# 2772. Apply Operations to Make All Array Elements Equal to Zero

## Description

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

## Solutions

Solution 1: Difference Array + Prefix Sum

First, let’s consider the first element of $nums$, $nums[0]$:

• If $nums[0] = 0$, we don’t need to do anything.
• If $nums[0] > 0$, we need to operate on $nums[0..k-1]$ for $nums[0]$ times, subtracting $nums[0]$ from all elements in $nums[0..k-1]$, so $nums[0]$ becomes $0$.

To perform the add and subtract operations on a contiguous segment of elements simultaneously, we can use a difference array to manage these operations. We represent the difference array with $d[i]$, and calculating the prefix sum of the difference array gives us the change of the value at each position.

Therefore, we iterate over $nums$. For each element $nums[i]$, the current position’s change quantity is $s = \sum_{j=0}^{i} d[j]$. We add $s$ to $nums[i]$ to get the actual value of $nums[i]$.

• If $nums[i] = 0$, there’s no need for any operation, and we can skip directly.
• If $nums[i]=0$ or $i + k > n$, it indicates that after the previous operations, $nums[i]$ has become negative, or $nums[i..i+k-1]$ is out of bounds. Therefore, it’s impossible to make all elements in $nums$ equal to $0$. We return false. Otherwise, we need to subtract $nums[i]$ from all elements in the interval $[i..i+k-1]$. Therefore, we subtract $nums[i]$ from $s$ and add $nums[i]$ to $d[i+k]$.
• We continue to iterate over the next element.

If the iteration ends, it means that all elements in $nums$ can be made equal to $0$, so we return true.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.

• 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;
}
}

• class Solution {
public:
bool checkArray(vector<int>& nums, int k) {
int n = nums.size();
vector<int> d(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;
}
};

• 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


• 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
}

• 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;
}