LeetCode 665 - Non-decreasing Array

https://leetcode.com/problems/non-decreasing-array

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:

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

https://discuss.leetcode.com/topic/101115/c-java-clean-code-6-liner-without-modifying-input

The strategy is to lower a[i-1] to match a[i] if possible - (a[i-2] not exist or no smaller than a[i]);
otherwise rise a[i] to match a[i-1].

Example:
 0 ..  i ...
 1 1 2[4]2 5   we can just raise a[i+1] to 4;
         4
 1 1 2[4]2 3   in this case lower a[i] is better;

    public boolean checkPossibility(int[] a) {
        int modified = 0;
        for (int i = 1; i < a.length; i++) {
            if (a[i] < a[i - 1]) {
                if (modified++ > 0) return false;
                if (i - 2 < 0 || a[i - 2] <= a[i]) a[i - 1] = a[i]; // lower a[i - 1]
                else a[i] = a[i - 1]; // rise a[i]
            }
        }
        return true;
    }

https://discuss.leetcode.com/topic/101144/java-c-simple-greedy-like-solution-with-explanation

This problem is like a greedy problem. When you find nums[i-1] > nums[i] for some i, you will prefer to change nums[i-1]'s value, since a larger nums[i] will give you more risks that you get inversion errors after position i. But, if you also find nums[i-2] > nums[i], then you have to change nums[i]'s value instead, or else you need to change both of nums[i-2]'s and nums[i-1]'s values.

 public boolean checkPossibility(int[] nums) {
        int cnt = 0;                                                                    //the number of changes
        for(int i = 1; i < nums.length && cnt<=1 ; i++){
            if(nums[i-1] > nums[i]){
                cnt++;
                if(i-2<0 || nums[i-2] <= nums[i])nums[i-1] = nums[i];                    //modify nums[i-1] of a priority
                else nums[i] = nums[i-1];                                                //have to modify nums[i]
            }
        }
        return cnt<=1; 
    }

https://discuss.leetcode.com/topic/101115/c-java-clean-code-6-liner-without-modifying-input

We can also do it without modifying the input by using a variable prev to hold the a[i-1]; if we have to lower a[i] to match a[i-1] instead of raising a[i-1], simply skip updating prev;
Java - Without Modifying Input

    public boolean checkPossibility(int[] a) {
        int modified = 0;
        for (int i = 1, prev = a[0]; i < a.length; i++) {
            if (a[i] < prev && modified++ > 0) return false;
            if (a[i] < prev && i - 2 >= 0 && a[i - 2] > a[i]) continue;
            prev = a[i];
        }
        return true;
    }