Apply Operations to Make All Array Elements Equal to Zero

Apply Operations to Make All Array Elements Equal to Zero - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,1,3,4], k = 3

› Output: true

💡 Note: Apply operation on [2,1,3] twice to get [0,0,1,4], then on [0,1,4] once to get [0,0,0,3], then on [0,0,3] three times to get [0,0,0,0]

Example 2 — Impossible Case

$ Input: nums = [1,3,1,1], k = 2

› Output: false

💡 Note: Cannot make all elements zero because the last element cannot be part of any k-sized subarray when operations are constrained

Example 3 — Edge Case

$ Input: nums = [0,0,0], k = 1

› Output: true

💡 Note: All elements are already zero, no operations needed

Constraints

1 ≤ nums.length ≤ 10⁵
0 ≤ nums[i] ≤ 10⁶
1 ≤ k ≤ nums.length

Visualization

Tap to expand

Understanding the Visualization

Input

Array nums and subarray size k

Apply Operations

Choose k consecutive elements and decrease by 1

Check Result

Determine if all elements can become 0

Key Takeaway

🎯 Key Insight: Use prefix sum to track operation effects efficiently rather than simulating each step

Asked in

G Google 15 M Microsoft 12

The key insight is to use a prefix sum approach to track operations efficiently rather than simulating each operation. We maintain a running count of active operations and check if each element can be reduced to zero. Best approach is prefix sum with difference array technique. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n × max(nums))	O(1)	Simulate the operation by repeatedly finding positive elements and applying k-sized decrements
Prefix Sum Difference Array	O(n)	O(1)	Use difference array technique to track operations efficiently without simulation

Brute Force Simulation — Algorithm Steps

Find first positive element in array
Apply operation (decrease k consecutive elements by 1)
Repeat until all elements are 0 or operation fails

Visualization

Tap to expand

Step-by-Step Walkthrough

Find First Positive

Scan array to find first element > 0

Apply Operation

Decrease k consecutive elements by 1

Repeat

Continue until all zeros or impossible

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(int* nums, int numsSize, int k) {
    // Create copy
    int* numsCopy = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        numsCopy[i] = nums[i];
    }
    
    while (true) {
        // Find first positive element
        int firstPos = -1;
        for (int i = 0; i < numsSize; i++) {
            if (numsCopy[i] > 0) {
                firstPos = i;
                break;
            }
        }
        
        // If no positive elements, we're done
        if (firstPos == -1) {
            free(numsCopy);
            return true;
        }
        
        // Check if we can apply operation
        if (firstPos + k > numsSize) {
            free(numsCopy);
            return false;
        }
        
        // Apply operation
        for (int i = firstPos; i < firstPos + k; i++) {
            numsCopy[i] -= 1;
            if (numsCopy[i] < 0) {
                free(numsCopy);
                return false;
            }
        }
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        if (num != 0 || strcmp(token, "0") == 0) {
            nums[numsSize++] = num;
        }
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    bool result = solution(nums, numsSize, k);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × max(nums))

For each element's value, we may need to scan the entire array

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

18.0K Views

Medium Frequency

~15 min Avg. Time

450 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler