Wednesday, July 18, 2012

Finding Kth Minimum (partial ordering) – Using Tournament Algorithm

Ok, so far in my previous blog entries about finding 2nd minimum (and for repeating elements) I wrote about efficient algorithm for finding second minimum element of an array. This optimized algorithm takes O(n + log(n)) time (worst case) to find the second minimum element.
So, the next obvious question is - how can we find Kth minimum element efficiently in an unsorted array where k ranges from 1 – n. How about partial sorting - efficiently returning first K minimum values from unsorted array. As you might have guessed it, we can achieve this by extending the logic used before and tweaking our implementation little bit. We can find Kth min (or return partially sorted list of K min elements) in O(n + klog(n)) time.

Design Details

 For simplicity we will assume that the input unsorted array consists of non-repeating, non-negative integer numbers only. Solution can be extended for arrays with repeating numbers using similar considerations as outlined in my blog for finding second min for repeating values.
As noted in the previous blog, using tournament method we can build output tree by comparing the adjacent elements and moving the lower of the two to next level. Repeating this process yields a tree with log(n) depth and the root element of this tree represent the minimum value. Also, the second minimum can then easily be obtained by finding the minimum of all the values that the root was compared against at each level.  Since the depth of tree is log(n), we have to do at most O(log(n)) comparisons. For example, Figure 1 below uses tournament method for finding the root element. For this particular case, the minimum value is 1 and second minimum 2.
Let’s say we want to find the 3rd minimum value. Extending the logic used to find second minimum, the third minimum value must have met the root (minimum) or the second minimum before ending its progression (just like winning a tournament round) to next level of the tree (as smaller of the two compared values move to next level). Thus 3rd minimum can be obtained by finding the least of all the values compared against the minimum and the second minimum, i.e. backtracking the root (minimum) and the second minimum up to the top of tree (original array) and noting all values each (root and second minimum) was compared against.
 Take the sample in Figure 1. Here is the list of values obtained by back-tracking the root (call it Set A).
Set A (for root) - [3, 2, 10, 8]
From this list we obtain the second minimum as 2. Back-tracking from 2, we obtain following list (Set B):
Set B (for Second minimum) [5, 14]
Thus the third minimum is the least value in union of set A and B, which is 3 (ignoring value 2, which was already established as second minimum). Note that the 3rd minimum must be in either Set A or Set B. In above case it was found from Set A. 
How about if we have to find the 4th minimum? Well, we have to back track the 3rd least value and collect adjacent values a (one level up). Then add this set to the values for root and second minimum. Then this set will contain the 2nd, 3rd and 4th minimum. Continuing our sample, let’s say Set C consists of such elements obtained by backtracking 3rd min – [4, 11, 16]. Thus the minimum value in union of Set A, B, C is 4 (ignoring already established 2 and 3 as second min and third min resp.), is fourth least value. Note that for calculating kth minimum we don’t backtrack and find the adjacency list for minimum through k -1. We can use already calculated results from any previous such computation. This is the reason why this technique falls under Dynamic Programming – we have optimal substructure and avoid repeated computations. 
Now it is time to formalize. To find Kth minimum, back track root, second min, third min up to k-1 min and collect the adjacent values for each - moving from the root (of sub-tree) all the way to the top. The resulting collection will contain the 2nd, 3rd … Kth minimum.
Implementation Details
To obtain Kth minimum, we need collection of all the comparisons from minimum (root) to K-1. Thus we need to make following changes to our previous code for finding second minimum.

  1. Return array of values (as indicated in rectangle boxes in Figure 2) while back-tracking root (could also be sub-tree root). Also, in addition to the values, we need the level and also index information for of each of these values. The reason why we need the level and index is because once we determine that a particular value from this list is minimum, we need to back-track the sub-tree from that value (hence we need to know the location of the element at that level) on order to obtain new collection to determine next lower value. Thus we will return a two-dimensional array with first index denoting the value and the other one the index. We can infer the level from the index of the pair (in 2-d array).For example, for the case above, for the first pass (to find the minimum value), the following will be returned (let’s call it adjacency list):Thus from above array we can conclude that the second minimum value is 2 (least of first value of all pairs), and this value appears two levels above the root (level 5), at 0th index (at this level).  We will use this info to identify the sub-tree for out next round of back-tracking.
  2. Change the api to take back-track a portion of tree. In order to achieve that we need to pass level and index to define the root of sub-tree to back track from.
  3. The 2-dimensional list obtained from each run will be added to a 3-dimensional list – full adjacency list for min to K-1 min elements. 3-D list is used rather than merging a bigger 2-D list to preserve and identify the results for a particular run. The reason being once we identify the minimum element for all k-1 runs, the level of that particular element is obtained from particular list where the minimum was found (alternatively our backtracking api could have returned a 3-d array with level info as one more dimension, either way we need this info).
  4. Api that take full adjacency list and min value and return the next min element info. The api should identify for which run it found the min and the index of element within that run. Refer Figure 4 below. When this list is passed to this api with value of second min (2), it returns (0,3), which is interpreted as the next min (third min) was found in first array (obtained from first backtrack) and 4th element within that array. Once we have this info, we can look into first array and locate the third min value as 3, at index 1 at level 3 (remember original array is level 1 – Figure 2). Note that for m elements in full adjacency list we are making m comparisons which seem not efficient at first glance, but since the size of the adjacency list is small O(log(n)), hence it is not significant.
  5. To find Kth min, we need to find min through k. Hence this algorithm can return partially sorted array up-to k elements. 

Can be downloaded from here - (remove the .txt extension)

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 * destruction.
 * I further grant you ("Licensee") a non-exclusive, royalty free, license to
 * use, modify and redistribute this software in source and binary code form,
 * provided that i) this copyright notice and license appear on all copies of
 * the software;
 * As with any code, ensure to test this code in a development environment
 * before attempting to run it in production.
 * @author Malkit S. Bhasin

public class KThMinimum {

  * @param inputArray
  *            unordered array of non-negative integers
  * @param k
  *            order of minimum value desired
  * @return kth minimum value
 public static int getKthMinimum(int[] inputArray, int k) {
  return findKthMinimum(inputArray, k)[k - 1];

  * @param inputArray
  *            unordered array of non-negative integers
  * @param k
  *            ordered number of minimum values
  * @return k ordered minimum values
 public static int[] getMinimumKSortedElements(int[] inputArray, int k) {
  return findKthMinimum(inputArray, k);

  * First output tree will be obtained using tournament method. For k
  * minimum, the output tree will be backtracked k-1 times for each sub tree
  * identified by the minimum value in the aggregate adjacency list obtained
  * from each run. The minimum value after each run will be recorded and
  * successive runs will produce next minimum value.
  * @param inputArray
  * @param k
  * @return ordered array of k minimum elements
 private static int[] findKthMinimum(int[] inputArray, int k) {
  int[] partiallySorted = new int[k];
  int[][] outputTree = getOutputTree(inputArray);
  int root = getRootElement(outputTree);
  partiallySorted[0] = root;
  int rootIndex = 0;
  int level = outputTree.length;
  int[][][] fullAdjacencyList = new int[k - 1][][];
  int[] kthMinIdx = null;
  for (int i = 1; i < k; i++) {
   fullAdjacencyList[i - 1] = getAdjacencyList(outputTree, root,
     level, rootIndex);
   kthMinIdx = getKthMinimum(fullAdjacencyList, i, root);
   int row = kthMinIdx[0];
   int column = kthMinIdx[1];
   root = fullAdjacencyList[row][column][0];
   partiallySorted[i] = root;
   level = column + 1;
   rootIndex = fullAdjacencyList[row][column][1];

  return partiallySorted;

  * Takes an input array and generated a two-dimensional array whose rows are
  * generated by comparing adjacent elements and selecting minimum of two
  * elements. Thus the output is inverse triangle (root at bottom)
  * @param values
  * @return
 public static int[][] getOutputTree(int[] values) {
  Integer size = new Integer(values.length);
  double treeDepth = Math.log(size.doubleValue()) / Math.log(2);
  // int intTreeDepth = getIntValue(Math.ceil(treeDepth)) + 1;
  int intTreeDepth = (int) (Math.ceil(treeDepth)) + 1;
  int[][] outputTree = new int[intTreeDepth][];

  // first row is the input
  outputTree[0] = values;

  int[] currentRow = values;
  int[] nextRow = null;
  for (int i = 1; i < intTreeDepth; i++) {
   nextRow = getNextRow(currentRow);
   outputTree[i] = nextRow;
   currentRow = nextRow;
  return outputTree;

  * Compares adjacent elements (starting from index 0), and construct a new
  * array with elements that are smaller of the adjacent elements.
  * For even sized input, the resulting array is half the size, for odd size
  * array, it is half + 1.
  * @param values
  * @return
 private static int[] getNextRow(int[] values) {
  int rowSize = getNextRowSize(values);
  int[] row = new int[rowSize];
  int i = 0;
  for (int j = 0; j < values.length; j++) {
   if (j == (values.length - 1)) {
    // this is the case where there are odd number of elements
    // in the array. Hence the last loop will have only one element.
    row[i++] = values[j];
   } else {
    row[i++] = getMin(values[j], values[++j]);
  return row;

  * From the passed full adjacency list and min value scans the list and
  * returns the information about next minimum value. It returns int array
  * with two values: first value: index of the back-track (the min value was
  * found in the Adjacency list for min value, second min etc.) second value:
  * index within the identified run.
  * @param fullAdjacencyList
  *            Adjacency list obtained after k-1 backtracks
  * @param kth
  *            Order of minimum value desired
  * @param kMinusOneMin
  *            value of k-1 min element
  * @return
 private static int[] getKthMinimum(int[][][] fullAdjacencyList, int kth,
   int kMinusOneMin) {
  int kThMin = Integer.MAX_VALUE;
  int[] minIndex = new int[2];
  int j = 0, k = 0;
  int temp = -1;

  for (int i = 0; i < kth; i++) {
   for (j = 0; j < fullAdjacencyList.length; j++) {
    int[][] row = fullAdjacencyList[j];
    if (row != null) {
     for (k = 0; k < fullAdjacencyList[j].length; k++) {
      temp = fullAdjacencyList[j][k][0];
      if (temp <= kMinusOneMin) {
      if ((temp > kMinusOneMin) && (temp < kThMin)) {
       kThMin = temp;
       minIndex[0] = j;
       minIndex[1] = k;
  return minIndex;

  * Back-tracks a sub-tree (specified by the level and index) parameter and
  * returns array of elements (during back-track path) along with their index
  * information. The order elements of output array indicate the level at
  * which these elements were found (with elements closest to the root at the
  * end of the list)
  * Starting from root element (which is minimum element), find the lower of
  * two adjacent element one row above. One of the two element must be root
  * element. If the root element is left adjacent, the root index (for one
  * row above) is two times the root index of any row. For right-adjacent, it
  * is two times plus one. Select the other element (of two adjacent
  * elements) as second minimum.
  * Then move to one row further up and find elements adjacent to lowest
  * element, again, one of the element must be root element (again, depending
  * upon the fact that it is left or right adjacent, you can derive the root
  * index for this row). Compare the other element with the second least
  * selected in previous step, select the lower of the two and update the
  * second lowest with this value.
  * Continue this till you exhaust all the rows of the tree.
  * @param tree
  *            output tree
  * @param rootElement
  *            root element (could be of sub-tree or outputtree)
  * @param level
  *            the level to find the root element. For the output tree the
  *            level is depth of the tree.
  * @param rootIndex
  *            index for the root element. For output tree it is 0
  * @return
 public static int[][] getAdjacencyList(int[][] tree, int rootElement,
   int level, int rootIndex) {
  int[][] adjacencyList = new int[level - 1][2];
  int adjacentleftElement = -1, adjacentRightElement = -1;
  int adjacentleftIndex = -1, adjacentRightIndex = -1;
  int[] rowAbove = null;

  // we have to scan in reverse order
  for (int i = level - 1; i > 0; i--) {
   // one row above
   rowAbove = tree[i - 1];
   adjacentleftIndex = rootIndex * 2;
   adjacentleftElement = rowAbove[adjacentleftIndex];

   // the root element could be the last element carried from row above
   // because of odd number of elements in array, you need to do
   // following
   // check. if you don't, this case will blow {8, 4, 5, 6, 1, 2}
   if (rowAbove.length >= ((adjacentleftIndex + 1) + 1)) {
    adjacentRightIndex = adjacentleftIndex + 1;
    adjacentRightElement = rowAbove[adjacentRightIndex];
   } else {
    adjacentRightElement = -1;

   // if there is no right adjacent value, then adjacent left must be
   // root continue the loop.
   if (adjacentRightElement == -1) {
    // just checking for error condition
    if (adjacentleftElement != rootElement) {
     throw new RuntimeException(
       "This is error condition. Since there "
         + " is only one adjacent element (last element), "
         + " it must be root element");
    } else {
     rootIndex = rootIndex * 2;
     adjacencyList[level - 1][0] = -1;
     adjacencyList[level - 1][1] = -1;

   // one of the adjacent number must be root (min value).
   // Get the other number and compared with second min so far
   if (adjacentleftElement == rootElement
     && adjacentRightElement != rootElement) {
    rootIndex = rootIndex * 2;
    adjacencyList[i - 1][0] = adjacentRightElement;
    adjacencyList[i - 1][1] = rootIndex + 1;
   } else if (adjacentleftElement != rootElement
     && adjacentRightElement == rootElement) {
    rootIndex = rootIndex * 2 + 1;
    adjacencyList[i - 1][0] = adjacentleftElement;
    adjacencyList[i - 1][1] = rootIndex - 1;
   } else if (adjacentleftElement == rootElement
     && adjacentRightElement == rootElement) {
    // This is case where the root element is repeating, we are not
    // handling this case.
    throw new RuntimeException(
      "Duplicate Elements. This code assumes no repeating elements in the input array");
   } else {
    throw new RuntimeException(
      "This is error condition. One of the adjacent "
        + "elements must be root element");

  return adjacencyList;

  * Returns minimum of two passed in values.
  * @param num1
  * @param num2
  * @return
 private static int getMin(int num1, int num2) {
  return Math.min(num1, num2);

  * following uses Math.ceil(double) to round to upper integer value..since
  * this function takes double value, diving an int by double results in
  * double.
  * Another way of achieving this is for number x divided by n would be -
  * (x+n-1)/n
  * @param values
  * @return
 private static int getNextRowSize(int[] values) {
  return (int) Math.ceil(values.length / 2.0);

  * Returns the root element of the two-dimensional array.
  * @param tree
  * @return
 public static int getRootElement(int[][] tree) {
  int depth = tree.length;
  return tree[depth - 1][0];

 private static void printRow(int[] values) {
  for (int i : values) {
   // System.out.print(i + " ");
  // System.out.println(" ");

 public static void main(String args[]) {
  int[] input = { 2, 14, 5, 13, 1, 8, 17, 10, 6, 12, 9, 4, 11, 15, 3, 16 };
  System.out.println("Fifth Minimum: " + getKthMinimum(input, 5));

  int minimumSortedElementSize = 10;
  int[] tenMinimum = getMinimumKSortedElements(input,
  System.out.print("Minimum " + minimumSortedElementSize + " Sorted: ");
  for (int i = 0; i < minimumSortedElementSize; i++) {
   System.out.print(tenMinimum[i] + " ");

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