# Solving Leetcode Interviews in Seconds with AI: Apply Operations to Make All Array Elements Equal to Zero


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2772" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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  

	# Explanation
	Here's the approach and the code:

*   **Key Idea:** Iterate through the array, and for each element `nums[i]`, if it's greater than 0, apply the operation (decrease the subarray of size `k` starting at `i`) by the value of `nums[i]`. Use a running sum to keep track of the net decrease applied to each index.
*   **Optimization:** Instead of actually modifying the array, maintain a running sum of changes applied to previous subarrays. This avoids unnecessary array manipulations and improves efficiency.
*   **Complexity:**
    *   Runtime: O(n)
    *   Storage: O(1)

	
	# Code
	```python
	def canMakeZeros(nums, k):
    n = len(nums)
    current_decrease = 0
    for i in range(n):
        if current_decrease > nums[i]:
            return False
        nums[i] -= current_decrease
        if nums[i] > 0:
            if i + k > n:
                return False
            current_decrease += nums[i]
            if i + k < n:
                current_decrease -= nums[i]
    return True
	```