LeetCode #665 — MEDIUM

Non-decreasing Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

Constraints:

n == nums.length
1 <= n <= 10⁴
-10⁵ <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[4,2,3]

Example 2

[4,2,1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #665: Non-decreasing Array
class Solution {
    public boolean checkPossibility(int[] nums) {
        for (int i = 0; i < nums.length - 1; ++i) {
            int a = nums[i], b = nums[i + 1];
            if (a > b) {
                nums[i] = b;
                if (isSorted(nums)) {
                    return true;
                }
                nums[i] = a;
                nums[i + 1] = a;
                return isSorted(nums);
            }
        }
        return true;
    }

    private boolean isSorted(int[] nums) {
        for (int i = 0; i < nums.length - 1; ++i) {
            if (nums[i] > nums[i + 1]) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #665: Non-decreasing Array
func checkPossibility(nums []int) bool {
	isSorted := func(nums []int) bool {
		for i, b := range nums[1:] {
			if nums[i] > b {
				return false
			}
		}
		return true
	}
	for i := 0; i < len(nums)-1; i++ {
		a, b := nums[i], nums[i+1]
		if a > b {
			nums[i] = b
			if isSorted(nums) {
				return true
			}
			nums[i] = a
			nums[i+1] = a
			return isSorted(nums)
		}
	}
	return true
}

# Accepted solution for LeetCode #665: Non-decreasing Array
class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        def is_sorted(nums: List[int]) -> bool:
            return all(a <= b for a, b in pairwise(nums))

        n = len(nums)
        for i in range(n - 1):
            a, b = nums[i], nums[i + 1]
            if a > b:
                nums[i] = b
                if is_sorted(nums):
                    return True
                nums[i] = nums[i + 1] = a
                return is_sorted(nums)
        return True

// Accepted solution for LeetCode #665: Non-decreasing Array
struct Solution;

impl Solution {
    fn check_possibility(nums: Vec<i32>) -> bool {
        let n = nums.len();
        if n < 2 {
            return true;
        }
        let mut p: Option<usize> = None;
        for i in 0..n - 1 {
            if nums[i] > nums[i + 1] {
                if p.is_some() {
                    return false;
                } else {
                    p = Some(i);
                }
            }
        }
        if let Some(p) = p {
            if p != 0 && p != n - 2 {
                if nums[p - 1] >= nums[p + 1] && nums[p] >= nums[p + 2] {
                    return false;
                }
            }
        }

        true
    }
}

#[test]
fn test() {
    assert_eq!(Solution::check_possibility(vec![4, 2, 3]), true);
    assert_eq!(Solution::check_possibility(vec![4, 2, 1]), false);
}

// Accepted solution for LeetCode #665: Non-decreasing Array
function checkPossibility(nums: number[]): boolean {
    const isSorted = (nums: number[]) => {
        for (let i = 0; i < nums.length - 1; ++i) {
            if (nums[i] > nums[i + 1]) {
                return false;
            }
        }
        return true;
    };
    for (let i = 0; i < nums.length - 1; ++i) {
        const a = nums[i],
            b = nums[i + 1];
        if (a > b) {
            nums[i] = b;
            if (isSorted(nums)) {
                return true;
            }
            nums[i] = a;
            nums[i + 1] = a;
            return isSorted(nums);
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.