Next Permutation — Leetcode

Topic: Array
Difficulty: Medium
Question Number: 31

Problem Statement:
A permutation of an array of integers is an arrangement of its members into a sequence or linear order.

• For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].

The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such an arrangement is impossible, the array must be rearranged in the lowest possible order (i.e., sorted in ascending order).

• For example, the next permutation of arr = [1,2,3] is [1,3,2].
• Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
• While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.

Given an array of integers nums, find the next permutation of nums.

The replacement must be in place and use only constant extra memory.

Example 1:

Input: nums = [1,2,3]
Output: [1,3,2]

Example 2:

Input: nums = [3,2,1]
Output: [1,2,3]

Explanation:
steps to follow-

• The number starts changing from the first decreasing sequence from the end
• find the number of substitution
• find the next highest after the number present at the point.
• last, rearrange the remaining array

Code:

in java-

class Solution {
    public void nextPermutation(int[] A) {
        if (A==null|| A.length<=1)return;
        int i= A.length-2;
        while(i>=0 && A[i]>=A[i+1])i--;
        if(i>=0){
            int j = A.length -1;
            while (A[j]<= A[i])j--;
            swap(A, i, j);
        }
        reverse (A, i+1,A.length-1);
    }
    public void swap(int[] A, int i, int j){
        int tmp= A[i];
        A[i]= A[j];
        A[j]= tmp;
    }
    public void reverse(int[] A, int i, int j){
        while(i<j) swap(A,i++, j--);
    }
}

in python by Vaishnav Jha

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        i=len(nums)-2
        j=len(nums)-1
        while nums[i] >= nums[i+1] and i>=0:
            i-=1
        if i<0:
            nums.sort()
        else:        
            while nums[j] <= nums[i] and j>=0:
                j-=1
            
            nums[i],nums[j] = nums[j],nums[i]
            j = len(nums)-1
            i+=1
            while i<=j:
                nums[i],nums[j] = nums[j],nums[i]
                i+=1
                j-=1


# Contributed By - Vaishnav Jha

Thank you
Oshi Raghav