0060. Permutation Sequence

https://leetcode.com/problems/permutation-sequence

Description

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 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!

ac: Math

class Solution {
    public String getPermutation(int n, int k) {
        int[] permutationCount = new int[n+1]; // How many permutations if there are i number: {1, 1, 2, 6, 24, ... n!}
        permutationCount[0] = 1; // 0! is 1.
        int sum = 1;
        for (int i = 1; i <= n; i++) {
            sum *= i;
            permutationCount[i] = sum;
        }
        
        List<Integer> nums = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            nums.add(i);
        }
        
        StringBuilder res = new StringBuilder();
        k--; // Find kth -> index is (k-1)
        for (int i = 0; i < n; i++) {
            int lenOfPermutationAfterITh = n - i - 1; // "-1" because we don't include ith number.
            // What is the number in ith position, in the answer string. 
            // There are n-i numbers make up of permutationCount[n-i] candidates.
            int ithNumIndex = k / permutationCount[lenOfPermutationAfterITh];
            res.append(nums.get(ithNumIndex));
            
            nums.remove(ithNumIndex); // This number won't appear in the later part of the permutation.
            
            k = k % permutationCount[lenOfPermutationAfterITh]; // Already found 
        }
        
        return res.toString();
    }
}

// list.remove() is O(n), so time complexity is O(N^2). The brute force is O(KNlogN), if K is large, it is bad.
// This is more of a math problem.