Given a set of ‘n’ elements, find their Kth permutation. Consider the following set of elements:

All permutations of the above elements are (with ordering):

Here we need to find the Kth permutation

- Recursion
- Factorial

void find_kth_permutation(vector<char>& v,int k,string& result) {//TODO: Write - Your - Code}

int factorial(int n) {if (n == 0 || n == 1) return 1;return n * factorial(n -1 );}void find_kth_permutation(vector<char>& v,int k,string& result) {if (v.empty()) {return;}int n = (int)(v.size());// count is number of permutations starting with each digitint count = factorial(n - 1);int selected = (k - 1) / count;result += v[selected];v.erase(v.begin() + selected);k = k - (count * selected);find_kth_permutation(v, k, result);}string get_permutation(int n, int k) {vector<char> v;for (char i = 1; i <= n; ++i) {v.push_back(i + '0');}string result;find_kth_permutation(v, k, result);return result;}int main(int argc, char* argv[]) {for (int i = 1; i <= factorial(4); ++i) {cout << i << "th permutation = " << get_permutation(4, i) << endl;}}

Linear, O(n).

Linear, O(n).

Recursive solution will consume memory on the stack.

Here is the algorithm that we will follow:

```
If input vector is empty return result vector
block_size = (n-1)! ['n' is the size of vector]
Figure out which block k will lie in and select the first element of that block
(this can be done by doing (k-1)/block_size )
Append selected element to result vector and remove it from original input vector
Deduce from k the blocks that are skipped i.e k = k - selected*block_size and goto step 1
```

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