60. 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!

Solutions

Solution 1: Enumeration

We know that the set $[1,2,..n]$ has a total of $n!$ permutations. If we determine the first digit, the number of permutations that the remaining digits can form is $(n-1)!$.

Therefore, we enumerate each digit $i$. If $k$ is greater than the number of permutations after the current position is determined, then we can directly subtract this number; otherwise, it means that we have found the number at the current position.

For each digit $i$, where $0 \leq i < n$, the number of permutations that the remaining digits can form is $(n-i-1)!$, which we denote as $fact$. The numbers used in the process are recorded in vis.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$.

Python3JavaC++GoRustC#TypeScript

class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        ans = []
        vis = [False] * (n + 1)
        for i in range(n):
            fact = 1
            for j in range(1, n - i):
                fact *= j
            for j in range(1, n + 1):
                if not vis[j]:
                    if k > fact:
                        k -= fact
                    else:
                        ans.append(str(j))
                        vis[j] = True
                        break
        return ''.join(ans)

class Solution {
    public String getPermutation(int n, int k) {
        StringBuilder ans = new StringBuilder();
        boolean[] vis = new boolean[n + 1];
        for (int i = 0; i < n; ++i) {
            int fact = 1;
            for (int j = 1; j < n - i; ++j) {
                fact *= j;
            }
            for (int j = 1; j <= n; ++j) {
                if (!vis[j]) {
                    if (k > fact) {
                        k -= fact;
                    } else {
                        ans.append(j);
                        vis[j] = true;
                        break;
                    }
                }
            }
        }
        return ans.toString();
    }
}

class Solution {
public:
    string getPermutation(int n, int k) {
        string ans;
        bitset<10> vis;
        for (int i = 0; i < n; ++i) {
            int fact = 1;
            for (int j = 1; j < n - i; ++j) fact *= j;
            for (int j = 1; j <= n; ++j) {
                if (vis[j]) continue;
                if (k > fact)
                    k -= fact;
                else {
                    ans += to_string(j);
                    vis[j] = 1;
                    break;
                }
            }
        }
        return ans;
    }
};

func getPermutation(n int, k int) string {
    ans := make([]byte, n)
    vis := make([]bool, n+1)
    for i := 0; i < n; i++ {
        fact := 1
        for j := 1; j < n-i; j++ {
            fact *= j
        }
        for j := 1; j <= n; j++ {
            if !vis[j] {
                if k > fact {
                    k -= fact
                } else {
                    ans[i] = byte('0' + j)
                    vis[j] = true
                    break
                }
            }
        }
    }
    return string(ans)
}

impl Solution {
    pub fn get_permutation(n: i32, k: i32) -> String {
        let mut k = k;
        let mut ans = String::new();
        let mut fact = vec![1; n as usize];
        for i in 1..n as usize {
            fact[i] = fact[i - 1] * (i as i32);
        }
        let mut vis = vec![false; n as usize + 1];

        for i in 0..n as usize {
            let cnt = fact[(n as usize) - i - 1];
            for j in 1..=n {
                if vis[j as usize] {
                    continue;
                }
                if k > cnt {
                    k -= cnt;
                } else {
                    ans.push_str(&j.to_string());
                    vis[j as usize] = true;
                    break;
                }
            }
        }

        ans
    }
}

public class Solution {
    public string GetPermutation(int n, int k) {
        var ans = new StringBuilder();
        int vis = 0;
        for (int i = 0; i < n; ++i) {
            int fact = 1;
            for (int j = 1; j < n - i; ++j) {
                fact *= j;
            }
            for (int j = 1; j <= n; ++j) {
                if (((vis >> j) & 1) == 0) {
                    if (k > fact) {
                        k -= fact;
                    } else {
                        ans.Append(j);
                        vis |= 1 << j;
                        break;
                    }
                }
            }
        }
        return ans.ToString();
    }
}

function getPermutation(n: number, k: number): string {
    let ans = '';
    const vis = Array.from({ length: n + 1 }, () => false);
    for (let i = 0; i < n; i++) {
        let fact = 1;
        for (let j = 1; j < n - i; j++) {
            fact *= j;
        }
        for (let j = 1; j <= n; j++) {
            if (!vis[j]) {
                if (k > fact) {
                    k -= fact;
                } else {
                    ans += j;
                    vis[j] = true;
                    break;
                }
            }
        }
    }
    return ans;
}

Comments