[LintCode] K Closest Numbers In Sorted Array

Given a target number, a non-negative integer k and an integer array A sorted in ascending order, find the k closest numbers to target in A, sorted in ascending order by the difference between the number and target. Otherwise, sorted in ascending order by number if the difference is same.

Example

Given A = [1, 2, 3], target = 2 and k = 3, return [2, 1, 3].

Given A = [1, 4, 6, 8], target = 3 and k = 3, return [4, 1, 6].

Challenge

O(logn + k) time complexity.

Algorithm

1. Find the index of the number closest to target.

2. Start from the number found in step 1, use two pointers to check both forward and backward, to get k closest numbers to the target.

 1 public class Solution {
 2     /**
 3      * @param A an integer array
 4      * @param target an integer
 5      * @param k a non-negative integer
 6      * @return an integer array
 7      */
 8     public int[] kClosestNumbers(int[] A, int target, int k) {
 9         if(k <= 0 || k > A.length) {
10             return new int[0];
11         }
12         int idx = closetNumberIdxIteration(A, target, 0, A.length - 1);
13         int left = idx - 1, right = idx + 1;
14         int[] result = new int[k];
15         result[0] = A[idx];
16         for(int i = 1; i < k; i++) {
17             if(left >= 0 && left < A.length && right >= 0 && right < A.length) {
18                 if(Math.abs(A[left] - target) <= Math.abs(A[right] - target)) {
19                     result[i] = A[left];
20                     left--;
21                 }
22                 else {
23                     result[i] = A[right];
24                     right++;
25                 }                
26             }
27             else if(left < 0 || left == A.length) {
28                 result[i] = A[right];
29                 right++;
30             }
31             else {
32                 result[i] = A[left];
33                 left--;
34             }
35         }
36         return result;
37     }
38     private int closestNumberIdxRecursion(int[] A, int target, int lo, int hi)
39     {
40         if(hi - lo <= 1)
41         {
42             if(Math.abs(A[lo] - target) <= Math.abs(A[hi] - target))
43             {
44                 return lo;
45             }
46             else
47             {
48                 return hi;
49             }
50         }
51         int mid = lo + (hi - lo) / 2;
52         if(A[mid] == target)
53         {
54             return mid;
55         }
56         else if(A[mid] > target)
57         {
58             return closestNumberIdxRecursion(A, target, lo, mid);
59         }
60         else
61         {
62             return closestNumberIdxRecursion(A, target, mid, hi);
63         }
64     }
65     private int closetNumberIdxIteration(int[] A, int target, int lo, int hi) {
66         while(lo + 1 < hi) {
67             int mid = lo + (hi - lo) / 2;
68             if(A[mid] == target) {
69                 return mid;
70             }
71             else if(A[mid] > target) {
72                 hi = mid;
73             }
74             else {
75                 lo = mid;
76             }
77         }
78         return Math.abs(A[lo] - target) <= Math.abs(A[hi] - target) ? lo : hi;
79     }
80 }