Created
July 26, 2011 04:30
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Write a program for finding the kth - smallest element in the array x[0..n-1] in O(n) expected time. Your algorithm may permute the elements of x.
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| package com.operations.search; | |
| public class Search{ | |
| private int[] arr=null; | |
| private int k=0; | |
| public Search(int[] arr){ | |
| this.arr = arr; | |
| if(this.arr == null) | |
| throw new SearchException("Cannot accept null array"); | |
| } | |
| /* | |
| * Total complexity O(n) | |
| */ | |
| public int findKthElement(int k){ | |
| if(k < 0 || k > this.arr.length-1){ | |
| throw new SearchException("k should be less than or equal to size of the array"); | |
| } | |
| this.k = k; | |
| return find(0,this.arr.length-1); | |
| } | |
| public int find(int pivot, int high){ | |
| int kth =0; | |
| if(pivot <= high){ | |
| int hold = 0; | |
| int i = pivot; | |
| for (int j = pivot+1; j <= high ; j ++){ | |
| if(this.arr[pivot] >= this.arr[j]){ | |
| hold = this.arr[i+1]; | |
| this.arr[i+1] = this.arr[j]; | |
| this.arr[j] = hold; | |
| i++; | |
| } | |
| } | |
| hold = this.arr[i]; | |
| this.arr[i] = this.arr[pivot]; | |
| this.arr[pivot] = hold; | |
| /* | |
| * make recursive calls | |
| */ | |
| if(i+1 == this.k) | |
| return this.arr[i]; | |
| else if(this.k < i+1) | |
| kth = find(pivot, i-1); | |
| else | |
| kth = find(i+1,high); | |
| } | |
| return kth; | |
| } | |
| } | |
| class SearchException extends RuntimeException{ | |
| public SearchException(String message){ | |
| super(message); | |
| } | |
| } |
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