LeetCode 2239. Find Closest Number to Zero Solution in Java, C++, Python & More | Explanation + Code

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 d to record the current minimum distance, initially d=∞. Then we traverse the array, for each element x, we calculate y=|x|. If y \lt d or y=d and x \gt ans, we update the answer ans=x and 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).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)