text: # Maximize given function by selecting equal length substrings from given Binary Strings
  
  • Last Updated : 03 Nov, 2021
  
  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) / (2xor(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/ (2xor(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++
  
   // C++ program for above approach#include 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 Codeint main(){    string s1 = "1110";    string s2 = "1101";     cout << lcs(s1, s2,                s1.size(), s2.size());     return 0;}
  
  ## Java
  
   // Java program for above approachclass 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
  
   # 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 Codeif __name__ == "__main__" :     s1 = "1110";    s2 = "1101";     print(lcs(s1, s2,len(s1), len(s2)));    # This code is contributed by AnkThon
  
  ## C#
  
   // C# program for above approachusing 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
  
  ## Javascript
  
  
  
  Output
  
  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.
  
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