Obsidian Notes

Next Permutation

3 min read

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 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

Approach

  • So basically we find a pivot which is basically the first element from right so that the next element is increasing so the element next to pivot the elements next to it would be decreasing because if not then that would be pivot
  • once we find the pivot then we just swap these two basically that is what is lexicographically we pick the next element
  • then we reverse that decreasing order because if we have a new element then next elements should be increasing and not decreasing
  • Time O(n) (at most 2 passes)
  • Space O(1) (in-place swap and reverse)
class Solution {
    public void nextPermutation(int[] nums) {
        int i = nums.length - 2;
        while (i >= 0 && nums[i] >= nums[i+1])
            i--;
 
        if (i >= 0) {
            int j = nums.length - 1;
            while (j > i && nums[i] >= nums[j])
                j--;
            
            swap(nums, i , j);
        }    
 
        reverse(nums, i + 1, nums.length - 1);
 
    }
 
    void swap(int[] nums, int i, int j) {
        int tmp = nums[j];
        nums[j] = nums[i];
        nums[i] = tmp;
    }
 
    void reverse(int[] nums, int start, int end) {
        while (start < end) {
            swap(nums, start, end);
            start++;
            end--;
        }
    }
}

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