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 $nums$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}