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Maximize [length(X)/2^(XOR(X, Y))] by choosing substrings X and Y from string A and B respectively

  • Last Updated : 04 Aug, 2021

Given two binary strings A and B of size N and M respectively, the task is to maximize the value of the length of (X) / 2XOR(X, Y) by choosing two substrings X and Y of equal length from the given string A and B respectively.

Examples:

Input: A = “0110”, B = “1101”
Output: 3
Explanation:
Choose the substring “110” and “110” from the string A and B respectively. The value of the expression of length(X) / 2XOR(X, Y) is 3 / 20 = 3, which is maximum among all possible combinations.

Input: A = “1111”, B = “0000”
Output: 0

Approach: The given problem can be solved by observing the expression that it needs to be maximized, therefore the denominator must be minimum, and to minimize it the value of Bitwise XOR of the substrings X and Y must be minimum i.e., zero and to make the value of Bitwise XOR as zero, the two substrings must be same. Therefore, the problem reduces to finding the Longest Common Substring of both the strings A and B. Follow the steps below to solve the problem:



  • Initialize a 2D array, say LCSuff[M + 1][N + 1] to store the lengths of the longest common suffixes of the substrings.
  • Initialize a variable, say result as 0 to store the result maximum value of the given expression.
  • Iterate over the range [0, M] using the variable i and nested iterate over the  range [0, N] using the variable j and perform the following steps:
    • If i equals 0 or j equals 0, then update the value of LCSSuff[i][j] equals 0.
    • Otherwise, if the value of A[i – 1] equals A[j – 1] then update the value of LCSSuff[i][j] as LCSSuff[i – 1][j – 1] + 1 and update the value of result as the maximum of result and LCSSuff[i][j].
    • Otherwise, update the value of LCSSuff[i][j] to 0.
  • After completing the above steps, print the value of result as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the
// longest common substring of the
// string X and Y
int LCSubStr(char* A, char* B, int m, int n)
{
    // LCSuff[i][j] stores the lengths
    // of the longest common suffixes
    // of substrings
    int LCSuff[m + 1][n + 1];
    int result = 0;
 
    // Itearate over strings A and B
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {
 
            // If first row or column
            if (i == 0 || j == 0)
                LCSuff[i][j] = 0;
 
            // If matching is found
            else if (A[i - 1] == B[j - 1]) {
                LCSuff[i][j]
                    = LCSuff[i - 1][j - 1]
                      + 1;
                result = max(result,
                             LCSuff[i][j]);
            }
 
            // Otherwise, if matching
            // is not found
            else
                LCSuff[i][j] = 0;
        }
    }
 
    // Finally, return the resultant
    // maximum value LCS
    return result;
}
 
// Driver Code
int main()
{
    char A[] = "0110";
    char B[] = "1101";
    int M = strlen(A);
    int N = strlen(B);
 
    // Function Call
    cout << LCSubStr(A, B, M, N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to find the length of the
// longest common substring of the
// string X and Y
static int lcsubtr(char a[], char b[], int length1,
                   int length2)
{
     
    // LCSuff[i][j] stores the lengths
    // of the longest common suffixes
    // of substrings
    int dp[][] = new int[length1 + 1][length2 + 1];
    int max = 0;
     
    // Itearate over strings A and B
    for(int i = 0; i <= length1; ++i)
    {
        for(int j = 0; j <= length2; ++j)
        {
             
            // If first row or column
            if (i == 0 || j == 0)
            {
                dp[i][j] = 0;
            }
             
            // If matching is found
            else if (a[i - 1] == b[j - 1])
            {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                max = Math.max(dp[i][j], max);
            }
             
            // Otherwise, if matching
            // is not found
            else
            {
                dp[i][j] = 0;
            }
        }
    }
     
    // Finally, return the resultant
    // maximum value LCS
    return max;
}
 
// Driver Code
public static void main(String[] args)
{
    String m = "0110";
    String n = "1101";
    char m1[] = m.toCharArray();
    char m2[] = n.toCharArray();
     
    // Function Call
    System.out.println(lcsubtr(m1, m2, m1.length,
                                       m2.length));
}
}
 
// This code is contributed by zack_aayush

Python3




# Python 3 program for the above approach
 
# Function to find the length of the
# longest common substring of the
# string X and Y
def LCSubStr(A, B, m, n):
   
    # LCSuff[i][j] stores the lengths
    # of the longest common suffixes
    # of substrings
    LCSuff = [[0 for i in range(n+1)] for j in range(m+1)]
    result = 0
 
    # Itearate over strings A and B
    for i in range(m + 1):
        for j in range(n + 1):
           
            # If first row or column
            if (i == 0 or j == 0):
                LCSuff[i][j] = 0
 
            # If matching is found
            elif(A[i - 1] == B[j - 1]):
                LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1
                result = max(result,LCSuff[i][j])
 
            # Otherwise, if matching
            # is not found
            else:
                LCSuff[i][j] = 0
 
    # Finally, return the resultant
    # maximum value LCS
    return result
 
# Driver Code
if __name__ == '__main__':
    A = "0110"
    B = "1101"
    M = len(A)
    N = len(B)
 
    # Function Call
    print(LCSubStr(A, B, M, N))
 
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the length of the
// longest common substring of the
// string X and Y
static int lcsubtr(char[] a, char[] b, int length1,
                   int length2)
{
     
    // LCSuff[i][j] stores the lengths
    // of the longest common suffixes
    // of substings
    int[,] dp = new int[length1 + 1, length2 + 1];
    int max = 0;
     
    // Itearate over strings A and B
    for(int i = 0; i <= length1; ++i)
    {
        for(int j = 0; j <= length2; ++j)
        {
             
            // If first row or column
            if (i == 0 || j == 0)
            {
                dp[i, j] = 0;
            }
             
            // If matching is found
            else if (a[i - 1] == b[j - 1])
            {
                dp[i, j] = dp[i - 1, j - 1] + 1;
                max = Math.Max(dp[i, j], max);
            }
             
            // Otherwise, if matching
            // is not found
            else
            {
                dp[i, j] = 0;
            }
        }
    }
     
    // Finally, return the resultant
    // maximum value LCS
    return max;
}
 
// Driver Code
public static void Main()
{
    string m = "0110";
    string n = "1101";
    char[] m1 = m.ToCharArray();
    char[] m2 = n.ToCharArray();
     
    // Function Call
    Console.Write(lcsubtr(m1, m2, m1.Length,
                                       m2.Length));
}
}
 
// This code is contributed by target_2.

Javascript




<script>
 
        // JavaScript program for the above approach
 
        // Function to find the length of the
        // longest common substring of the
        // string X and Y
        function LCSubStr(A, B, m, n)
        {
         
            // LCSuff[i][j] stores the lengths
            // of the longest common suffixes
            // of substrings
            let LCSuff = Array(m + 1).fill(Array(n + 1));
            let result = 0;
 
            // Itearate over strings A and B
            for (let i = 0; i <= m; i++) {
                for (let j = 0; j <= n; j++) {
 
                    // If first row or column
                    if (i == 0 || j == 0)
                        LCSuff[i][j] = 0;
 
                    // If matching is found
                    else if (A.charAt(i - 1) == B.charAt(j - 1)) {
                        LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1;
                        if (LCSuff[i][j] > result) {
                            result = LCSuff[i][j];
                        }
                    }
 
                    // Otherwise, if matching
                    // is not found
                    else
                        LCSuff[i][j] = 0;
                }
            }
            result++;
            // Finally, return the resultant
            // maximum value LCS
            return result;
        }
 
        // Driver Code
 
        let A = "0110";
        let B = "1101";
        let M = A.length;
        let N = B.length;
 
        // Function Call
        document.write(LCSubStr(A, B, M, N));
 
    // This code is contributed by Potta Lokesh
    </script>
Output: 
3

 

Time Complexity: O(M*N)
Auxiliary Space: O(M*N)

 

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