Question

In the text box below, write down a divide and conquer algorithm for counting the number of entries in a sorted array of ints that are smaller than a given value.  In other words, the function takes as input an array A and an int value and returns the number of ints in A that are less than value.  To get any credit, your solution must use the divide and conquer technique.  To get full credit, your solution should run in  time in the worst case (where n is the size of the array) and must work on arrays that contain duplicate entries. You can still get half credit for a solution that runs in  time in the worst case or that works only on arrays without duplicate entries.
Hint1:  Remember that every divide and conquer algorithm we’ve done on arrays requires a recursive function that has extra parameters to indicate the sub range of the array that you are focusing on.  You may add extra input parameters below.
Hint2: If you are unsure how to start, you might begin by ignoring the fact that the array is sorted and coming up with a divide and conquer algorithm that isn’t fast, but that works on an unsorted array first, and then, improve your algorithm by modifying it to take advantage of the fact that the array is sorted.

Input parameters:  A : array of int,  value : int Output:  the number of ints in A that are smaller than value

Answer 1

package search;

public class findNumber {

public static int mergeSearch(int ar[],int l,int r,int k){

if(l<r){

int mid=(l+r)/2; //take mid index

int f=mergeSearch(ar,l,mid,k); //call mergesort on first half and return smaller in first half subaaray

int s=mergeSearch(ar,mid+1,r,k); //call mergesort on second half and return smaller in second half subaaray

int res=marge(ar,l,r,mid,k); //find the max in both side

return res; //now return max of all three

}

return -1;

}

public static int marge(int ar[],int l,int r,int mid,int k){

int res=0;

for(int i=l;i<=mid;i++){ //find the smaller in first half

if(ar[i]<k){

res++;

}

}

for(int i=mid+1;i<=r;i++){ //find the smaller in second half

if(ar[i]<k){

res++;

}

}

return res;

}

//driver program to test

public static void main(String[] args) {

int ar[]={1,2,3,4,5,6,7};

int k=mergeSearch(ar,0,ar.length-1,5);

System.out.println(k);

ar=new int[]{1,2,2,4,5,5,7};

k=mergeSearch(ar,0,ar.length-1,10);

System.out.println(k);

ar=new int[]{1,4,8,9,10,11};

k=mergeSearch(ar,0,ar.length-1,0);

System.out.println(k);

}

}

output

4

7

0