Question

A) Use the partitioning idea of quicksort to give an algorithm that finds the median element of an array of n integers in expected O(n) time. (Hint: must you look at both sides of the partition)?

B) Write a Java program to demonstrate how the above algorithm can find the median element of the array A. in O(n) time. A= [1,5,9,7,20,22,22,6,8,3,2,1,4,9]

Answer 1

The question asks how to find the median element of an array using the partitioning concept of quicksort in expected O(n) time. The explanation provided is a Java program that uses a recursive function to partition the array and find the median without sorting the entire array.

Step-by-step explanation:

To find the median element of an array using the partitioning idea of quicksort, the algorithm doesn't require looking at both sides of the partition. Similar to quicksort, we select a pivot and partition the array such that elements smaller than the pivot are on the left and elements greater than the pivot are on the right. However, unlike quicksort, we only recurse into the side of the partition where the median is located, depending on the pivot's position. If the pivot is at index (n/2), being an odd array, or average of indices (n/2) and (n/2 - 1), for an even-sized array, we have found our median.

Here's a simplified version of such an algorithm in Java that looks for the median in an unordered array:

public class MedianFinder {
public static void main(String[] args) {
int[] A = {1,5,9,7,20,22,22,6,8,3,2,1,4,9};
int median = findMedian(A, 0, A.length - 1, A.length/2);
System.out.println("The median is: " + median);
}

public static int findMedian(int[] A, int left, int right, int medianIndex) {
if(left == right) return A[left];
int pivotIndex = partition(A, left, right);
if(pivotIndex == medianIndex) {
return A[medianIndex];
} else if(pivotIndex < medianIndex) {
return findMedian(A, pivotIndex + 1, right, medianIndex);
} else {
return findMedian(A, left, pivotIndex - 1, medianIndex);
}
}

private static int partition(int[] A, int left, int right) {
int pivot = A[right], i = left;
for(int j = left; j < right; j++) {
if(A[j] <= pivot) {
swap(A, i, j);
i++;
}
}
swap(A, i, right);
return i;
}

private static void swap(int[] A, int i, int j) {
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
}

The program above first calls the findMedian function, which recursively partitions the array and compares the pivot index with the required median index to determine the direction to continue the search. This process is repeated until the median is found.