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# Find K Closest Elements LeetCode Solution

## Problem – Find K Closest Elements

Given a sorted integer array `arr`, two integers `k` and `x`, return the `k` closest integers to `x` in the array. The result should also be sorted in ascending order.

An integer `a` is closer to `x` than an integer `b` if:

• `|a - x| < |b - x|`, or
• `|a - x| == |b - x|` and `a < b`

Example 1:

``````Input: arr = [1,2,3,4,5], k = 4, x = 3
Output: [1,2,3,4]``````

Example 2:

``````Input: arr = [1,2,3,4,5], k = 4, x = -1
Output: [1,2,3,4]``````

Constraints:

• `1 <= k <= arr.length`
• `1 <= arr.length <= 104`
• `arr` is sorted in ascending order.
• `-104 <= arr[i], x <= 104`

### Find K Closest Elements LeetCode Solution in Java

``````    public List<Integer> findClosestElements(int[] A, int k, int x) {
int left = 0, right = A.length - k;
while (left < right) {
int mid = (left + right) / 2;
if (x - A[mid] > A[mid + k] - x)
left = mid + 1;
else
right = mid;
}
return Arrays.stream(A, left, left + k).boxed().collect(Collectors.toList());
}
``````

### Find K Closest Elements LeetCode Solution in C++

``````    vector<int> findClosestElements(vector<int>& A, int k, int x) {
int left = 0, right = A.size() - k;
while (left < right) {
int mid = (left + right) / 2;
if (x - A[mid] > A[mid + k] - x)
left = mid + 1;
else
right = mid;
}
return vector<int>(A.begin() + left, A.begin() + left + k);
}
``````

### Find K Closest Elements LeetCode Solution in Python

``````    def findClosestElements(self, A, k, x):
left, right = 0, len(A) - k
while left < right:
mid = (left + right) / 2
if x - A[mid] > A[mid + k] - x:
left = mid + 1
else:
right = mid
return A[left:left + k]
``````
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