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# 658. Find K Closest Elements

Given a sorted array, two integers`k`and`x`, find the`k`closest elements to`x`in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

**Example 1:**

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

Output:
 [1,2,3,4]
```

**Example 2:**

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

Output:
 [1,2,3,4]
```

**Note:**

1. The value k is positive and will always be smaller than the length of the sorted array.
2. Length of the given array is positive and will not exceed 10^4
3. Absolute value of elements in the array and x will not exceed 10^4

**Thoughts:**

1. Binary -searching the first index i that arr\[i] is closer or equally close to arr\[i+k]  (Original [Post](https://leetcode.com/problems/find-k-closest-elements/discuss/106419/O%28log-n%29-Java-1-line-O%28log%28n%29-+-k%29-Ruby))
2. Binary-searching for`x`and then expanding to the left and to the right: The idea is to find the first number which is equal to or greater than`x`in`arr`. Then, we determine the indices of the start and the end of a subarray in`arr`, where the subarray is our result. The time complexity is `O(logn + k)`

   .

**Code:** Binary -searching for the first index i **T: O(log(n-k))**

```java
class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int i = 0, j = arr.length - k;
        while (i < j){
            int mid = (i & j) + ((i ^ j) >> 1);
            // int mid = i + (j - i >> 1);
            if(x - arr[mid] > arr[mid + k] - x)
                i = mid + 1;
            else
                j = mid;
        }
        List<Integer> res = new ArrayList<>();
        for (int b = i; b < i + k; ++b) res.add(arr[b]);
        return res;
        // return Arrays.stream(Arrays.copyOfRange(arr, i, i + k)).boxed().collect(Collectors.toList()); 

    }
}
```

**Python: T: O(log(n-k))**

```python
class Solution(object):
    def findClosestElements(self, arr, k, x):
        """
        :type arr: List[int]
        :type k: int
        :type x: int
        :rtype: List[int]
        """
        i, j = 0, len(arr) - k
        while i < j:
            mid = (i & j) + ((i^j)>> 1)
            if x - arr[mid] > arr[mid + k] - x: # arr is sorted
                i = mid + 1
            else:
                j = mid

        return arr[i:i+k]
```

**Code 2:** [**Reference**](https://leetcode.com/problems/find-k-closest-elements/discuss/106439/JavaC++-Very-simple-binary-search-solution) **T: O(log(n)+k)**

```cpp
  public List<Integer> findClosestElements(List<Integer> arr, int k, int x) {
        int index = Collections.binarySearch(arr, x);
        if(index < 0) index = -(index + 1);
        int i = index - 1, j = index;                                    
        while(k-- > 0){
            if(i < 0 || (j < arr.size() && Math.abs(arr.get(i) - x) > Math.abs(arr.get(j) - x)))j++;
            else i--;
        }
        return arr.subList(i+1, j);
    }
```


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