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2239. Find Closest Number to Zero

Description

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
  • -105 <= nums[i] <= 105

Solutions

Solution 1: Single Pass

We define a variable \(\textit{d}\) to record the current minimum distance, initially \(\textit{d}=\infty\). Then we traverse the array, for each element \(x\), we calculate \(y=|x|\). If \(y \lt d\) or \(y=d\) and \(x \gt \textit{ans}\), we update the answer \(\textit{ans}=x\) and \(\textit{d}=y\).

After the traversal, return the answer.

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

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class Solution:
    def findClosestNumber(self, nums: List[int]) -> int:
        ans, d = 0, inf
        for x in nums:
            if (y := abs(x)) < d or (y == d and x > ans):
                ans, d = x, y
        return ans

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class Solution {
    public int findClosestNumber(int[] nums) {
        int ans = 0, d = 1 << 30;
        for (int x : nums) {
            int y = Math.abs(x);
            if (y < d || (y == d && x > ans)) {
                ans = x;
                d = y;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int findClosestNumber(vector<int>& nums) {
        int ans = 0, d = 1 << 30;
        for (int x : nums) {
            int y = abs(x);
            if (y < d || (y == d && x > ans)) {
                ans = x;
                d = y;
            }
        }
        return ans;
    }
};

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func findClosestNumber(nums []int) int {
    ans, d := 0, 1<<30
    for _, x := range nums {
        if y := abs(x); y < d || (y == d && x > ans) {
            ans, d = x, y
        }
    }
    return ans
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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function findClosestNumber(nums: number[]): number {
    let [ans, d] = [0, 1 << 30];
    for (const x of nums) {
        const y = Math.abs(x);
        if (y < d || (y == d && x > ans)) {
            [ans, d] = [x, y];
        }
    }
    return ans;
}

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