Make Array Zero by Subtracting Equal Amounts

Make Array Zero by Subtracting Equal Amounts - Problem

Array Hash Table Greedy Sorting Heap (Priority Queue) Simulation Easy

You are given a non-negative integer array nums. In one operation, you must:

Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.
Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,5,0,3,5]

› Output: 3

💡 Note: Operation 1: x=1, subtract from all positives → [0,4,0,2,4]. Operation 2: x=2, subtract from all positives → [0,2,0,0,2]. Operation 3: x=2, subtract from all positives → [0,0,0,0,0]. Total: 3 operations.

Example 2 — With Zeros

$ Input: nums = [0]

› Output: 0

💡 Note: Array already contains all zeros, so no operations needed.

Example 3 — All Same Values

$ Input: nums = [2,2,2,2]

› Output: 1

💡 Note: All elements are the same non-zero value, so only one operation needed to make all zeros.

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 M Meta 12

The key insight is that each unique non-zero value requires exactly one operation to eliminate. Best approach is to count unique non-zero values using a set. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(n²) Space: O(1)

For each operation, find the smallest non-zero element, then subtract it from all positive elements. Continue until all elements become zero.

Set-Based Count

⏱️ Time: O(n) Space: O(n)

The key insight is that each unique non-zero value requires exactly one operation. Use a set to count distinct non-zero elements.

Brute Force Simulation — Algorithm Steps

Find smallest non-zero element
Subtract it from all positive elements
Increment operation count
Repeat until all zeros

Visualization

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Step-by-Step Walkthrough

Find Minimum

Find smallest non-zero element

Subtract

Subtract from all positive elements

Repeat

Continue until all zeros

Code -

solution.c — C

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

int solution(int* nums, int size) {
    int operations = 0;
    while (1) {
        int minNonzero = -1;
        for (int i = 0; i < size; i++) {
            if (nums[i] > 0) {
                if (minNonzero == -1 || nums[i] < minNonzero) {
                    minNonzero = nums[i];
                }
            }
        }
        if (minNonzero == -1) break;
        
        for (int i = 0; i < size; i++) {
            if (nums[i] > 0) {
                nums[i] -= minNonzero;
            }
        }
        operations++;
    }
    return operations;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* ptr = strchr(line, '[') + 1;
    while (*ptr != ']') {
        nums[size++] = strtol(ptr, &ptr, 10);
        while (*ptr == ',' || *ptr == ' ') ptr++;
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Outer loop runs for each unique value, inner loop processes all n elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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