665. Non-decreasing Array #

Problem #

Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element.

We define an array is non-decreasing if nums[i] <= nums[i + 1] holds for every i (0-based) such that (0 <= i <= n - 2).

Example 1:

Input: nums = [4,2,3]
Output: true
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

Example 2:

Input: nums = [4,2,1]
Output: false
Explanation: You can't get a non-decreasing array by modify at most one element.

Constraints:

n == nums.length
1 <= n <= 104
-10^5 <= nums[i] <= 10^5

Problem Summary #

Given an integer array of length n, determine whether the array can become a non-decreasing sequence by changing at most 1 element. We define a non-decreasing sequence as follows: for any i (0 <= i <= n-2) in the array, nums[i] <= nums[i + 1] is always satisfied.

Solution Approach #

Easy problem. Loop through the array and find decreasing pairs where nums[i] > nums[i+1]. Once there are more than 2 such pairs, return false directly. When the first decreasing pair is found, one manual adjustment is needed. If nums[i + 1] < nums[i - 1], then even if nums[i+1] and nums[i] are swapped, after the swap, nums[i - 1] may still be greater than nums[i + 1], which does not satisfy the requirement. The correct approach should be to make the smaller number larger, namely nums[i + 1] = nums[i]. Two equal elements satisfy the requirement of being non-decreasing.

Code #

package leetcode

func checkPossibility(nums []int) bool {
	count := 0
	for i := 0; i < len(nums)-1; i++ {
		if nums[i] > nums[i+1] {
			count++
			if count > 1 {
				return false
			}
			if i > 0 && nums[i+1] < nums[i-1] {
				nums[i+1] = nums[i]
			}
		}
	}
	return true
}