Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Example:
Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True
There is a subset (4, 5) with sum 9.
Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
There is no subset that add up to 30.
Approach:
For the recursive approach we will consider two cases.
- Consider the last element and now the required sum = target sum – value of ‘last’ element and number of elements = total elements – 1
- Leave the ‘last’ element and now the required sum = target sum and number of elements = total elements – 1
Following is the recursive formula for isSubsetSum() problem.
isSubsetSum(set, n, sum)
= isSubsetSum(set, n-1, sum) ||
isSubsetSum(set, n-1, sum-set[n-1])
Base Cases:
isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0
isSubsetSum(set, n, sum) = true, if sum == 0
Let’s take a look at the simulation of above approach-:
set[]={3, 4, 5, 2}
sum=9
(x, y)= 'x' is the left number of elements,
'y' is the required sum
(4, 9)
{True}
/ \
(3, 6) (3, 9)
/ \ / \
(2, 2) (2, 6) (2, 5) (2, 9)
{True}
/ \
(1, -3) (1, 2)
{False} {True}
/ \
(0, 0) (0, 2)
{True} {False}
// A recursive solution for subset sum problem
#include <iostream>
using namespace std;
// Returns true if there is a subset
// of set[] with sum equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
/* else, check if sum can be obtained by any
of the following:
(a) including the last element
(b) excluding the last element */
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
int main()
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = sizeof(set) / sizeof(set[0]);
if (isSubsetSum(set, n, sum) == true)
cout <<"Found a subset with given sum";
else
cout <<"No subset with given sum";
return 0;
}
// A Dynamic Programming solution for subset
// sum problem
class Solution {
// Returns true if there is a subset of
// set[] with sum equal to given sum
static boolean isSubsetSum(int set[],
int n, int sum)
{
// The value of subset[i][j] will be
// true if there is a subset of
// set[0..j-1] with sum equal to i
boolean subset[][] = new boolean[sum + 1][n + 1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
subset[0][i] = true;
// If sum is not 0 and set is empty,
// then answer is false
for (int i = 1; i <= sum; i++)
subset[i][0] = false;
// Fill the subset table in bottom
// up manner
for (int i = 1; i <= sum; i++) {
for (int j = 1; j <= n; j++) {
subset[i][j] = subset[i][j - 1];
if (i >= set[j - 1])
subset[i][j] = subset[i][j]
|| subset[i - set[j - 1]][j - 1];
}
}
/* // uncomment this code to print table
for (int i = 0; i <= sum; i++)
{
for (int j = 0; j <= n; j++)
System.out.println (subset[i][j]);
} */
return subset[sum][n];
}
/* Driver code*/
public static void main(String args[])
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = set.length;
if (isSubsetSum(set, n, sum) == true)
System.out.println("Found a subset"
+ " with given sum");
else
System.out.println("No subset with"
+ " given sum");
}
}
def isSubsetSum(set, n, sum) :
# Base Cases
if (sum == 0) :
return True
if (n == 0 and sum != 0) :
return False
# If last element is greater than
# sum, then ignore it
if (set[n - 1] > sum) :
return isSubsetSum(set, n - 1, sum);
# else, check if sum can be obtained
# by any of the following
# (a) including the last element
# (b) excluding the last element
return isSubsetSum(set, n-1, sum) or isSubsetSum(set, n-1, sum-set[n-1])
# Driver program to test above function
set = [3, 34, 4, 12, 5, 2]
sum = 9
n = len(set)
if (isSubsetSum(set, n, sum) == True) :
print("Found a subset with given sum")
else :
print("No subset with given sum")
#include <stdio.h>
// Returns true if there is a subset
// of set[] with sum equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
/* else, check if sum can be obtained by any
of the following:
(a) including the last element
(b) excluding the last element */
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
int main()
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = sizeof(set) / sizeof(set[0]);
if (isSubsetSum(set, n, sum) == true)
printf("Found a subset with given sum");
else
printf("No subset with given sum");
return 0;
}
function isSubsetSum(set, n, sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
/* else, check if sum can be obtained
by any of the following
(a) including the last element
(b) excluding the last element */
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
let set = [ 3, 34, 4, 12, 5, 2 ];
let sum = 9;
let n = set.length;
if (isSubsetSum(set, n, sum) == true)
document.write("Found a subset with given sum");
else
document.write("No subset with given sum");
<?php
// A recursive solution for subset sum problem
// Returns true if there is a subset of set
// with sun equal to given sum
function isSubsetSum($set, $n, $sum)
{
// Base Cases
if ($sum == 0)
return true;
if ($n == 0)
return false;
// If last element is greater
// than sum, then ignore it
if ($set[$n - 1] > $sum)
return isSubsetSum($set, $n - 1, $sum);
/* else, check if sum can be
obtained by any of the following
(a) including the last element
(b) excluding the last element */
return isSubsetSum($set, $n - 1, $sum) ||
isSubsetSum($set, $n - 1,
$sum - $set[$n - 1]);
}
// Driver Code
$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = 6;
if (isSubsetSum($set, $n, $sum) == true)
echo"Found a subset with given sum";
else
echo "No subset with given sum";
?>
using System;
class solution {
// Returns true if there is a subset of set[] with sum
// equal to given sum
static bool isSubsetSum(int[] set, int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
/* else, check if sum can be obtained
by any of the following
(a) including the last element
(b) excluding the last element */
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
public static void Main()
{
int[] set = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = set.Length;
if (isSubsetSum(set, n, sum) == true)
Console.WriteLine("Found a subset with given sum");
else
Console.WriteLine("No subset with given sum");
}
}
**Output:**
Found a subset with given sum