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Dorchester Center, MA 02124

Given an integer array `nums`

of size `n`

, return *the number with the value closest to *

`0`

`nums`

. If there are multiple answers, return **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}

```
int findClosestNumber(vector<int>& nums) {
return *min_element(begin(nums), end(nums), [](int a, int b) {
return abs(a) < abs(b) || (abs(a) == abs(b) && a > b);
});
}
```

```
def findClosestNumber(self, A):
return max([-abs(a), a] for a in A)[1]
```

```
// If absolute of n is less than min, update the closest_num
// If absolute of n is same of as min, update the bigger closest_num
class Solution {
public int findClosestNumber(int[] nums) {
int min = Integer.MAX_VALUE, closest_num = 0;
for(int n : nums) {
if(min > Math.abs(n)) {
min = Math.abs(n);
closest_num = n;
} else if(min == Math.abs(n) && closest_num < n) {
closest_num = n;
}
}
return closest_num;
}
}
```

```
var findClosestNumber = function(nums) {
let closest = Infinity;
for (let num of nums) {
if (Math.abs(num) < Math.abs(closest)) {
closest = num
} else if (Math.abs(num) === Math.abs(closest)) {
closest = Math.max(num, closest)
}
}
return closest;
};
```

```
import kotlin.math.abs
class Solution {
fun findClosestNumber(nums: IntArray): Int {
var closest: Int? = null
for (num in nums) {
if (closest == null) {
closest = num
} else {
if (abs(num) < abs(closest)) {
closest = num
} else if (abs(num) == abs(closest)) {
if (num > closest) {
closest = num
}
}
}
}
return closest!!
}
}
```

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