Arrays: K Closest ELements

Problem Statement:
       Given a number in a sorted array , find K closest elements to that number.

Closest Integers in the vicinity of 8 are : 5,6,7,10

Solution: 
Run Source Code

     Since the given array is sorted , binary search can be applied to get the index of the given number.
We will then place two pointers viz left and right , which will decremented and incremented respectively based on the minimum difference with the given number.

Below is the logic implementation.

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public static int[] KClosestElements(int[] arr, int index, int k){ int left = index -1; int right = index +1; int[] result = new int[k]; while(right - left -2 < k ) { if(arr[right] - arr[index] > arr[index]- arr[left]) { if(left>0) result[right - left - 2] = arr[left--]; else result[right - left - 2]=arr[right++]; } else { if(right < arr.length) result[right - left - 2]=arr[right++]; else result[right - left - 2] = arr[left--]; } //result[right - left] = arr[right] - arr[index] > arr[index]- arr[left] ? if(left>0){arr[left--]} : arr[right++] ; } System.out.println(Arrays.toString(result)); return result; }

Binary Search Sub-routine

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private static int binarySearch(int[]arr,int num,int low,int high) { while(low<=high) { int mid=(low+high)/2; if(arr[mid]==num) return mid; else if(arr[mid] < num) low=mid+1; else if (arr[mid] > num) high=mid-1; } return -1; }

Please post comments and suggestions.
Happy Coding !! :)