Maximize given function by selecting equal length substrings from given Binary Strings

Given two binary strings s1 and s2. The task is to choose substring from s1 and s2 say sub1 and sub2 of equal length such that it maximizes the function:

fun(s1, s2) = len(sub1) / (2^{xor(sub1, sub2)})

Examples:

Input: s1= "1101", s2= "1110"
Output: 3
Explanation: Below are the substrings chosen from s1 and s2
Substring chosen from s1 -> "110"
Substring chosen from s2 -> "110"
Therefore, fun(s1, s2) = 3/ (2^{xor(110, 110)}) = 3, which is maximum possible. 

Input: s1= "1111", s2= "1000"
Output: 1

Approach: In order to maximize the given function large substrings needed to be chosen with minimum XOR. To minimize the denominator, choose substrings in a way such that XOR of sub1 and sub2 is always 0 so that the denominator term will always be 1 (2⁰). So for that, find the longest common substring from the two strings s1 and s2, and print its length that would be the required answer. 

Below is the implementation of above approach:

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

int dp[1000][1000];

// Function to find longest common substring.
int lcs(string s, string k, int n, int m)
{
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            if (i == 0 or j == 0) {
                dp[i][j] = 0;
            }
            else if (s[i - 1] == k[j - 1]) {
                dp[i][j] = 1 + dp[i - 1][j - 1];
            }
            else {
                dp[i][j] = max(dp[i - 1][j],
                               dp[i][j - 1]);
            }
        }
    }

    // Return the result
    return dp[n][m];
}

// Driver Code
int main()
{
    string s1 = "1110";
    string s2 = "1101";

    cout << lcs(s1, s2,
                s1.size(), s2.size());

    return 0;
}

// Java program for above approach
class GFG{
  
  static int dp[][] = new int[1000][1000];

  // Function to find longest common substring.
  static int lcs(String s, String k, int n, int m)
  {
      for (int i = 0; i <= n; i++) {
          for (int j = 0; j <= m; j++) {
              if (i == 0 || j == 0) {
                  dp[i][j] = 0;
              }
              else if (s.charAt(i - 1) == k.charAt(j - 1)) {
                  dp[i][j] = 1 + dp[i - 1][j - 1];
              }
              else {
                  dp[i][j] = Math.max(dp[i - 1][j],
                                 dp[i][j - 1]);
              }
          }
      }

      // Return the result
      return dp[n][m];
  }

  // Driver Code
  public static void main(String [] args)
  {
      String s1 = "1110";
      String s2 = "1101";

      System.out.print(lcs(s1, s2,
                  s1.length(), s2.length()));

      
  }
}

// This code is contributed by AR_Gaurav

# Python3 program for above approach

import numpy as np;

dp = np.zeros((1000,1000));

# Function to find longest common substring.
def lcs( s,  k,  n,  m) :

    for i in range(n + 1) :
        for j in range(m + 1) :
            if (i == 0 or j == 0) :
                dp[i][j] = 0;
            
            elif (s[i - 1] == k[j - 1]) :
                dp[i][j] = 1 + dp[i - 1][j - 1];
            
            else :
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

    # Return the result
    return dp[n][m];

# Driver Code
if __name__ == "__main__" :

    s1 = "1110";
    s2 = "1101";

    print(lcs(s1, s2,len(s1), len(s2)));

   # This code is contributed by AnkThon

// C# program for above approach
using System;
public class GFG{
  
  static int [,]dp = new int[1000,1000];

  // Function to find longest common substring.
  static int lcs(string s, string k, int n, int m)
  {
      for (int i = 0; i <= n; i++) {
          for (int j = 0; j <= m; j++) {
              if (i == 0 || j == 0) {
                  dp[i, j] = 0;
              }
              else if (s[i - 1] == k[j - 1]) {
                  dp[i, j] = 1 + dp[i - 1, j - 1];
              }
              else {
                  dp[i, j] = Math.Max(dp[i - 1, j],
                                 dp[i, j - 1]);
              }
          }
      }

      // Return the result
      return dp[n, m];
  }

  // Driver Code
  public static void Main(string [] args)
  {
      string s1 = "1110";
      string s2 = "1101";

      Console.Write(lcs(s1, s2, s1.Length, s2.Length));
  }
}

// This code is contributed by AnkThon

<script>
// JavaScript program for above approach
var dp = new Array(1000); 
for (var i = 0; i < 1000; i++) {
  dp[i] = new Array(1000); 
}
    
// Function to find longest common substring.
function lcs( s,  k,  n,  m)
{
    for (var i = 0; i <= n; i++) {
        for (var j = 0; j <= m; j++) {
            if (i == 0 || j == 0) {
                dp[i][j] = 0;
            }
            else if (s[i - 1] == k[j - 1]) {
                dp[i][j] = 1 + dp[i - 1][j - 1];
            }
            else {
                dp[i][j] = Math.max(dp[i - 1][j],
                               dp[i][j - 1]);
            }
        }
    }

    // Return the result
    return dp[n][m];
}

// Driver Code
var s1 = "1110";
var s2 = "1101";

document.write(lcs(s1, s2, s1.length, s2.length))

// This code is contributed by AnkThon
</script>

3

Time Complexity: O(N*M), where N is the size of s1 and M is the size of s2.

Auxiliary Space: O(N*M), where N is the size of s1 and M is the size of s2.

Approach2: Using memoised version of dynamic programming

In order to maximize the given function large substrings needed to be chosen with minimum XOR. To minimize the denominator, choose substrings in a way such that XOR of sub1 and sub2 is always 0 so that the denominator term will always be 1 (20). So for that, find the longest common substring from the two strings s1 and s2, and print its length that would be the required answer.

To find longest common substring we will go with memoisation approach.

Algorithm:

1.    Take two strings as input: s1 and s2.
2.    Initialize a 2D array dp of size n+1 by m+1 with -1, where n is the length of s1 and m is the length of s2.
3.    Define a function lcs(s1, s2, n, m) that takes s1, s2, n, and m as input.
4.    If n or m is equal to 0, return 0 as the base case.
5.    If dp[n][m] is not equal to -1, return dp[n][m].
6.    If the last characters of s1 and s2 match, then return 1 + lcs(s1, s2, n-1, m-1).
7.    Otherwise, return the maximum of lcs(s1, s2, n-1, m) and lcs(s1, s2, n, m-1).
8.    In the main function, call lcs(s1, s2, n, m) and print the result.

Below is the implementation of above approach:

// C++ program for above approach

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

int dp[1000][1000];

// Function to find longest common substring.
int lcs(string s, string k, int n, int m)
{
      // base case
    if (n == 0 or m == 0) {
        return 0;
    }
    
      // if value is already computed
      // return that value
      if(dp[n][m] != -1)
          return dp[n][m];
  
      // if characters at (n-1) and (m-1)th position
      // of the strings are equal
      if (s[n - 1] == k[m - 1]) {
        return dp[n][m] = 1 + lcs(s, k, n - 1, m - 1);
    }
  
      // if characters at (n-1) and (m-1)th position
      // of the strings are not equal,
    // return maximum of LCS of two substrings after
    // excluding last character of each string
    return dp[n][m] = max(lcs(s, k, n - 1, m),
                    lcs(s, k, n, m - 1));
}

// Driver Code
int main()
{
    string s1 = "1110";
    string s2 = "1101";
    
      // initialise dp with -1
      memset(dp, -1, sizeof(dp));
  
    cout << lcs(s1, s2,
                s1.size(), s2.size());

    return 0;
}

// This code is contributed by Chandramani Kumar

import java.util.Arrays;

public class GFG {
    static int[][] dp;

    // Function to find longest common substring.
    static int lcs(String s, String k, int n, int m) {
        // base case
        if (n == 0 || m == 0) {
            return 0;
        }

        // if value is already computed, return that value
        if (dp[n][m] != -1) {
            return dp[n][m];
        }

        // if characters at (n-1) and (m-1)th position
        // of the strings are equal
        if (s.charAt(n - 1) == k.charAt(m - 1)) {
            return dp[n][m] = 1 + lcs(s, k, n - 1, m - 1);
        }

        // if characters at (n-1) and (m-1)th position
        // of the strings are not equal,
        // return the maximum of LCS of two substrings after
        // excluding the last character of each string
        return dp[n][m] = Math.max(lcs(s, k, n - 1, m), lcs(s, k, n, m - 1));
    }

    // Driver code
    public static void main(String[] args) {
        String s1 = "1110";
        String s2 = "1101";

        // initialize dp with -1
        dp = new int[s1.length() + 1][s2.length() + 1];
        for (int[] row : dp) {
            Arrays.fill(row, -1);
        }

        System.out.println(lcs(s1, s2, s1.length(), s2.length()));
    }
}

# Function to find the longest common substring
def lcs(s, k, n, m):
    # Base case
    if n == 0 or m == 0:
        return 0

    # If value is already computed, return that value
    if dp[n][m] != -1:
        return dp[n][m]

    # If characters at (n-1) and (m-1) positions of the strings are equal
    if s[n - 1] == k[m - 1]:
        dp[n][m] = 1 + lcs(s, k, n - 1, m - 1)
        return dp[n][m]

    # If characters at (n-1) and (m-1) positions of the strings are not equal,
    # return the maximum of LCS of two substrings after excluding the last character of each string
    dp[n][m] = max(lcs(s, k, n - 1, m), lcs(s, k, n, m - 1))
    return dp[n][m]


# Driver Code
s1 = "1110"
s2 = "1101"

# Initialize dp with -1
dp = [[-1 for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]

print(lcs(s1, s2, len(s1), len(s2)))

using System;

public class GFG
{
    static int[,] dp;

    // Function to find longest common substring.
    public static int LCS(string s, string k, int n, int m)
    {
        // Base case
        if (n == 0 || m == 0)
        {
            return 0;
        }

        // If value is already computed, return that value
        if (dp[n, m] != -1)
            return dp[n, m];

        // If characters at (n-1) and (m-1)th position of the strings are equal
        if (s[n - 1] == k[m - 1])
        {
            return dp[n, m] = 1 + LCS(s, k, n - 1, m - 1);
        }

        // If characters at (n-1) and (m-1)th position of the strings are not equal,
        // return maximum of LCS of two substrings after excluding the last character of each string
        return dp[n, m] = Math.Max(LCS(s, k, n - 1, m), LCS(s, k, n, m - 1));
    }

    public static void Main(string[] args)
    {
        string s1 = "1110";
        string s2 = "1101";

        // Initialize dp with -1
        dp = new int[s1.Length + 1, s2.Length + 1];
        for (int i = 0; i <= s1.Length; i++)
        {
            for (int j = 0; j <= s2.Length; j++)
            {
                dp[i, j] = -1;
            }
        }

        Console.WriteLine(LCS(s1, s2, s1.Length, s2.Length));
    }
}

// Function to find longest common substring.
function lcs(s, k, n, m) {
    // Initialize dp with -1
    const dp = new Array(n + 1).fill().map(() => new Array(m + 1).fill(-1));

    // Base case
    if (n === 0 || m === 0) {
        return 0;
    }
    
    // If value is already computed, return that value
    if (dp[n][m] !== -1) {
        return dp[n][m];
    }
  
    // If characters at (n-1) and (m-1)th position of the strings are equal
    if (s[n - 1] === k[m - 1]) {
        return dp[n][m] = 1 + lcs(s, k, n - 1, m - 1);
    }
  
    // If characters at (n-1) and (m-1)th position of the strings are not equal,
    // return the maximum of LCS of two substrings after excluding the last character of each string
    return dp[n][m] = Math.max(lcs(s, k, n - 1, m),
                               lcs(s, k, n, m - 1));
}

// Driver Code
function main() {
    const s1 = "1110";
    const s2 = "1101";
    
    const n = s1.length;
    const m = s2.length;
  
    console.log(lcs(s1, s2, n, m));
}

main();

3

Time Complexity: O(N*M), where N is the size of s1 and M is the size of s2.
Auxiliary Space: O(N*M), where N is the size of s1 and M is the size of s2.