Question

You are given a function median(A,p,r) that finds the index corresponding to the median of an array ? with starting index p and ending index r, in worst-case complexity Θ(?) where ? is the length of ?. Making use of the given median function, write an algorithm with complexity Θ(?) to partition the array ? using its median as the pivot. You may call the functions discussed in class.

Using your algorithm in Qn 1, write an algorithm that selects the ?-th smallest element of ? in worst-case complexity ?(?) . Prove that your algorithm indeed has complexity ?(?), justifying every step clearly. Note that the select algorithm given in class has worst-case complexity ?(? 2 ).

Answer 1

// Java program to find 
// median of an array 
// within given range 
import java.io.*; 
import java.util.Arrays;
  
class Main  
{ 
    // function to get elements within given range 
    static int[] getElementsWithInRange(int arr[], int n, int x, int y) 
    { 
        // initialize new array
        int newLength = y - x + 1;
        int j = 0;
        int[] newArr = new int[newLength]; 
        for (int i = x; i < n && i <= y; i++) { 
      
            // get element within the range 
            newArr[j] = arr[i];
            j++;
        } 
        return newArr; 
    } 
    
    // Function for calculating median
    public static double findMedian(int a[], int n)
    {
        // First we sort the array
        Arrays.sort(a);
 
        // check for even case
        if (n % 2 != 0)
            return (double)a[n / 2];
 
        return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;
    }
  
    // driver function 
    public static void main (String[] args) 
    { 
        int arr[] = { 1, 3, 4, 9, 10, 3 }; 
        int n = arr.length; 
  
        // Get The Array WithIn the Starting And Ending Index 
        int i = 1, j = 4; 
        int newArr[] =getElementsWithInRange(arr,n,i,j);
        System.out.println("Median = " + findMedian(newArr, newArr.length)); 
    } 
} 

Output:

Above Program is to calculate the median

Time Complexity to find median = O(n Log n) as we need to sort the array first.

Program to find the ith Smallest element:

The idea in this new method is similar to quickSelect(), we get worst-case linear time by selecting a pivot that divides array in a balanced way (there are not very few elements on one side and many on another side). After the array is divided in a balanced way, we apply the same steps as used in quickSelect() to decide whether to go left or right of the pivot.

Psuedo Code:

kthSmallest(arr[0..n-1], k)
1) Divide arr[] into ⌈n/5⌉ groups where size of each group is 5 except possibly the last group which may have less than 5 elements.

2) Sort the above created ⌈n/5⌉ groups and find median of all groups. Create an auxiliary array ‘median[]’ and store medians of all ⌈n/5⌉ groups in this median array.

// Recursively call this method to find median of median[0..⌈n/5⌉-1]
3) medOfMed = kthSmallest(median[0..⌈n/5⌉-1], ⌈n/10⌉)

4) Partition arr[] around medOfMed and obtain its position.
pos = partition(arr, n, medOfMed)

5) If pos == k return medOfMed
6) If pos > k return kthSmallest(arr[l..pos-1], k)
7) If pos < k return kthSmallest(arr[pos+1..r], k-pos+l-1)

// Java implementation of worst  
// case linear time algorithm 
// to find k'th smallest element 
import java.util.*; 
  
class Main  
{ 
  
// int partition(int arr[], int l, int r, int k); 
  
// A simple function to find median of arr[]. This is called 
// only for an array of size 5 in this program. 
static int findMedian(int arr[], int i,int n) 
{ 
    if(i <= n) 
        Arrays.sort(arr, i, n); // Sort the array 
    else
        Arrays.sort(arr, n, i); 
    return arr[n/2]; // Return middle element 
} 
  
// Returns k'th smallest element 
// in arr[l..r] in worst case 
// linear time. ASSUMPTION: ALL  
// ELEMENTS IN ARR[] ARE DISTINCT 
static int kthSmallest(int arr[], int l, int r, int k) 
{ 
    // If k is smaller than  
    // number of elements in array 
    if (k > 0 && k <= r - l + 1) 
    { 
        int n = r - l + 1 ; // Number of elements in arr[l..r] 
  
        // Divide arr[] in groups of size 5,  
        // calculate median of every group 
        //  and store it in median[] array. 
        int i; 
          
         // There will be floor((n+4)/5) groups; 
        int []median = new int[(n + 4) / 5]; 
        for (i = 0; i < n/5; i++) 
            median[i] = findMedian(arr,l + i * 5, 5); 
              
        // For last group with less than 5 elements 
        if (i*5 < n)  
        { 
            median[i] = findMedian(arr,l + i * 5, n % 5);  
            i++; 
        }  
  
        // Find median of all medians using recursive call. 
        // If median[] has only one element, then no need 
        // of recursive call 
        int medOfMed = (i == 1)? median[i - 1]: 
                                kthSmallest(median, 0, i - 1, i / 2); 
  
        // Partition the array around a random element and 
        // get position of pivot element in sorted array 
        int pos = partition(arr, l, r, medOfMed); 
  
        // If position is same as k 
        if (pos-l == k - 1) 
            return arr[pos]; 
        if (pos-l > k - 1) // If position is more, recur for left 
            return kthSmallest(arr, l, pos - 1, k); 
  
        // Else recur for right subarray 
        return kthSmallest(arr, pos + 1, r, k - pos + l - 1); 
    } 
  
    // If k is more than number of elements in array 
    return Integer.MAX_VALUE; 
} 
  
static int[] swap(int []arr, int i, int j) 
{ 
    int temp = arr[i]; 
    arr[i] = arr[j]; 
    arr[j] = temp; 
    return arr; 
} 
  
// It searches for x in arr[l..r], and  
// partitions the array around x. 
static int partition(int arr[], int l, 
                        int r, int x) 
{ 
    // Search for x in arr[l..r] and move it to end 
    int i; 
    for (i = l; i < r; i++) 
        if (arr[i] == x) 
        break; 
    swap(arr, i, r); 
  
    // Standard partition algorithm 
    i = l; 
    for (int j = l; j <= r - 1; j++) 
    { 
        if (arr[j] <= x) 
        { 
            swap(arr, i, j); 
            i++; 
        } 
    } 
    swap(arr, i, r); 
    return i; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int arr[] = {12, 3, 5, 7, 4, 19, 26}; 
    int n = arr.length, k = 3; 
    System.out.println("K'th smallest element is "
        + kthSmallest(arr, 0, n - 1, k)); 
} 
} 

Output:

Time Complexity:
The worst case time complexity of the above algorithm is O(n). Let us analyze all steps.

The steps 1) and 2) take O(n) time as finding median of an array of size 5 takes O(1) time and there are n/5 arrays of size 5.
The step 3) takes T(n/5) time. The step 4 is standard partition and takes O(n) time.
The interesting steps are 6) and 7). At most, one of them is executed. These are recursive steps. What is the worst case size of these recursive calls. The answer is maximum number of elements greater than medOfMed (obtained in step 3) or maximum number of elements smaller than medOfMed.
How many elements are greater than medOfMed and how many are smaller?
At least half of the medians found in step 2 are greater than or equal to medOfMed. Thus, at least half of the n/5 groups contribute 3 elements that are greater than medOfMed, except for the one group that has fewer than 5 elements. Therefore, the number of elements greater than medOfMed is at least.

Similarly, the number of elements that are less than medOfMed is at least 3n/10 – 6. In the worst case, the function recurs for at most n – (3n/10 – 6) which is 7n/10 + 6 elements.

Note that 7n/10 + 6 20 20 and that any input of 80 or fewer elements requires O(1) time. We can therefore obtain the recurrence

We show that the running time is linear by substitution. Assume that T(n) cn for some constant c and all n > 80. Substituting this inductive hypothesis into the right-hand side of the recurrence yields

T(n) <= cn/5 + c(7n/10 + 6) + O(n)
     <= cn/5 + c + 7cn/10 + 6c + O(n)
     <= 9cn/10 + 7c + O(n)
     <= cn, 

since we can pick c large enough so that c(n/10 – 7) is larger than the function described by the O(n) term for all n > 80. The worst-case running time of is therefore linear