Last updated on October 5th, 2024 at 03:57 pm

Here, We see Non-decreasing Array LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Non-decreasing Array LeetCode Solution

Problem Statement

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 cannot get a non-decreasing array by modifying at most one element.

1. Non-decreasing Array Leetcode Solution C++

class Solution {
public:
    bool checkPossibility(vector<int>& nums) 
    {
        for(int i=0;i<nums.size()-1;i++)
        {
            if(nums[i]>nums[i+1])
            {
                 if(i-1>=0&&nums[i-1]>nums[i+1])
                {
                    nums[i+1]=nums[i];
                }
                else
                {
                    nums[i]=nums[i+1];
                }
                break;
            }
        }
        for(int i=0;i<nums.size()-1;i++)
        {
            if(nums[i]>nums[i+1])
            {
                return false;
            }
        }
        return true;  
    }
};

2. Non-decreasing Array Leetcode Solution Java

class Solution {
    public boolean checkPossibility(int[] nums) {
        for (int i = 1, err = 0; i < nums.length; i++)
            if (nums[i] < nums[i-1])
                if (err++ > 0 || (i > 1 && i < nums.length - 1 && nums[i-2] > nums[i] && nums[i+1] < nums[i-1]))
                    return false;
        return true;
    }
}

3. Non-decreasing Array Leetcode Solution JavaScript

var checkPossibility = function(nums) {
    for (let i = 1, err = 0; i < nums.length; i++)
        if (nums[i] < nums[i-1])
            if (err++ || (i > 1 && i < nums.length - 1 && nums[i-2] > nums[i] && nums[i+1] < nums[i-1]))
                return false 
    return true
};

4. Non-decreasing Array Leetcode Solution Python

class Solution(object):
    def checkPossibility(self, nums):
        err = 0
        for i in range(1, len(nums)):
            if nums[i] < nums[i-1]:
                if err or (i > 1 and i < len(nums) - 1 and nums[i-2] > nums[i] and nums[i+1] < nums[i-1]):
                    return False
                err = 1
        return True