60. Permutation Sequence

60. Permutation Sequence (Medium)

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:

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the k^th permutation sequence.

Note:

• Given n will be between 1 and 9 inclusive.
• Given k will be between 1 and n! inclusive.

Example 1:

Input: n = 3, k = 3
Output: "213"

Example 2:

Input: n = 4, k = 9
Output: "2314"

Companies:
Google

Related Topics:
Math, Backtracking

Similar Questions:

• Next Permutation (Medium)
• Permutations (Medium)

Solution 1. Brute Force

// OJ: https://leetcode.com/explore/challenge/card/june-leetcoding-challenge/541/week-3-june-15th-june-21st/3366/
// Author: github.com/lzl124631x
// Time: O(NK)
// Space: O(N)
class Solution {
public:
    string getPermutation(int n, int k) {
        string s;
        for (int i = 0; i < n; ++i) s += ('1' + i);
        while (--k) next_permutation(s.begin(), s.end());
        return s;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/permutation-sequence/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    string getPermutation(int n, int k) {
        int fac = 1;
        string ans;
        for (int i = 1; i <= n; ++i) {
            fac *= i;
            ans += '0' + i;
        }
        --k;
        for (int i = 0; i < n; ++i) {
            fac /= n - i;
            int j = k / fac + i, tmp = ans[j];
            for (; j > i; --j) ans[j] = ans[j - 1];
            ans[j] = tmp;
            k %= fac;
        }
        return ans;
    }
};