java - Return the k elements of an array farthest from val -

method needs return k elements a[i] such abs(a[i] - val) k largest evaluation. code works integers greater val. fail if integers less val. can without importing other java.util.arrays? enlighten me? appreciated!

 public static int[] farthestk(int[] a, int val, int k) {// line should not change   int[] b = new int[a.length];   (int = 0; < b.length; i++) {      b[i] = math.abs(a[i] - val);   }   arrays.sort(b);   int[] c = new int[k];   int w = 0;   (int = b.length-1; > b.length-k-1; i--) {             c[w] = b[i] + val;      w++;        }   return c;     } 

test case:

  @test public void farthestktest() {          int[] = {-2, 4, -6, 7, 8, 13, 15};          int[] expected = {15, -6, 13, -2};          int[] actual = selector.farthestk(a, 4, 4);          assert.assertarrayequals(expected, actual);        }   there 1 failure:  1) farthestktest(selectortest)  arrays first differed @ element [1]; expected:<-6> was:<14>  failures!!!  tests run: 1,  failures: 1 

the top k problem can solved in many ways. in case add new parameter, doesn't matter.

the first , easiest one: sort array. time complexity: o(nlogn)

public static int[] farthestk(integer[] a, final int val, int k) {     arrays.sort(a, new java.util.comparator<integer>() {         @override         public int compare(integer o1, integer o2) {             return -math.abs(o1 - val) + math.abs(o2 - val);         }     });     int[] c = new int[k];     (int = 0; < k; i++) {         c[i] = a[i];     }     return c; } 

the second way: use heap save max k values, time complexity: o(nlogk)

/**  * use min heap save max k values. time complexity: o(nlogk)  */ public static int[] farthestkwithheap(integer[] a, final int val, int k) {     priorityqueue<integer> minheap = new priorityqueue<integer>(4,             new java.util.comparator<integer>() {                 @override                 public int compare(integer o1, integer o2) {                     return math.abs(o1 - val) - math.abs(o2 - val);                 }             });     (int : a) {         minheap.add(i);         if (minheap.size() > k) {             minheap.poll();         }     }     int[] c = new int[k];     (int = 0; < k; i++) {         c[i] = minheap.poll();     }     return c; } 

the third way: divide , conquer, quicksort. partition array 2 part, , find kth in 1 of them. time complexity: o(n + klogk) code little long, provide link here.

selection problem.