LeetCode #2772 — MEDIUM

Apply Operations to Make All Array Elements Equal to Zero

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

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed integer array nums and a positive integer k.

You can apply the following operation on the array any number of times:

Choose any subarray of size k from the array and decrease all its elements by 1.

Return true if you can make all the array elements equal to 0, or false otherwise.

A subarray is a contiguous non-empty part of an array.

Example 1:

Input: nums = [2,2,3,1,1,0], k = 3
Output: true
Explanation: We can do the following operations:
- Choose the subarray [2,2,3]. The resulting array will be nums = [1,1,2,1,1,0].
- Choose the subarray [2,1,1]. The resulting array will be nums = [1,1,1,0,0,0].
- Choose the subarray [1,1,1]. The resulting array will be nums = [0,0,0,0,0,0].

Example 2:

Input: nums = [1,3,1,1], k = 2
Output: false
Explanation: It is not possible to make all the array elements equal to 0.

Constraints:

1 <= k <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums and a positive integer k. You can apply the following operation on the array any number of times: Choose any subarray of size k from the array and decrease all its elements by 1. Return true if you can make all the array elements equal to 0, or false otherwise. A subarray is a contiguous non-empty part of an array.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,2,3,1,1,0]
3

Example 2

[1,3,1,1]
2

Core Insight

What unlocks the optimal approach

In case it is possible, then how can you do the operations? which subarrays do you choose and in what order?
The order of the chosen subarrays should be from the left to the right of the array

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 #2772: Apply Operations to Make All Array Elements Equal to Zero
class Solution {
    public boolean checkArray(int[] nums, int k) {
        int n = nums.length;
        int[] d = new int[n + 1];
        int s = 0;
        for (int i = 0; i < n; ++i) {
            s += d[i];
            nums[i] += s;
            if (nums[i] == 0) {
                continue;
            }
            if (nums[i] < 0 || i + k > n) {
                return false;
            }
            s -= nums[i];
            d[i + k] += nums[i];
        }
        return true;
    }
}

// Accepted solution for LeetCode #2772: Apply Operations to Make All Array Elements Equal to Zero
func checkArray(nums []int, k int) bool {
	n := len(nums)
	d := make([]int, n+1)
	s := 0
	for i, x := range nums {
		s += d[i]
		x += s
		if x == 0 {
			continue
		}
		if x < 0 || i+k > n {
			return false
		}
		s -= x
		d[i+k] += x
	}
	return true
}

# Accepted solution for LeetCode #2772: Apply Operations to Make All Array Elements Equal to Zero
class Solution:
    def checkArray(self, nums: List[int], k: int) -> bool:
        n = len(nums)
        d = [0] * (n + 1)
        s = 0
        for i, x in enumerate(nums):
            s += d[i]
            x += s
            if x == 0:
                continue
            if x < 0 or i + k > n:
                return False
            s -= x
            d[i + k] += x
        return True

// Accepted solution for LeetCode #2772: Apply Operations to Make All Array Elements Equal to Zero
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2772: Apply Operations to Make All Array Elements Equal to Zero
// class Solution {
//     public boolean checkArray(int[] nums, int k) {
//         int n = nums.length;
//         int[] d = new int[n + 1];
//         int s = 0;
//         for (int i = 0; i < n; ++i) {
//             s += d[i];
//             nums[i] += s;
//             if (nums[i] == 0) {
//                 continue;
//             }
//             if (nums[i] < 0 || i + k > n) {
//                 return false;
//             }
//             s -= nums[i];
//             d[i + k] += nums[i];
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #2772: Apply Operations to Make All Array Elements Equal to Zero
function checkArray(nums: number[], k: number): boolean {
    const n = nums.length;
    const d: number[] = Array(n + 1).fill(0);
    let s = 0;
    for (let i = 0; i < n; ++i) {
        s += d[i];
        nums[i] += s;
        if (nums[i] === 0) {
            continue;
        }
        if (nums[i] < 0 || i + k > n) {
            return false;
        }
        s -= nums[i];
        d[i + k] += nums[i];
    }
    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(n)

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.