60. Permutation Sequence LeetCode Solution

In this guide, you will get 60. Permutation Sequence LeetCode Solution with the best time and space complexity. The solution to Permutation Sequence problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Permutation Sequence solution in C++
4. Permutation Sequence solution in Java
5. Permutation Sequence solution in Python
6. Additional Resources

Problem Statement of Permutation Sequence

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:

“123”
“132”
“213”
“231”
“312”
“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!

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

60. Permutation Sequence LeetCode Solution in C++

class Solution {
 public:
  string getPermutation(int n, int k) {
    string ans;
    vector<int> nums(n);
    vector<int> fact(n + 1, 1);  // fact[i] := i!

    iota(nums.begin(), nums.end(), 1);

    for (int i = 2; i <= n; ++i)
      fact[i] = fact[i - 1] * i;

    --k;  // 0-indexed

    for (int i = n - 1; i >= 0; --i) {
      const int j = k / fact[i];
      k %= fact[i];
      ans += to_string(nums[j]);
      nums.erase(nums.begin() + j);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

60. Permutation Sequence LeetCode Solution in Java

class Solution {
  public String getPermutation(int n, int k) {
    StringBuilder sb = new StringBuilder();
    List<Integer> nums = new ArrayList<>();
    int[] fact = new int[n + 1]; // fact[i] := i!

    for (int i = 1; i <= n; ++i)
      nums.add(i);

    Arrays.fill(fact, 1);
    for (int i = 2; i <= n; ++i)
      fact[i] = fact[i - 1] * i;

    --k; // 0-indexed

    for (int i = n - 1; i >= 0; --i) {
      final int j = k / fact[i];
      k %= fact[i];
      sb.append(nums.get(j));
      nums.remove(j);
    }

    return sb.toString();
  }
}
// code provided by PROGIEZ

60. Permutation Sequence LeetCode Solution in Python

class Solution:
  def getPermutation(self, n: int, k: int) -> str:
    ans = ''
    nums = [i + 1 for i in range(n)]
    fact = [1] * (n + 1)  # fact[i] := i!

    for i in range(2, n + 1):
      fact[i] = fact[i - 1] * i

    k -= 1  # 0-indexed

    for i in reversed(range(n)):
      j = k // fact[i]
      k %= fact[i]
      ans += str(nums[j])
      nums.pop(j)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.