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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Understanding the Visualization

Input Array

Array with positive and zero values

Find Unique

Count distinct non-zero values

Operations

Each unique value needs one operation

Key Takeaway

🎯 Key Insight: Count unique non-zero values - each requires exactly one operation to eliminate

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

Approach	Time	Space	Notes
✓ Set-Based Count	O(n)	O(n)	Use set to count unique non-zero values
Brute Force Simulation	O(n²)	O(1)	Simulate each operation by repeatedly finding minimum and subtracting

Set-Based Count — Algorithm Steps

Add all non-zero values to set
Return size of set

Visualization

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Step-by-Step Walkthrough

Process Array

Add non-zero values to set

Count Unique

Set automatically handles duplicates

Result

Set size equals operations needed

Code -

solution.c — C

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

int solution(int* nums, int size) {
    int unique[1001] = {0}; // Assuming values <= 1000
    int count = 0;
    
    for (int i = 0; i < size; i++) {
        if (nums[i] > 0 && !unique[nums[i]]) {
            unique[nums[i]] = 1;
            count++;
        }
    }
    return count;
}

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(n)

Single pass through array to add elements to set

✓ Linear Growth

Space Complexity

O(n)

Set can store up to n unique values in worst case

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Set-Based Count — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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