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Formatted question description: https://leetcode.ca/all/2239.html
2239. Find Closest Number to Zero (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.
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
10^{5} <= nums[i] <= 10^{5}
Similar Questions:
Solution 1.

// OJ: https://leetcode.com/problems/findclosestnumbertozero/ // Time: O(N) // Space: O(1) class Solution { public: int findClosestNumber(vector<int>& A) { int diff = INT_MAX, ans = INT_MAX; for (int n : A) { if (abs(n) < diff  (abs(n) == diff && n > ans)) { diff = abs(n); ans = n; } } return ans; } };

// Todo

class Solution: def findClosestNumber(self, nums: List[int]) > int: ans, d = 0, 1000000 for v in nums: if (t := abs(v)) < d or (t == d and v > ans): ans, d = v, t return ans ############ # 2239. Find Closest Number to Zero # https://leetcode.com/problems/findclosestnumbertozero/ class Solution: def findClosestNumber(self, nums: List[int]) > int: m = float('inf') for x in nums: if abs(x) < abs(m): m = x elif abs(x) == abs(m): m = max(m, x) return m
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