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Find the kth Smallest Element in Array in Linear Time

Posted by sanstechbytes on July 24, 2011

Finding the kth smallest element in array is an application related to Order Statistics problem. For details on the algo used here, watch the video lecture embedded below.

Robert Sedgewick, whose lecture this approach has been inspired from, proposes random shuffle mechanism and partitioning strategy to select the kth smallest element. The strategy requires maintaining variables lo and hi to delimit the subarray that contains the index k of the element to be selected, using partitioning to shrink the size of the subarray (Step 2). The assumption made is that the input array to the algo doesn’t contain duplicates (This can be carried out as a preprocessing step, which is performed as first step in the algo below, though). Runs in linear time, O(n).

1) Preprocess array to remove duplicates.
2) Rearrange the array from a[lo] through a[hi] and return an integer j such that a[lo] through a[j-1] are less than a[j], and a[j+1] through a[hi] are greater than a[j].
a) If k is equal to j, done.
b) Else If k is less than j, then continue working in the right subarray (by changing the value of hi to j-1).
c) Else If k is greater than j, then continue working in the right subarray (by changing lo to j+1).
3) Repeat step (2), until the array consists just of k.

MIT Video Lecture on Order Statistics:

Java Code:

/** Sedgewick's partitioning approach */
 public class QuickSelect {

private static Random random = new Random();

public static Comparable select(Comparable[] a, int k) {
 if (k = a.length) {
 throw new RuntimeException("Selected element out of bounds");


int lo = 0, hi = a.length - 1;
 while (hi > lo) {
 int i = partition(a, lo, hi);
 if (i > k)
 hi = i - 1;
 else if (i < k)
 lo = i + 1;
 return a[i];
 return a[lo];

private static int partition(Comparable[] a, int lo, int hi) {
 int i = lo;
 int j = hi + 1;
 Comparable v = a[lo];
 while (true) {
 // find item on lo to swap
 while (less(a[++i], v))
 if (i == hi)
 // find item on hi to swap
 while (less(v, a[--j]))
 if (j == lo)
 break; // redundant since a[lo] acts as sentinel
 // check if pointers cross
 if (i >= j)
 exch(a, i, j);
 // put v = a[j] into position
 exch(a, lo, j);
 // with a[lo .. j-1] <= a[j] <= a[j+1 .. hi]
 return j;

public static void shuffle(Object[] a) {
 int n = a.length;
 for (int i = 0; i < n; i++) {
 int r = i + uniform(n - i); // between i and N-1
 Object temp = a[i];
 a[i] = a[r];
 a[r] = temp;

 * Return an integer uniformly between 0 and N-1.
 public static int uniform(int N) {
 return random.nextInt(N);

// is v < w ?
 private static boolean less(Comparable v, Comparable w) {
 return (v.compareTo(w) < 0);

// exchange a[i] and a[j]
 private static void exch(Object[] a, int i, int j) {
 Object swap = a[i];
 a[i] = a[j];
 a[j] = swap;

// Read strings from standard input, sort them, and print.
 public static void main(String[] args) {
 // Read input from StdIn here
 // String[] a = new String[]{"aa", "aaa", "bb", "ab", "ca", "bc",
 // "cab", "cc", "cad" };
 Integer[] a = new Integer[] { 2, 11, 10, 6, 8, 99, 7, 100, 101, 28,
 9 };
 int k = 2;
 Object kth =, k - 1); // returns a[k-1]
 System.out.println(k + "th smallest element in given array = "
 + kth);


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