Leetcode 2772: Apply Operations to Make All Array Elements Equal to Zero

Input: A 0-indexed integer array nums and an integer k.

Example: nums = [3,3,2,1,0], k = 2

Constraints:

• 1 <= k <= nums.length <= 100,000

• 0 <= nums[i] <= 1,000,000

Output: Return a boolean value: true if all elements can be made 0, and false otherwise.

Example: Output: true

Constraints:

• Output is either true or false.

Goal: Check if all elements in the array can be reduced to 0 using the specified operation.

Steps:

• Iterate through the array while keeping track of the cumulative decrease required to meet the operation conditions.

• At each index, ensure the required decrease does not exceed the current element.

• Adjust the cumulative decrease using a sliding window technique to manage subarrays of size k.

• Return true if all elements can be made 0; otherwise, return false.

Goal: Limits and conditions for input and output validity.

Steps:

• The array nums has at least 1 element and at most 100,000 elements.

• k is a positive integer no greater than the length of nums.

• Elements of nums are non-negative integers.

Assumptions:

• The operation can be performed any number of times.

• The array elements are non-negative.

• Input: nums = [3,2,1,0], k = 2

• Explanation: Select the subarray [3,2] to reduce it by 1. Repeat until all elements are 0. The output is true.

• Input: nums = [1,3,1,1], k = 3

• Explanation: It is not possible to reduce all elements to 0. The output is false.

Approach: Use a greedy algorithm combined with a sliding window to manage the operations efficiently.

Observations:

• At each step, only subarrays of size k can be reduced.

• The operation is cumulative and impacts the next elements.

• Keep track of the cumulative decrease to efficiently determine if the operation is valid for each index.

Steps:

• Initialize a variable to track the cumulative decrease.

• Iterate through the array, applying the cumulative decrease to each element.

• If the cumulative decrease exceeds the current element, return false.

• Use a sliding window approach to remove the cumulative impact of elements outside the current subarray of size k.

• After processing all elements, verify if the cumulative decrease has reached 0.

Empty Inputs:

• An empty array is invalid as per constraints.

Large Inputs:

• Test arrays with the maximum size (100,000) and values to ensure efficiency.

Special Values:

• Test arrays with all elements as 0, which should immediately return true.

Constraints:

• Ensure k is less than or equal to the array size.

bool checkArray(vector<int>& A, int k) {
    int cur = 0, n = A.size();
    for (int i = 0; i < n; ++i) {
        if (cur > A[i])
            return false;
        A[i] -= cur;
        cur += A[i];
        if (i >= k - 1)
            cur -= A[i - k + 1];
    }
    return cur == 0;
}

1 : Variable Initialization

bool checkArray(vector<int>& A, int k) {

This line declares the function `checkArray` that accepts a reference to an integer vector `A` and an integer `k`, which represents the size of the subarrays we are considering.

2 : Variable Initialization

    int cur = 0, n = A.size();

The variable `cur` is initialized to 0, which will be used to keep track of the sum of the elements in the sliding window. `n` stores the size of the array `A`.

3 : Looping

    for (int i = 0; i < n; ++i) {

A loop is initiated to iterate through the elements of the array `A` from index `0` to `n-1`.

4 : Conditional Function Call

        if (cur > A[i])

This condition checks if the current value `cur` exceeds the current element `A[i]`. If true, it means the transformation is not possible, so the function returns `false`.

5 : Return Statement

            return false;

If the condition `cur > A[i]` is met, the function immediately returns `false`, indicating that the desired transformation is not possible.

6 : Variable Manipulation

        A[i] -= cur;

The current element `A[i]` is adjusted by subtracting the value of `cur`, essentially modifying the array during the iteration.

7 : Variable Manipulation

        cur += A[i];

The modified value of `A[i]` is added to `cur`, updating the sum of the current window.

8 : Conditional Function Call

        if (i >= k - 1)

This condition checks if the index `i` has reached or exceeded `k - 1`, which marks the point where we begin reducing the window size.

9 : Variable Manipulation

            cur -= A[i - k + 1];

When the condition `i >= k - 1` is true, the value at `A[i - k + 1]` is subtracted from `cur` to shrink the sliding window by removing the effect of the element that has moved out of the window.

10 : Return Statement

    return cur == 0;

After completing the loop, the function returns `true` if `cur` is equal to 0, indicating that the transformation is possible, otherwise `false`.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The solution involves a single pass through the array, with constant-time operations per index.

Best Case: O(1)

Worst Case: O(1)

Description: The solution uses a constant amount of extra space.

Leetcode 2772: Apply Operations to Make All Array Elements Equal to Zero

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