31. Next Permutation

Description

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

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

Constraints

• 1 <= nums.length <= 100

• 0 <= nums[i] <= 100

Approach

Links

• GeeksforGeeks

• Leetcode
• ProgramCreek
• YouTube

Examples

Example 1

Input: nums = [1, 2, 3]

Output: [1, 3, 2]

Example 2

Input: nums = [3, 2, 1]

Output: [1, 2, 3]

Example 3

Input: nums = [1, 1, 5]

Output: [1, 5, 1]

Example 4

Input: nums = [1]

Output: [1]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

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

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