Find Closest Number to Zero

Find Closest Number to Zero - Problem

Array Easy

Given an integer array nums of size n, return the number with the value closest to 0 in nums.

If there are multiple answers, return the number with the largest value.

Input & Output

Example 1 — Basic Case

$ Input: nums = [-4,-2,1,4,8]

› Output: 1

💡 Note: The distances are: |-4| = 4, |-2| = 2, |1| = 1, |4| = 4, |8| = 8. The closest to zero is 1.

Example 2 — Tie Breaking

$ Input: nums = [2,-1,1]

› Output: 1

💡 Note: Both -1 and 1 have distance 1 from zero. We return 1 since it's the larger value.

Example 3 — Single Element

$ Input: nums = [5]

› Output: 5

💡 Note: Only one element, so 5 is the closest to zero.

Constraints

1 ≤ nums.length ≤ 1000
-10⁵ ≤ nums[i] ≤ 10⁵

Visualization

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Asked in

G Google 32 M Microsoft 28 A Amazon 25 🍎 Apple 18

The key insight is to track the closest value while handling tie-breaking by preferring positive numbers. Best approach uses single pass comparison: Time O(n), Space O(1).

Common Approaches

✓ Brute Force - Calculate All Distances

⏱️ Time: O(n) Space: O(1)

Iterate through all elements, calculate |num - 0| for each, and track the number with minimum distance. When distances are equal, prefer the larger number.

Single Pass Optimized

⏱️ Time: O(n) Space: O(1)

Use a single loop to find the closest number by maintaining the current closest value and immediately handling tie-breaking logic when distances are equal.

Brute Force - Calculate All Distances — Algorithm Steps

Initialize closest as first element
For each number, calculate absolute distance to zero
Update closest if current distance is smaller, or same distance but larger value

Visualization

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Step-by-Step Walkthrough

Initialize

Start with first element as closest

Compare

Calculate distance for each element

Update

Keep closest value, prefer positive for ties

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int closest = nums[0];
    int minDistance = abs(nums[0]);
    
    for (int i = 0; i < numsSize; i++) {
        int distance = abs(nums[i]);
        if (distance < minDistance || (distance == minDistance && nums[i] > closest)) {
            closest = nums[i];
            minDistance = distance;
        }
    }
    
    return closest;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through all n elements

✓ Linear Growth

Space Complexity

O(1)

Only using variables for tracking closest value and distance

✓ Linear Space

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Input & Output

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Related Problems

Common Approaches

Brute Force - Calculate All Distances — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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