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 Yint 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 Codeint 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 approachclass GFG{// Function to find the length of the// longest common substring of the// string X and Ystatic 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 Codepublic 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 Ydef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the length of the// longest common substring of the// string X and Ystatic 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 Codepublic 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; // Iterate 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)
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 <bits/stdc++.h>using namespace std;// Function to find the length of the// longest common substring of the// string X and Yint 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 codeint 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 approachimport 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 codeA = "0110"B = "1101"M = len(A)N = len(B)# function callprint(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 codeconst A = "0110";const B = "1101";const M = A.length;const N = B.length;// Function callconsole.log(LCSubStr(A, B, M, N)); |
Output:
3
Time Complexity: O(M*N)
Auxiliary Space: O(N)
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