Permutation Sequence LeetCode Solution

Problem – Permutation Sequence LeetCode Solution

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 Solution in Java

public class Solution {
public String getPermutation(int n, int k) {
    int pos = 0;
    List<Integer> numbers = new ArrayList<>();
    int[] factorial = new int[n+1];
    StringBuilder sb = new StringBuilder();
    
    // create an array of factorial lookup
    int sum = 1;
    factorial[0] = 1;
    for(int i=1; i<=n; i++){
        sum *= i;
        factorial[i] = sum;
    }
    // factorial[] = {1, 1, 2, 6, 24, ... n!}
    
    // create a list of numbers to get indices
    for(int i=1; i<=n; i++){
        numbers.add(i);
    }
    // numbers = {1, 2, 3, 4}
    
    k--;
    
    for(int i = 1; i <= n; i++){
        int index = k/factorial[n-i];
        sb.append(String.valueOf(numbers.get(index)));
        numbers.remove(index);
        k-=index*factorial[n-i];
    }
    
    return String.valueOf(sb);
}

Permutation Sequence LeetCode Solution in C++

class Solution {
public:
    // Our recursive function that will complete the ans string.
	// v - is our current array = [1,2,3,4]
	// ans is the answer string, n and k are current values of n and k
	// factVal is an array containing the factorial of all integers from 0-9 to get factorial in O(1) time.
	// That means I have stored all the factorials in this array before hand to avoid calculation. You can also write factorial funciton if you want.
	
    void setPerm(vector<int>& v,string& ans,int n,int k,vector<int>& factVal){
       // if there is only one element left append it to our ans (Base case)
	   if(n==1){
            ans+=to_string(v.back());
            return;
        }
		
		// We are calculating the required index.  factVal[n-1] means for n =  4 => factVal[3] = 6.
        // 15 / 6 = 2 will the index for k =15 and n = 4.
		int index = (k/factVal[n-1]);
        // if k is a multiple of elements of partition then decrement the index (Corner case I was talking about)
		if(k % factVal[n-1] == 0){
            index--;
        }
		
		ans+= to_string(v[index]);  // add value to string
        v.erase(v.begin() + index);  // remove element from array
        k -= factVal[n-1] * index;   // adjust value of k; k = 15 - 6*2 = 3.
		// Recursive call with n=n-1 as one element is added we need remaing.
        setPerm(v,ans,n-1,k,factVal);
    }
    
    string getPermutation(int n, int k) {
        if(n==1) return "1";
		//Factorials of 0-9 stored in the array. factVal[3] = 6. (3! = 6)
        vector<int>factVal = {1,1,2,6,24,120,720,5040,40320,362880};
        string ans = "";
        vector<int> v;
		// Fill the array with all elements
        for(int i=1;i<=n;i++) v.emplace_back(i);
        setPerm(v,ans,n,k,factVal);
        return ans;
    }
};

Permutation Sequence LeetCode Solution in Python

import math
class Solution:
    # @param {integer} n
    # @param {integer} k
    # @return {string}
    def getPermutation(self, n, k):
        numbers = range(1, n+1)
        permutation = ''
        k -= 1
        while n > 0:
            n -= 1
            # get the index of current digit
            index, k = divmod(k, math.factorial(n))
            permutation += str(numbers[index])
            # remove handled number
            numbers.remove(numbers[index])

        return permutation
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