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permutation seq..cpp
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/*
//////////////////////////////////////////////////////
//Question/Info
60. Permutation Sequence
Hard
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The set [1, 2, 3, ..., n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Example 1:
Input: n = 3, k = 3
Output: "213"
Example 2:
Input: n = 4, k = 9
Output: "2314"
Example 3:
Input: n = 3, k = 1
Output: "123"
Constraints:
1 <= n <= 9
1 <= k <= n!
author: srj_v
//////////////////////////////////////////////////////
*/
#include <bits/stdc++.h>
using namespace std;
#define _IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long double ld;
typedef long long int lli;
#pragma GCC optimize("Ofast")
void c_p_c()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
//////////////////////////////////////////////////////
int32_t main() {
///////////
// c_p_c();
///////////
_IOS
//////////
// code
/*
int t ; cin >> t; while(t--){}
*/
class Solution {
public:
string getPermutation(int n, int k) {
// complexity of recursion is o(n!)..~finding sets.. and finding... exponential...
// hence we optimise to o(n^2)
int ft = 1;
vector<int> v;
for (int i = 1; i < n; i++) {
ft = ft * i; // for factorial...
v.push_back(i); // stores 1,2,3,4....n
}
v.push_back(n);
string ans = "";
k--; // decrementing, as 0 based indexing... like in josephus problem...
while (1) {
// note that 'n' and 'k' is given in the question...
ans = ans + to_string(v[k / ft]); // 16/6...
v.erase(v.begin() + k / ft);
if (v.size() == 0) break;
k = k % ft;
ft = ft / v.size();
}
return ans;
/*
say n = 4, you have {1, 2, 3, 4}
If you were to list out all the permutations you have
1 + (permutations of 2, 3, 4) - its total permutations possible is 3! = 6...
2 + (permutations of 1, 3, 4) - ... 3! = 6 ...
3 + (permutations of 1, 2, 4) -...6...
4 + (permutations of 1, 2, 3) -...6...
*/
}
};
// cerr << "time: " << clock() << " ms" << '\n';
return 0;
}
//////////////////////////////////////////////////////