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+---
+id: permutation-sequence
+title: Permutation Sequence(LeetCode)
+sidebar_label: 0060-Permutation Sequence
+tags:
+  - Math
+  - Recursion
+description: The set [1, 2, 3, ..., n] contains a total of n! unique permutations. Given n and k, return the kth permutation sequence.
+---
+
+## Problem Statement
+
+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.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: n = 3, k = 3
+Output: "213"
+```
+
+**Example 2:**
+
+```plaintext
+Input: n = 4, k = 9
+Output: "2314"
+```
+
+**Example 3:**
+
+```plaintext
+Input: n = 3, k = 1
+Output: "123"
+```
+
+### Constraints
+
+- `1 <= n <= 9`
+- `1 <= k <= n!`
+
+## Solution
+
+### Approach
+
+#### Algorithm
+
+1. Initialize Factorial Values:
+* Precompute and store the factorial values of integers from 0 to 9 in a vector `factVal` to get factorials in O(1) time.
+2. Initialize Array:
+* Create a vector `v` containing elements from 1 to `n`.
+3. Recursive Function `setPerm`:
+* Base Case: If `n == 1`, append the last remaining element in `v` to `ans` and return.
+* Calculate the required index using `k / factVal[n-1]`.
+* Handle the corner case where if `k` is a multiple of `(n-1)!`, decrement the index by 1.
+* Append the element at the calculated index from `v` to `ans`.
+* Remove the selected element from `v`.
+* Adjust the value of `k` to be the remainder after dividing by `factVal[n-1]`.
+* Make a recursive call with updated values of `n`, `k`, `v`, and `ans`.
+4. Construct Result:
+* Initialize an empty string `ans`.
+* Call the recursive function setPerm with initial values of `v`, `ans`, `n`, `k`, and `factVal`.
+* Return the result `ans`.
+
+#### Implementation
+
+```C++
+class Solution {
+public:
+    void setPerm(vector<int>& v, string& ans, int n, int k, vector<int>& factVal) {
+        if (n == 1) {
+            ans += to_string(v.back());
+            return;
+        }
+        
+        int index = (k / factVal[n-1]);
+        if (k % factVal[n-1] == 0) {
+            index--;
+        }
+        
+        ans += to_string(v[index]);
+        v.erase(v.begin() + index);
+        k -= factVal[n-1] * index;
+        setPerm(v, ans, n-1, k, factVal);
+    }
+    
+    string getPermutation(int n, int k) {
+        if (n == 1) return "1";
+        
+        vector<int> factVal = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
+        string ans = "";
+        vector<int> v;
+        for (int i = 1; i <= n; i++) v.emplace_back(i);
+        
+        setPerm(v, ans, n, k, factVal);
+        return ans;
+    }
+};
+```
+
+### Complexity Analysis
+
+- **Time complexity**: $O(N^2)$
+- **Space complexity**: $O(N)$
+
+### Conclusion
+
+This algorithm efficiently finds the k-th permutation of a set of n elements by using a combination of precomputed factorials and recursive selection. The approach ensures that the computation is done in a systematic manner without generating all permutations.