Leetcode #2772: Apply Operations to Make All Array Elements Equal to Zero
In this guide, we solve Leetcode #2772 Apply Operations to Make All Array Elements Equal to Zero in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Prefix Sum
Intuition
Each operation decreases a length-k window by 1, so we only need to know how much total decrement is currently affecting index i.
If we track that running decrement with a difference array, we can greedily decide how many operations must start at i to make nums[i] become zero.
Approach
Maintain a difference array d and a running sum s of active decrements.
At each index, compute the effective value x = nums[i] + s. If x is negative, it is impossible. If x is positive, we must start x operations at i, which decreases s by x and ends those operations at i + k.
Steps:
- Initialize a difference array
dand running sums. - For each index
i, updatesand compute the effective value. - If the value is positive, apply that many operations starting at
i; if invalid, returnFalse.
Example
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].
Python Solution
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
Complexity
The time complexity is , and the space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.