Next Permutation

This article is for self practice 

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 considered permutations of arr: [1,2,3], [1,3,2], [3,1,2], [2,3,1].

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 arrangement is not possible, the array must be rearranged as 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]

Example 3:

Input: nums = [1,1,5]
Output: [1,5,1]

 

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 100

package practice;

public class NextPermutation {
    public static void main(String[] args) {
        NextPermutation permu = new NextPermutation();
        int[] nums = {3, 2, 1};
        permu.nextPermutation(nums);
    }


    public void nextPermutation(int[] nums) {
        //from right to left, find the point where it starts decresing
        // as the next permutation is the next immediate bigger
        int i = nums.length - 2;
        while (i >= 0 && nums[i + 1] <= nums[i]) {
            i--;
        }

        //traverse right to left from  to find out immediate bigger number than replaceDigit
        if (i >= 0) {
            int j = nums.length - 1;
            while (nums[j] <= nums[i]) {
                j--;
            }
            swap(nums, i, j);
        }
        reverse(nums, i + 1, nums.length - 1);
    }

    void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }

    void reverse(int[] nums, int i, int j) {
        while (i < j) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
            i++;
            j--;

        }
    }

}