Find Closest Number to Zero

📄 Problem Statement

✅ Problem: Find Closest Number to Zero

Given an integer array nums, return the number with the value closest to 0.

Note:

• If there are multiple answers, return the number with the largest value.
• Use absolute value to determine closeness.

Example 1:

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

Example 2:

• Input: nums = [2, -1, 1]
• Output: 1

Constraints:

• 1 <= nums.length <= 1000
• -10^5 <= nums[i] <= 10^5

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🧠 Problem Explanation

Given an integer array nums, return the number with the value closest to 0.

If there are multiple answers (e.g., -x and x), return the number with the larger value (i.e., the positive one).

💡 Approach

• Use a linear scan to go through the array.
• Track the number with the smallest absolute value.
• If two numbers are equally close to 0, pick the positive one.

📊 Complexities

• Time Complexity: O(n)
• Space Complexity: O(1)

🧾 Pseudocode

function: findClosestNumber
params:
  - nums: List[int]
logic:
  - initialize closest with nums[0]
  - for each num in nums:
      - if abs(num) < abs(closest):
          - set closest = num
      - else if abs(num) == abs(closest) and num > closest:
          - set closest = num
  - return closest

✅ Code

class Solution(object):
    def findClosestNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        number = nums[0]
        for i in nums:
            if abs(i) < abs(number):
                number = i

        if abs(number) in nums:
            return abs(number)
        else:
            return number

🧪 Example Tests

• Input: [-4, -2, 1, 4, 8]
  Expected Output: 1

• Input: [2, -1, 1]
  Expected Output: 1

📝 Notes

• Pattern Used: Linear Scan
• Insights:
  • I came up with a solution intuitively using a brute-force approach. Since the time complexity was acceptable, I decided to proceed with it.
  • How to use abs() for comparison and handle tie-breaking effectively.