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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`

:

`"123"`

`"132"`

`"213"`

`"231"`

`"312"`

`"321"`

Given `n`

and `k`

, return the `k`

permutation sequence.^{th}

**Example 1:**

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

**Example 2:**

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

**Example 3:**

Input:n = 3, k = 1Output:"123"

**Constraints:**

Permutation Sequence– LeetCode Solutions

Permutation Sequence in C++:

class Solution { public: string getPermutation(int n, int k) { string ans; vectornums(n); vector 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(); Listnums = 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