Find Closest Number to Zero

Problem

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.

Examples

Example 1:

Input: nums = [-4,-2,1,4,8] Output: 1 Explanation: The distance from -4 to 0 is |-4| = 4. The distance from -2 to 0 is |-2| = 2. The distance from 1 to 0 is |1| = 1. The distance from 4 to 0 is |4| = 4. The distance from 8 to 0 is |8| = 8. Thus, the closest number to 0 in the array is 1.

Example 2:

Input: nums = [2,-1,1] Output: 1 Explanation: 1 and -1 are both the closest numbers to 0, so 1 being larger is returned.

Constraints

• 1 <= n <= 1000
• 105 <= nums[i] <= 105

Approach

Iterate through the array and keep track of the number with the smallest absolute value (closest to zero). If two numbers have the same absolute value, the larger number is preferred.

Solution

Code in Different Languages

C++ Solution

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int closestToZero(vector<int>& nums) {
    int closest = nums[0];
    for (int i = 1; i < nums.size(); ++i) {
        if (abs(nums[i]) < abs(closest) || (abs(nums[i]) == abs(closest) && nums[i] > closest)) {
            closest = nums[i];
        }
    }
    return closest;
}

int main() {
    vector<int> nums = {-4, -2, 1, 4, 8};
    int result = closestToZero(nums);
    cout << "Output: " << result << endl;  // Output: 1
    
    vector<int> nums2 = {2, -1, 1};
    int result2 = closestToZero(nums2);
    cout << "Output: " << result2 << endl;  // Output: 1
    
    return 0;
}

Java Solution

import java.util.Arrays;

public class ClosestToZero {
    public static int closestToZero(int[] nums) {
        int closest = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if (Math.abs(nums[i]) < Math.abs(closest) || (Math.abs(nums[i]) == Math.abs(closest) && nums[i] > closest)) {
                closest = nums[i];
            }
        }
        return closest;
    }
    
    public static void main(String[] args) {
        int[] nums1 = {-4, -2, 1, 4, 8};
        System.out.println("Output: " + closestToZero(nums1));  // Output: 1
        
        int[] nums2 = {2, -1, 1};
        System.out.println("Output: " + closestToZero(nums2));  // Output: 1
    }
}

Python Solution

def closest_to_zero(nums):
    closest = nums[0]
    for num in nums[1:]:
        if abs(num) < abs(closest) or (abs(num) == abs(closest) and num > closest):
            closest = num
    return closest

# Example usage:
nums1 = [-4, -2, 1, 4, 8]
print("Output:", closest_to_zero(nums1))  # Output: 1

nums2 = [2, -1, 1]
print("Output:", closest_to_zero(nums2))  # Output: 1

Complexity Analysis

Time Complexity: $O(n)$

Reason:Iteration through the array where n is the length of the array.

Space Complexity: $O(n)$

Reason: for storing elements of the array.