Non-decreasing Array

Given an array nums 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 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

Approach

Java

public class NonDecreasingArray {
    public static void main(String[] args) {
        int[] nums = { 4, 2, 3 };
        System.out.println(checkPossibility(nums));
    }
    static boolean checkPossibility(int[] nums) {
        int n = nums.length, cnt = 0;
        for (int i = 0; i < n - 1 && cnt <= 1; i++) {
            if (nums[i] > nums[i + 1]) {
                if (i > 0) {
                    if (nums[i - 1] <= nums[i + 1])
                        nums[i] = nums[i - 1];
                    else
                        nums[i + 1] = nums[i];
                }
                ++cnt;
            }
        }
        return cnt <= 1;
    }
}

C++

#include <bits/stdc++.h>
using namespace std;
bool checkPossibility(vector<int>& nums)
{
        int n=nums.size(),cnt=0;
        for (int i = 0; i <n-1 && cnt<=1;i++) 
        {
            if (nums[i] > nums[i + 1]) 
            {
                if (i > 0)
                {
                    if (nums[i - 1] <= nums[i + 1])
                        nums[i] = nums[i - 1];  
                    else           
                        nums[i + 1] = nums[i]; 
                }
                ++cnt;
            }
        }
        return cnt<=1;
}
int main()
{
     vector<int> nums ={4,2,3};
     if(checkPossibility(nums))
       cout<<"true";
    else
      cout<<"false";
    return  0;
}