Permutation Sequence LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Permutation Sequence in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Permutation Sequence– LeetCode Problem

Permutation Sequence– LeetCode Problem

Problem:

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!

Permutation Sequence– LeetCode Solutions

Permutation Sequence in C++:

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

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

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

    --k;  // 0-indexed

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

    return ans;
  }
};

Permutation Sequence in Java:

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

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

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

    --k; // 0-indexed

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

    return sb.toString();
  }
}

Permutation Sequence in Python:

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

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

    k -= 1  # 0-indexed

    for i in range(n - 1, -1, -1):
      j = k // factorial[i]
      k %= factorial[i]
      ans += str(nums[j])
      nums.pop(j)

    return ans