), \"x\" );\n", "printTest( \"changePi(\\\"\\\") \u2192 \\\"\\\"\", changePi(\"\"), \"\" );\n", "printTest( \"changePi(\\\"pixx\\\") \u2192 \\\"3.14xx\\\"\", changePi(\"pixx\"), \"3.14xx\" );\n", "printTest( \"changePi(\\\"xyzzy\\\") \u2192 \\\"xyzzy\\\"\", changePi(\"xyzzy\"), \"xyzzy\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > noX"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively a new string where all the 'x' chars have\n", "been removed.\n", "\n", "noX(\"xaxb\") \u2192 \"ab\"\n", "noX(\"abc\") \u2192 \"abc\"\n", "noX(\"xx\") \u2192 \"\"\n", "*/\n", "\n", "string noX(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"noX(\\\"xaxb\\\") \u2192 \\\"ab\\\"\", noX(\"xaxb\"), \"ab\" );\n", "printTest( \"noX(\\\"abc\\\") \u2192 \\\"abc\\\"\", noX(\"abc\"), \"abc\" );\n", "printTest( \"noX(\\\"xx\\\") \u2192 \\\"\\\"\", noX(\"xx\"), \"\" );\n", "printTest( \"noX(\\\"\\\") \u2192 \\\"\\\"\", noX(\"\"), \"\" );\n", "printTest( \"noX(\\\"axxbxx\\\") \u2192 \\\"ab\\\"\", noX(\"axxbxx\"), \"ab\" );\n", "printTest( \"noX(\\\"Hellox\\\") \u2192 \\\"Hello\\\"\", noX(\"Hellox\"), \"Hello\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > array6"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given an array of ints, compute recursively if the array contains a 6. We\u02bcll use\n", "the convention of considering only the part of the array that begins at the\n", "given index. In this way, a recursive call can pass index+1 to move down the\n", "array. The initial call will pass in index as 0.\n", "\n", "array6([1, 6, 4], 0) \u2192 true\n", "array6([1, 4], 0) \u2192 false\n", "array6([6], 0) \u2192 true\n", "*/\n", "\n", "bool array6(vector<int> nums, int index) {\n", "    return false;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"array6([1, 6, 4], 0) \u2192 true\", array6({1, 6, 4}, 0), true );\n", "printTest( \"array6([1, 4], 0) \u2192 false\", array6({1, 4}, 0), false );\n", "printTest( \"array6([6], 0) \u2192 true\", array6({6}, 0), true );\n", "printTest( \"array6([], 0) \u2192 false\", array6({}, 0), false );\n", "printTest( \"array6([6, 2, 2], 0) \u2192 true\", array6({6, 2, 2}, 0), true );\n", "printTest( \"array6([2, 5], 0) \u2192 false\", array6({2, 5}, 0), false );\n", "printTest( \"array6([1, 9, 4, 6, 6], 0) \u2192 true\", array6({1, 9, 4, 6, 6}, 0), true );\n", "printTest( \"array6([2, 5, 6], 0) \u2192 true\", array6({2, 5, 6}, 0), true );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > array11"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given an array of ints, compute recursively the number of times that the value\n", "11 appears in the array. We\u02bcll use the convention of considering only the part\n", "of the array that begins at the given index. In this way, a recursive call can\n", "pass index+1 to move down the array. The initial call will pass in index as 0.\n", "\n", "array11([1, 2, 11], 0) \u2192 1\n", "array11([11, 11], 0) \u2192 2\n", "array11([1, 2, 3, 4], 0) \u2192 0\n", "*/\n", "\n", "int array11(vector<int> nums, int index) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"array11([1, 2, 11], 0) \u2192 1\", array11({1, 2, 11}, 0), 1 );\n", "printTest( \"array11([11, 11], 0) \u2192 2\", array11({11, 11}, 0), 2 );\n", "printTest( \"array11([1, 2, 3, 4], 0) \u2192 0\", array11({1, 2, 3, 4}, 0), 0 );\n", "printTest( \"array11([1, 11, 3, 11, 11], 0) \u2192 3\", array11({1, 11, 3, 11, 11}, 0), 3 );\n", "printTest( \"array11([11], 0) \u2192 1\", array11({11}, 0), 1 );\n", "printTest( \"array11([1], 0) \u2192 0\", array11({1}, 0), 0 );\n", "printTest( \"array11([], 0) \u2192 0\", array11({}, 0), 0 );\n", "printTest( \"array11([11, 2, 3, 4, 11, 5], 0) \u2192 2\", array11({11, 2, 3, 4, 11, 5}, 0), 2 );\n", "printTest( \"array11([11, 5, 11], 0) \u2192 2\", array11({11, 5, 11}, 0), 2 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > array220"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given an array of ints, compute recursively if the array contains somewhere a\n", "value followed in the array by that value times 10. We\u02bcll use the convention of\n", "considering only the part of the array that begins at the given index. In this\n", "way, a recursive call can pass index+1 to move down the array. The initial call\n", "will pass in index as 0.\n", "\n", "array220([1, 2, 20], 0) \u2192 true\n", "array220([3, 30], 0) \u2192 true\n", "array220([3], 0) \u2192 false\n", "*/\n", "\n", "bool array220(vector<int> nums, int index) {\n", "    return false;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"array220([1, 2, 20], 0) \u2192 true\", array220({1, 2, 20}, 0), true );\n", "printTest( \"array220([3, 30], 0) \u2192 true\", array220({3, 30}, 0), true );\n", "printTest( \"array220([3], 0) \u2192 false\", array220({3}, 0), false );\n", "printTest( \"array220([], 0) \u2192 false\", array220({}, 0), false );\n", "printTest( \"array220([3, 3, 30, 4], 0) \u2192 true\", array220({3, 3, 30, 4}, 0), true );\n", "printTest( \"array220([2, 19, 4], 0) \u2192 false\", array220({2, 19, 4}, 0), false );\n", "printTest( \"array220([20, 2, 21], 0) \u2192 false\", array220({20, 2, 21}, 0), false );\n", "printTest( \"array220([20, 2, 21, 210], 0) \u2192 true\", array220({20, 2, 21, 210}, 0), true );\n", "printTest( \"array220([2, 200, 2000], 0) \u2192 true\", array220({2, 200, 2000}, 0), true );\n", "printTest( \"array220([0, 0], 0) \u2192 true\", array220({0, 0}, 0), true );\n", "printTest( \"array220([1, 2, 3, 4, 5, 6], 0) \u2192 false\", array220({1, 2, 3, 4, 5, 6}, 0), false );\n", "printTest( \"array220([1, 2, 3, 4, 5, 50, 6], 0) \u2192 true\", array220({1, 2, 3, 4, 5, 50, 6}, 0), true );\n", "printTest( \"array220([1, 2, 3, 4, 5, 51, 6], 0) \u2192 false\", array220({1, 2, 3, 4, 5, 51, 6}, 0), false );\n", "printTest( \"array220([1, 2, 3, 4, 4, 50, 500, 6], 0) \u2192 true\", array220({1, 2, 3, 4, 4, 50, 500, 6}, 0), true );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > allStar"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively a new string where all the adjacent chars\n", "are now separated by a \"*\".\n", "\n", "allStar(\"hello\") \u2192 \"h*e*l*l*o\"\n", "allStar(\"abc\") \u2192 \"a*b*c\"\n", "allStar(\"ab\") \u2192 \"a*b\"\n", "*/\n", "\n", "string allStar(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"allStar(\\\"hello\\\") \u2192 \\\"h*e*l*l*o\\\"\", allStar(\"hello\"), \"h*e*l*l*o\" );\n", "printTest( \"allStar(\\\"abc\\\") \u2192 \\\"a*b*c\\\"\", allStar(\"abc\"), \"a*b*c\" );\n", "printTest( \"allStar(\\\"ab\\\") \u2192 \\\"a*b\\\"\", allStar(\"ab\"), \"a*b\" );\n", "printTest( \"allStar(\\\"a\\\") \u2192 \\\"a\\\"\", allStar(\"a\"), \"a\" );\n", "printTest( \"allStar(\\\"\\\") \u2192 \\\"\\\"\", allStar(\"\"), \"\" );\n", "printTest( \"allStar(\\\"3.14\\\") \u2192 \\\"3*.*1*4\\\"\", allStar(\"3.14\"), \"3*.*1*4\" );\n", "printTest( \"allStar(\\\"Chocolate\\\") \u2192 \\\"C*h*o*c*o*l*a*t*e\\\"\", allStar(\"Chocolate\"), \"C*h*o*c*o*l*a*t*e\" );\n", "printTest( \"allStar(\\\"1234\\\") \u2192 \\\"1*2*3*4\\\"\", allStar(\"1234\"), \"1*2*3*4\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > pairStar"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively a new string where identical chars that are\n", "adjacent in the original string are separated from each other by a \"*\".\n", "\n", "pairStar(\"hello\") \u2192 \"hel*lo\"\n", "pairStar(\"xxyy\") \u2192 \"x*xy*y\"\n", "pairStar(\"aaaa\") \u2192 \"a*a*a*a\"\n", "*/\n", "\n", "string pairStar(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"pairStar(\\\"hello\\\") \u2192 \\\"hel*lo\\\"\", pairStar(\"hello\"), \"hel*lo\" );\n", "printTest( \"pairStar(\\\"xxyy\\\") \u2192 \\\"x*xy*y\\\"\", pairStar(\"xxyy\"), \"x*xy*y\" );\n", "printTest( \"pairStar(\\\"aaaa\\\") \u2192 \\\"a*a*a*a\\\"\", pairStar(\"aaaa\"), \"a*a*a*a\" );\n", "printTest( \"pairStar(\\\"aaab\\\") \u2192 \\\"a*a*ab\\\"\", pairStar(\"aaab\"), \"a*a*ab\" );\n", "printTest( \"pairStar(\\\"aa\\\") \u2192 \\\"a*a\\\"\", pairStar(\"aa\"), \"a*a\" );\n", "printTest( \"pairStar(\\\"a\\\") \u2192 \\\"a\\\"\", pairStar(\"a\"), \"a\" );\n", "printTest( \"pairStar(\\\"\\\") \u2192 \\\"\\\"\", pairStar(\"\"), \"\" );\n", "printTest( \"pairStar(\\\"noadjacent\\\") \u2192 \\\"noadjacent\\\"\", pairStar(\"noadjacent\"), \"noadjacent\" );\n", "printTest( \"pairStar(\\\"abba\\\") \u2192 \\\"ab*ba\\\"\", pairStar(\"abba\"), \"ab*ba\" );\n", "printTest( \"pairStar(\\\"abbba\\\") \u2192 \\\"ab*b*ba\\\"\", pairStar(\"abbba\"), \"ab*b*ba\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > endX"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively a new string where all the lowercase 'x'\n", "chars have been moved to the end of the string.\n", "\n", "endX(\"xxre\") \u2192 \"rexx\"\n", "endX(\"xxhixx\") \u2192 \"hixxxx\"\n", "endX(\"xhixhix\") \u2192 \"hihixxx\"\n", "*/\n", "\n", "string endX(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"endX(\\\"xxre\\\") \u2192 \\\"rexx\\\"\", endX(\"xxre\"), \"rexx\" );\n", "printTest( \"endX(\\\"xxhixx\\\") \u2192 \\\"hixxxx\\\"\", endX(\"xxhixx\"), \"hixxxx\" );\n", "printTest( \"endX(\\\"xhixhix\\\") \u2192 \\\"hihixxx\\\"\", endX(\"xhixhix\"), \"hihixxx\" );\n", "printTest( \"endX(\\\"hiy\\\") \u2192 \\\"hiy\\\"\", endX(\"hiy\"), \"hiy\" );\n", "printTest( \"endX(\\\"h\\\") \u2192 \\\"h\\\"\", endX(\"h\"), \"h\" );\n", "printTest( \"endX(\\\"x\\\") \u2192 \\\"x\\\"\", endX(\"x\"), \"x\" );\n", "printTest( \"endX(\\\"xx\\\") \u2192 \\\"xx\\\"\", endX(\"xx\"), \"xx\" );\n", "printTest( \"endX(\\\"\\\") \u2192 \\\"\\\"\", endX(\"\"), \"\" );\n", "printTest( \"endX(\\\"bxx\\\") \u2192 \\\"bxx\\\"\", endX(\"bxx\"), \"bxx\" );\n", "printTest( \"endX(\\\"bxax\\\") \u2192 \\\"baxx\\\"\", endX(\"bxax\"), \"baxx\" );\n", "printTest( \"endX(\\\"axaxax\\\") \u2192 \\\"aaaxxx\\\"\", endX(\"axaxax\"), \"aaaxxx\" );\n", "printTest( \"endX(\\\"xxhxi\\\") \u2192 \\\"hixxx\\\"\", endX(\"xxhxi\"), \"hixxx\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > countPairs"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "We\u02bcll say that a \"pair\" in a string is two instances of a char separated by a\n", "char. So \"AxA\" the A\u02bcs make a pair. Pair\u02bcs can overlap, so \"AxAxA\" contains 3\n", "pairs -- 2 for A and 1 for x. Recursively compute the number of pairs in the\n", "given string.\n", "\n", "countPairs(\"axa\") \u2192 1\n", "countPairs(\"axax\") \u2192 2\n", "countPairs(\"axbx\") \u2192 1\n", "*/\n", "\n", "int countPairs(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"countPairs(\\\"axa\\\") \u2192 1\", countPairs(\"axa\"), 1 );\n", "printTest( \"countPairs(\\\"axax\\\") \u2192 2\", countPairs(\"axax\"), 2 );\n", "printTest( \"countPairs(\\\"axbx\\\") \u2192 1\", countPairs(\"axbx\"), 1 );\n", "printTest( \"countPairs(\\\"hi\\\") \u2192 0\", countPairs(\"hi\"), 0 );\n", "printTest( \"countPairs(\\\"hihih\\\") \u2192 3\", countPairs(\"hihih\"), 3 );\n", "printTest( \"countPairs(\\\"ihihhh\\\") \u2192 3\", countPairs(\"ihihhh\"), 3 );\n", "printTest( \"countPairs(\\\"ihjxhh\\\") \u2192 0\", countPairs(\"ihjxhh\"), 0 );\n", "printTest( \"countPairs(\\\"\\\") \u2192 0\", countPairs(\"\"), 0 );\n", "printTest( \"countPairs(\\\"a\\\") \u2192 0\", countPairs(\"a\"), 0 );\n", "printTest( \"countPairs(\\\"aa\\\") \u2192 0\", countPairs(\"aa\"), 0 );\n", "printTest( \"countPairs(\\\"aaa\\\") \u2192 1\", countPairs(\"aaa\"), 1 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > countAbc"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Count recursively the total number of \"abc\" and \"aba\" substrings that appear in\n", "the given string.\n", "\n", "countAbc(\"abc\") \u2192 1\n", "countAbc(\"abcxxabc\") \u2192 2\n", "countAbc(\"abaxxaba\") \u2192 2\n", "*/\n", "\n", "int countAbc(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"countAbc(\\\"abc\\\") \u2192 1\", countAbc(\"abc\"), 1 );\n", "printTest( \"countAbc(\\\"abcxxabc\\\") \u2192 2\", countAbc(\"abcxxabc\"), 2 );\n", "printTest( \"countAbc(\\\"abaxxaba\\\") \u2192 2\", countAbc(\"abaxxaba\"), 2 );\n", "printTest( \"countAbc(\\\"ababc\\\") \u2192 2\", countAbc(\"ababc\"), 2 );\n", "printTest( \"countAbc(\\\"abxbc\\\") \u2192 0\", countAbc(\"abxbc\"), 0 );\n", "printTest( \"countAbc(\\\"aaabc\\\") \u2192 1\", countAbc(\"aaabc\"), 1 );\n", "printTest( \"countAbc(\\\"hello\\\") \u2192 0\", countAbc(\"hello\"), 0 );\n", "printTest( \"countAbc(\\\"\\\") \u2192 0\", countAbc(\"\"), 0 );\n", "printTest( \"countAbc(\\\"ab\\\") \u2192 0\", countAbc(\"ab\"), 0 );\n", "printTest( \"countAbc(\\\"aba\\\") \u2192 1\", countAbc(\"aba\"), 1 );\n", "printTest( \"countAbc(\\\"aca\\\") \u2192 0\", countAbc(\"aca\"), 0 );\n", "printTest( \"countAbc(\\\"aaa\\\") \u2192 0\", countAbc(\"aaa\"), 0 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > count11"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively (no loops) the number of \"11\" substrings in\n", "the string. The \"11\" substrings should not overlap.\n", "\n", "count11(\"11abc11\") \u2192 2\n", "count11(\"abc11x11x11\") \u2192 3\n", "count11(\"111\") \u2192 1\n", "*/\n", "\n", "int count11(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"count11(\\\"11abc11\\\") \u2192 2\", count11(\"11abc11\"), 2 );\n", "printTest( \"count11(\\\"abc11x11x11\\\") \u2192 3\", count11(\"abc11x11x11\"), 3 );\n", "printTest( \"count11(\\\"111\\\") \u2192 1\", count11(\"111\"), 1 );\n", "printTest( \"count11(\\\"1111\\\") \u2192 2\", count11(\"1111\"), 2 );\n", "printTest( \"count11(\\\"1\\\") \u2192 0\", count11(\"1\"), 0 );\n", "printTest( \"count11(\\\"\\\") \u2192 0\", count11(\"\"), 0 );\n", "printTest( \"count11(\\\"hi\\\") \u2192 0\", count11(\"hi\"), 0 );\n", "printTest( \"count11(\\\"11x111x1111\\\") \u2192 4\", count11(\"11x111x1111\"), 4 );\n", "printTest( \"count11(\\\"1x111\\\") \u2192 1\", count11(\"1x111\"), 1 );\n", "printTest( \"count11(\\\"1Hello1\\\") \u2192 0\", count11(\"1Hello1\"), 0 );\n", "printTest( \"count11(\\\"Hello\\\") \u2192 0\", count11(\"Hello\"), 0 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > stringClean"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, return recursively a \"cleaned\" string where adjacent chars that\n", "are the same have been reduced to a single char. So \"yyzzza\" yields \"yza\".\n", "\n", "stringClean(\"yyzzza\") \u2192 \"yza\"\n", "stringClean(\"abbbcdd\") \u2192 \"abcd\"\n", "stringClean(\"Hello\") \u2192 \"Helo\"\n", "*/\n", "\n", "string stringClean(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"stringClean(\\\"yyzzza\\\") \u2192 \\\"yza\\\"\", stringClean(\"yyzzza\"), \"yza\" );\n", "printTest( \"stringClean(\\\"abbbcdd\\\") \u2192 \\\"abcd\\\"\", stringClean(\"abbbcdd\"), \"abcd\" );\n", "printTest( \"stringClean(\\\"Hello\\\") \u2192 \\\"Helo\\\"\", stringClean(\"Hello\"), \"Helo\" );\n", "printTest( \"stringClean(\\\"XXabcYY\\\") \u2192 \\\"XabcY\\\"\", stringClean(\"XXabcYY\"), \"XabcY\" );\n", "printTest( \"stringClean(\\\"112ab445\\\") \u2192 \\\"12ab45\\\"\", stringClean(\"112ab445\"), \"12ab45\" );\n", "printTest( \"stringClean(\\\"Hello Bookkeeper\\\") \u2192 \\\"Helo Bokeper\\\"\", stringClean(\"Hello Bookkeeper\"), \"Helo Bokeper\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > countHi2"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, compute recursively the number of times lowercase \"hi\" appears\n", "in the string, however do not count \"hi\" that have an 'x' immedately before\n", "them.\n", "\n", "countHi2(\"ahixhi\") \u2192 1\n", "countHi2(\"ahibhi\") \u2192 2\n", "countHi2(\"xhixhi\") \u2192 0\n", "*/\n", "\n", "int countHi2(string str) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"countHi2(\\\"ahixhi\\\") \u2192 1\", countHi2(\"ahixhi\"), 1 );\n", "printTest( \"countHi2(\\\"ahibhi\\\") \u2192 2\", countHi2(\"ahibhi\"), 2 );\n", "printTest( \"countHi2(\\\"xhixhi\\\") \u2192 0\", countHi2(\"xhixhi\"), 0 );\n", "printTest( \"countHi2(\\\"hixhi\\\") \u2192 1\", countHi2(\"hixhi\"), 1 );\n", "printTest( \"countHi2(\\\"hixhhi\\\") \u2192 2\", countHi2(\"hixhhi\"), 2 );\n", "printTest( \"countHi2(\\\"hihihi\\\") \u2192 3\", countHi2(\"hihihi\"), 3 );\n", "printTest( \"countHi2(\\\"hihihix\\\") \u2192 3\", countHi2(\"hihihix\"), 3 );\n", "printTest( \"countHi2(\\\"xhihihix\\\") \u2192 2\", countHi2(\"xhihihix\"), 2 );\n", "printTest( \"countHi2(\\\"xxhi\\\") \u2192 0\", countHi2(\"xxhi\"), 0 );\n", "printTest( \"countHi2(\\\"hixxhi\\\") \u2192 1\", countHi2(\"hixxhi\"), 1 );\n", "printTest( \"countHi2(\\\"hi\\\") \u2192 1\", countHi2(\"hi\"), 1 );\n", "printTest( \"countHi2(\\\"xxxx\\\") \u2192 0\", countHi2(\"xxxx\"), 0 );\n", "printTest( \"countHi2(\\\"h\\\") \u2192 0\", countHi2(\"h\"), 0 );\n", "printTest( \"countHi2(\\\"x\\\") \u2192 0\", countHi2(\"x\"), 0 );\n", "printTest( \"countHi2(\\\"\\\") \u2192 0\", countHi2(\"\"), 0 );\n", "printTest( \"countHi2(\\\"Hellohi\\\") \u2192 1\", countHi2(\"Hellohi\"), 1 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > parenBit"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string that contains a single pair of parenthesis, compute recursively a\n", "new string made of only of the parenthesis and their contents, so \"xyz(abc)123\"\n", "yields \"(abc)\".\n", "\n", "parenBit(\"xyz(abc)123\") \u2192 \"(abc)\"\n", "parenBit(\"x(hello)\") \u2192 \"(hello)\"\n", "parenBit(\"(xy)1\") \u2192 \"(xy)\"\n", "*/\n", "\n", "string parenBit(string str) {\n", "    return \"\";\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"parenBit(\\\"xyz(abc)123\\\") \u2192 \\\"(abc)\\\"\", parenBit(\"xyz(abc)123\"), \"(abc)\" );\n", "printTest( \"parenBit(\\\"x(hello)\\\") \u2192 \\\"(hello)\\\"\", parenBit(\"x(hello)\"), \"(hello)\" );\n", "printTest( \"parenBit(\\\"(xy)1\\\") \u2192 \\\"(xy)\\\"\", parenBit(\"(xy)1\"), \"(xy)\" );\n", "printTest( \"parenBit(\\\"not really (possible)\\\") \u2192 \\\"(possible)\\\"\", parenBit(\"not really (possible)\"), \"(possible)\" );\n", "printTest( \"parenBit(\\\"(abc)\\\") \u2192 \\\"(abc)\\\"\", parenBit(\"(abc)\"), \"(abc)\" );\n", "printTest( \"parenBit(\\\"(abc)xyz\\\") \u2192 \\\"(abc)\\\"\", parenBit(\"(abc)xyz\"), \"(abc)\" );\n", "printTest( \"parenBit(\\\"(abc)x\\\") \u2192 \\\"(abc)\\\"\", parenBit(\"(abc)x\"), \"(abc)\" );\n", "printTest( \"parenBit(\\\"(x)\\\") \u2192 \\\"(x)\\\"\", parenBit(\"(x)\"), \"(x)\" );\n", "printTest( \"parenBit(\\\"()\\\") \u2192 \\\"()\\\"\", parenBit(\"()\"), \"()\" );\n", "printTest( \"parenBit(\\\"res (ipsa) loquitor\\\") \u2192 \\\"(ipsa)\\\"\", parenBit(\"res (ipsa) loquitor\"), \"(ipsa)\" );\n", "printTest( \"parenBit(\\\"hello(not really)there\\\") \u2192 \\\"(not really)\\\"\", parenBit(\"hello(not really)there\"), \"(not really)\" );\n", "printTest( \"parenBit(\\\"ab(ab)ab\\\") \u2192 \\\"(ab)\\\"\", parenBit(\"ab(ab)ab\"), \"(ab)\" );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > nestParen"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string, return true if it is a nesting of zero or more pairs of\n", "parenthesis, like \"(())\" or \"((()))\". Suggestion: check the first and last\n", "chars, and then recur on what\u02bcs inside them.\n", "\n", "nestParen(\"(())\") \u2192 true\n", "nestParen(\"((()))\") \u2192 true\n", "nestParen(\"(((x))\") \u2192 false\n", "*/\n", "\n", "bool nestParen(string str) {\n", "    return false;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"nestParen(\\\"(())\\\") \u2192 true\", nestParen(\"(())\"), true );\n", "printTest( \"nestParen(\\\"((()))\\\") \u2192 true\", nestParen(\"((()))\"), true );\n", "printTest( \"nestParen(\\\"(((x))\\\") \u2192 false\", nestParen(\"(((x))\"), false );\n", "printTest( \"nestParen(\\\"((())\\\") \u2192 false\", nestParen(\"((())\"), false );\n", "printTest( \"nestParen(\\\"((()()\\\") \u2192 false\", nestParen(\"((()()\"), false );\n", "printTest( \"nestParen(\\\"()\\\") \u2192 true\", nestParen(\"()\"), true );\n", "printTest( \"nestParen(\\\"\\\") \u2192 true\", nestParen(\"\"), true );\n", "printTest( \"nestParen(\\\"(yy)\\\") \u2192 false\", nestParen(\"(yy)\"), false );\n", "printTest( \"nestParen(\\\"(())\\\") \u2192 true\", nestParen(\"(())\"), true );\n", "printTest( \"nestParen(\\\"(((y))\\\") \u2192 false\", nestParen(\"(((y))\"), false );\n", "printTest( \"nestParen(\\\"((y)))\\\") \u2192 false\", nestParen(\"((y)))\"), false );\n", "printTest( \"nestParen(\\\"((()))\\\") \u2192 true\", nestParen(\"((()))\"), true );\n", "printTest( \"nestParen(\\\"(())))\\\") \u2192 false\", nestParen(\"(())))\"), false );\n", "printTest( \"nestParen(\\\"((yy())))\\\") \u2192 false\", nestParen(\"((yy())))\"), false );\n", "printTest( \"nestParen(\\\"(((())))\\\") \u2192 true\", nestParen(\"(((())))\"), true );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > strCount"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string and a non-empty substring sub, compute recursively the number of\n", "times that sub appears in the string, without the sub strings overlapping.\n", "\n", "strCount(\"catcowcat\", \"cat\") \u2192 2\n", "strCount(\"catcowcat\", \"cow\") \u2192 1\n", "strCount(\"catcowcat\", \"dog\") \u2192 0\n", "*/\n", "\n", "int strCount(string str, string sub) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"strCount(\\\"catcowcat\\\", \\\"cat\\\") \u2192 2\", strCount(\"catcowcat\", \"cat\"), 2 );\n", "printTest( \"strCount(\\\"catcowcat\\\", \\\"cow\\\") \u2192 1\", strCount(\"catcowcat\", \"cow\"), 1 );\n", "printTest( \"strCount(\\\"catcowcat\\\", \\\"dog\\\") \u2192 0\", strCount(\"catcowcat\", \"dog\"), 0 );\n", "printTest( \"strCount(\\\"cacatcowcat\\\", \\\"cat\\\") \u2192 2\", strCount(\"cacatcowcat\", \"cat\"), 2 );\n", "printTest( \"strCount(\\\"xyx\\\", \\\"x\\\") \u2192 2\", strCount(\"xyx\", \"x\"), 2 );\n", "printTest( \"strCount(\\\"iiiijj\\\", \\\"i\\\") \u2192 4\", strCount(\"iiiijj\", \"i\"), 4 );\n", "printTest( \"strCount(\\\"iiiijj\\\", \\\"ii\\\") \u2192 2\", strCount(\"iiiijj\", \"ii\"), 2 );\n", "printTest( \"strCount(\\\"iiiijj\\\", \\\"iii\\\") \u2192 1\", strCount(\"iiiijj\", \"iii\"), 1 );\n", "printTest( \"strCount(\\\"iiiijj\\\", \\\"j\\\") \u2192 2\", strCount(\"iiiijj\", \"j\"), 2 );\n", "printTest( \"strCount(\\\"iiiijj\\\", \\\"jj\\\") \u2192 1\", strCount(\"iiiijj\", \"jj\"), 1 );\n", "printTest( \"strCount(\\\"aaabababab\\\", \\\"ab\\\") \u2192 4\", strCount(\"aaabababab\", \"ab\"), 4 );\n", "printTest( \"strCount(\\\"aaabababab\\\", \\\"aa\\\") \u2192 1\", strCount(\"aaabababab\", \"aa\"), 1 );\n", "printTest( \"strCount(\\\"aaabababab\\\", \\\"a\\\") \u2192 6\", strCount(\"aaabababab\", \"a\"), 6 );\n", "printTest( \"strCount(\\\"aaabababab\\\", \\\"b\\\") \u2192 4\", strCount(\"aaabababab\", \"b\"), 4 );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > strCopies"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string and a non-empty substring sub, compute recursively if at least n\n", "copies of sub appear in the string somewhere, possibly with overlapping. N will\n", "be non-negative.\n", "\n", "strCopies(\"catcowcat\", \"cat\", 2) \u2192 true\n", "strCopies(\"catcowcat\", \"cow\", 2) \u2192 false\n", "strCopies(\"catcowcat\", \"cow\", 1) \u2192 true\n", "*/\n", "\n", "bool strCopies(string str, string sub, int n) {\n", "    return false;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"strCopies(\\\"catcowcat\\\", \\\"cat\\\", 2) \u2192 true\", strCopies(\"catcowcat\", \"cat\", 2), true );\n", "printTest( \"strCopies(\\\"catcowcat\\\", \\\"cow\\\", 2) \u2192 false\", strCopies(\"catcowcat\", \"cow\", 2), false );\n", "printTest( \"strCopies(\\\"catcowcat\\\", \\\"cow\\\", 1) \u2192 true\", strCopies(\"catcowcat\", \"cow\", 1), true );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"i\\\", 3) \u2192 true\", strCopies(\"iiijjj\", \"i\", 3), true );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"i\\\", 4) \u2192 false\", strCopies(\"iiijjj\", \"i\", 4), false );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"ii\\\", 2) \u2192 true\", strCopies(\"iiijjj\", \"ii\", 2), true );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"ii\\\", 3) \u2192 false\", strCopies(\"iiijjj\", \"ii\", 3), false );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"x\\\", 3) \u2192 false\", strCopies(\"iiijjj\", \"x\", 3), false );\n", "printTest( \"strCopies(\\\"iiijjj\\\", \\\"x\\\", 0) \u2192 true\", strCopies(\"iiijjj\", \"x\", 0), true );\n", "printTest( \"strCopies(\\\"iiiiij\\\", \\\"iii\\\", 3) \u2192 true\", strCopies(\"iiiiij\", \"iii\", 3), true );\n", "printTest( \"strCopies(\\\"iiiiij\\\", \\\"iii\\\", 4) \u2192 false\", strCopies(\"iiiiij\", \"iii\", 4), false );\n", "printTest( \"strCopies(\\\"ijiiiiij\\\", \\\"iiii\\\", 2) \u2192 true\", strCopies(\"ijiiiiij\", \"iiii\", 2), true );\n", "printTest( \"strCopies(\\\"ijiiiiij\\\", \\\"iiii\\\", 3) \u2192 false\", strCopies(\"ijiiiiij\", \"iiii\", 3), false );\n", "printTest( \"strCopies(\\\"dogcatdogcat\\\", \\\"dog\\\", 2) \u2192 true\", strCopies(\"dogcatdogcat\", \"dog\", 2), true );"]},
{"cell_type": "markdown", "metadata": {}, "source": ["# Recursion-1 > strDist"]},
{"cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": ["/*\n", "Given a string and a non-empty substring sub, compute recursively the largest\n", "substring which starts and ends with sub and return its length.\n", "\n", "strDist(\"catcowcat\", \"cat\") \u2192 9\n", "strDist(\"catcowcat\", \"cow\") \u2192 3\n", "strDist(\"cccatcowcatxx\", \"cat\") \u2192 9\n", "*/\n", "\n", "int strDist(string str, string sub) {\n", "    return 0;\n", "}\n", "\n", "\n", "// tests:\n", "printTest( \"strDist(\\\"catcowcat\\\", \\\"cat\\\") \u2192 9\", strDist(\"catcowcat\", \"cat\"), 9 );\n", "printTest( \"strDist(\\\"catcowcat\\\", \\\"cow\\\") \u2192 3\", strDist(\"catcowcat\", \"cow\"), 3 );\n", "printTest( \"strDist(\\\"cccatcowcatxx\\\", \\\"cat\\\") \u2192 9\", strDist(\"cccatcowcatxx\", \"cat\"), 9 );\n", "printTest( \"strDist(\\\"abccatcowcatcatxyz\\\", \\\"cat\\\") \u2192 12\", strDist(\"abccatcowcatcatxyz\", \"cat\"), 12 );\n", "printTest( \"strDist(\\\"xyx\\\", \\\"x\\\") \u2192 3\", strDist(\"xyx\", \"x\"), 3 );\n", "printTest( \"strDist(\\\"xyx\\\", \\\"y\\\") \u2192 1\", strDist(\"xyx\", \"y\"), 1 );\n", "printTest( \"strDist(\\\"xyx\\\", \\\"z\\\") \u2192 0\", strDist(\"xyx\", \"z\"), 0 );\n", "printTest( \"strDist(\\\"z\\\", \\\"z\\\") \u2192 1\", strDist(\"z\", \"z\"), 1 );\n", "printTest( \"strDist(\\\"x\\\", \\\"z\\\") \u2192 0\", strDist(\"x\", \"z\"), 0 );\n", "printTest( \"strDist(\\\"\\\", \\\"z\\\") \u2192 0\", strDist(\"\", \"z\"), 0 );\n", "printTest( \"strDist(\\\"hiHellohihihi\\\", \\\"hi\\\") \u2192 13\", strDist(\"hiHellohihihi\", \"hi\"), 13 );\n", "printTest( \"strDist(\\\"hiHellohihihi\\\", \\\"hih\\\") \u2192 5\", strDist(\"hiHellohihihi\", \"hih\"), 5 );\n", "printTest( \"strDist(\\\"hiHellohihihi\\\", \\\"o\\\") \u2192 1\", strDist(\"hiHellohihihi\", \"o\"), 1 );\n", "printTest( \"strDist(\\\"hiHellohihihi\\\", \\\"ll\\\") \u2192 2\", strDist(\"hiHellohihihi\", \"ll\"), 2 );"]}
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