Question

Suppose we are given two sorted arrays (nondecreasing from index 1 to index n) X[1] · · · X[n] and Y [1] · · · Y [n] of integers. For simplicity, assume that n is a power of 2. Problem is to design an algorithm that determines if there is a number p in X and a number q in Y such that p + q is zero. If such numbers exist, the algorithm returns true; otherwise, it returns false. We can consider every pair of numbers, one taken from X and other from Y , using a double-nested loop and solve this in O(n 2 ) time (a faster algorithm than the ones proposed here exists, but the homework is about divide and conquer). (1) We want to design a better algorithm using divide and conquer. Let (X, Y ) refer to the given problem. Let T(n) be the complexity of your algorithm when each of X, Y has n elements. Consider each of X, Y divided into two equal halves as shown below: X = X1 X2 Y = Y1 Y2 If such a pair p, q of numbers existed, then p has to be in one of X1, X2, and q has to be in one of Y1, Y2. Thus, we can determine the answer to problem (X, Y ) from answers to the following four smaller problems of half the size: (X1, Y1), (X1, Y2), (X2, Y1), and (X2, Y2)

(d) (15 pts) Present complete pseudocode for solving the problem (X[i...j], Y [k...l]) using divide and conquer by specifying details of the following. Invoking the algorithm as zeroSum (X, 1, n, Y, 1, n) solves the problem (X, Y ).

// Solves the problem (X[i ... j], Y[k ... l])

// Returns true if some p in X[i ... j] and some q in Y[k ... l]

// add up to zero; returns false otherwise

// Pre: (1) X and Y are in nondecreasing order

// (2) (j-i+1) = (l-k+1), i.e., X, Y have equal number of elements

boolean zeroSum (X[], i, j, Y[], k, l)

{

}

You must clearly specify the base case(s).

Answer 1

/*steps to proceed:

1)write a function with arguments X array,Y array and with their lengths and 0.

2) implement that function as follows:

checksum(X,Y,m,n,p)

{

flag;

for loop from 0 to m;

value=p-X[i];

check a value in Y array(we can solve this in o(logn) time because array is sorted)

{

if(value )

flag=1;

return true;

}

return false;

} */

//full code in c++;

#include <bits/stdc++.h>
using namespace std;

// because array is sorted,we can search a element on (O(logn)) time.
bool isPresent(int arr[], int low,
int high, int value)
{
while (low <= high)
{
int mid = (low + high) / 2;

// value found
if (arr[mid] == value)
return true;

else if (arr[mid] > value)
high = mid - 1;
else
low = mid + 1;
}

// value not found
return false;
}

int checksumpair(int arr1[], int arr2[],
int m, int n, int x)
{
int flag = 0;
for (int i = 0; i < m; i++)
{
int value = x - arr1[i];
if(flag==1)
{
return true;
}
if (isPresent(arr2, 0, n - 1, value))
{
flag=1;
}
}

return false;
}

// Driver Code
int main()
{
int arr1[] = {1, 3, 5, 7};
int arr2[] = {2, 3, 5, 8};
int m = arr1.length();
int n = arr2.length();
int p = 0;
cout << checksumpair(arr1, arr2, m, n, p);

return 0;

}

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