# Find Kth Permutation

## Problem Statement

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

## Try it yourself

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

## Solution

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 digit
int 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;
}
}

## Solution Explanation

Linear, O(n).

### Memory Complexity

Linear, O(n).

Recursive solution will consume memory on the stack.

### Solution Breakdown

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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