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

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 ``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;` `    ``// Iterate 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``;``    ` `    ``// Iterate 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` `    ``# Iterate 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 substrings``    ``int``[,] dp = ``new` `int``[length1 + 1, length2 + 1];``    ``int` `max = 0;``    ` `    ``// Iterate 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

 ``

Output
```3

```

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

Efficient approach : Space optimization

In previous approach the current value LCSuff[i][j] is only depend upon the current and previous row values of matrix. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector LCSuff of size n+1.
• Set a base case by initializing the values of LCSuff.
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• Now Create a temporary variables prev used to store previous computations and temp for current value.
• After every iteration assign the value of temp to prev for further iteration.
• Initialize a variable result to store the final answer and update it by iterating through the LCSuff.
• At last return and print the final answer stored in result.

Implementation:

## C++

 `// C++ code for above approach` `#include ``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[] stores the lengths``    ``// of the longest common suffixes``    ``// of substrings``    ``int` `LCSuff[n + 1];``    ``int` `result = 0;` `    ``// Iterate over strings A and B``    ``for` `(``int` `i = 0; i <= m; i++) {``        ` `        ``// previous element``        ``int` `prev = 0;``        ``for` `(``int` `j = 0; j <= n; j++) {``            ``int` `temp = LCSuff[j];``            ` `             ``// If first row or column``            ``if` `(i == 0 || j == 0)``                ``LCSuff[j] = 0;``             ``// If matching is found``            ``else` `if` `(A[i - 1] == B[j - 1]) {``                ``LCSuff[j] = prev + 1;``                ``result = max(result, LCSuff[j]);``            ``}``            ``// Otherwise, if matching``            ``// is not found``            ``else``                ``LCSuff[j] = 0;``            ``prev = temp;``        ``}``    ``}``    ` `    ``// 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;``}` `// --- by bhardwajji`

## Java

 `// Java code for above approach` `import` `java.util.*;` `public` `class` `Main ``{` `  ``// Function to find the length of the``  ``// longest common substring of the``  ``// string X and Y``  ``public` `static` `int` `LCSubStr(``char``[] A, ``char``[] B, ``int` `m, ``int` `n) ``  ``{` `    ``// LCSuff[] stores the lengths``    ``// of the longest common suffixes``    ``// of substrings``    ``int``[] LCSuff = ``new` `int``[n + ``1``];``    ``int` `result = ``0``;``    ``// Iterate over strings A and B``    ``for` `(``int` `i = ``0``; i <= m; i++) {``      ``// previous element``      ``int` `prev = ``0``;``      ``for` `(``int` `j = ``0``; j <= n; j++) {``        ``int` `temp = LCSuff[j];` `        ``// If first row or column``        ``if` `(i == ``0` `|| j == ``0``)``          ``LCSuff[j] = ``0``;``        ``// If matching is found``        ``else` `if` `(A[i - ``1``] == B[j - ``1``]) {``          ``LCSuff[j] = prev + ``1``;``          ``result = Math.max(result, LCSuff[j]);``        ``}``        ``// Otherwise, if matching``        ``// is not found``        ``else``          ``LCSuff[j] = ``0``;``        ``prev = temp;``      ``}``    ``}` `    ``// Finally, return the resultant``    ``// maximum value LCS``    ``return` `result;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args) {``    ``char``[] A = ``"0110"``.toCharArray();``    ``char``[] B = ``"1101"``.toCharArray();``    ``int` `M = A.length;``    ``int` `N = B.length;` `    ``// function call``    ``System.out.println(LCSubStr(A, B, M, N));``  ``}``}`

## Python

 `def` `LCSubStr(A, B, m, n):``    ``# LCSuff[] stores the lengths``    ``# of the longest common suffixes``    ``# of substrings``    ``LCSuff ``=` `[``0``] ``*` `(n ``+` `1``)``    ``result ``=` `0` `    ``# Iterate over strings A and B``    ``for` `i ``in` `range``(m``+``1``):``        ``# previous element``        ``prev ``=` `0``        ``for` `j ``in` `range``(n``+``1``):``            ``temp ``=` `LCSuff[j]``            ` `            ``# If first row or column``            ``if` `i ``=``=` `0` `or` `j ``=``=` `0``:``                ``LCSuff[j] ``=` `0``            ``# If matching is found``            ``elif` `A[i ``-` `1``] ``=``=` `B[j ``-` `1``]:``                ``LCSuff[j] ``=` `prev ``+` `1``                ``result ``=` `max``(result, LCSuff[j])``            ``# Otherwise, if matching``            ``# is not found``            ``else``:``                ``LCSuff[j] ``=` `0``            ``prev ``=` `temp``    ` `    ``# Finally, return the resultant``    ``# maximum value LCS``    ``return` `result` `# Driver code``A ``=` `"0110"``B ``=` `"1101"``M ``=` `len``(A)``N ``=` `len``(B)` `# function call``print``(LCSubStr(A, B, M, N))`

## C#

 `using` `System;` `public` `class` `GFG``{``    ``// Function to find the length of the``    ``// longest common substring of the``    ``// string X and Y``    ``public` `static` `int` `LCSubStr(``char``[] A, ``char``[] B, ``int` `m, ``int` `n)``    ``{``        ``// LCSuff[] stores the lengths``        ``// of the longest common suffixes``        ``// of substrings``        ``int``[] LCSuff = ``new` `int``[n + 1];``        ``int` `result = 0;` `        ``// Iterate over strings A and B``        ``for` `(``int` `i = 0; i <= m; i++)``        ``{``            ``// previous element``            ``int` `prev = 0;``            ``for` `(``int` `j = 0; j <= n; j++)``            ``{``                ``int` `temp = LCSuff[j];` `                ``// If first row or column``                ``if` `(i == 0 || j == 0)``                    ``LCSuff[j] = 0;``                ``// If matching is found``                ``else` `if` `(A[i - 1] == B[j - 1])``                ``{``                    ``LCSuff[j] = prev + 1;``                    ``result = Math.Max(result, LCSuff[j]);``                ``}``                ``// Otherwise, if matching``                ``// is not found``                ``else``                    ``LCSuff[j] = 0;` `                ``prev = temp;``            ``}``        ``}` `        ``// Finally, return the resultant``        ``// maximum value LCS``        ``return` `result;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``char``[] A = ``"0110"``.ToCharArray();``        ``char``[] B = ``"1101"``.ToCharArray();``        ``int` `M = A.Length;``        ``int` `N = B.Length;` `        ``// function call``        ``Console.WriteLine(LCSubStr(A, B, M, N));``    ``}``}`

## Javascript

 `function` `LCSubStr(A, B, m, n) {``    ``// LCSuff[] stores the lengths``    ``// of the longest common suffixes``    ``// of substrings``    ``const LCSuff = ``new` `Array(n + 1).fill(0);``    ``let result = 0;` `    ``// Iterate over strings A and B``    ``for` `(let i = 0; i <= m; i++) {``        ``// previous element``        ``let prev = 0;``        ``for` `(let j = 0; j <= n; j++) {``            ``let temp = LCSuff[j];` `            ``// If first row or column``            ``if` `(i === 0 || j === 0)``                ``LCSuff[j] = 0;``            ``// If matching is found``            ``else` `if` `(A[i - 1] === B[j - 1]) {``                ``LCSuff[j] = prev + 1;``                ``result = Math.max(result, LCSuff[j]);``            ``}``            ``// Otherwise, if matching``            ``// is not found``            ``else``                ``LCSuff[j] = 0;``            ``prev = temp;``        ``}``    ``}` `    ``// Finally, return the resultant``    ``// maximum value LCS``    ``return` `result;``}` `// Driver code``const A = ``"0110"``;``const B = ``"1101"``;``const M = A.length;``const N = B.length;` `// Function call``console.log(LCSubStr(A, B, M, N));`

Output:

`3`

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

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