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) / (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++
#include <bits/stdc++.h>
using namespace std;
int dp[1000][1000];
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 dp[n][m];
}
int main()
{
string s1 = "1110" ;
string s2 = "1101" ;
cout << lcs(s1, s2,
s1.size(), s2.size());
return 0;
}
|
Java
class GFG{
static int dp[][] = new int [ 1000 ][ 1000 ];
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 dp[n][m];
}
public static void main(String [] args)
{
String s1 = "1110" ;
String s2 = "1101" ;
System.out.print(lcs(s1, s2,
s1.length(), s2.length()));
}
}
|
Python3
import numpy as np;
dp = np.zeros(( 1000 , 1000 ));
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 dp[n][m];
if __name__ = = "__main__" :
s1 = "1110" ;
s2 = "1101" ;
print (lcs(s1, s2, len (s1), len (s2)));
|
C#
using System;
public class GFG{
static int [,]dp = new int [1000,1000];
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 dp[n, m];
}
public static void Main( string [] args)
{
string s1 = "1110" ;
string s2 = "1101" ;
Console.Write(lcs(s1, s2, s1.Length, s2.Length));
}
}
|
Javascript
<script>
var dp = new Array(1000);
for ( var i = 0; i < 1000; i++) {
dp[i] = new Array(1000);
}
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 dp[n][m];
}
var s1 = "1110" ;
var s2 = "1101" ;
document.write(lcs(s1, s2, s1.length, s2.length))
</script>
|
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:
- Take two strings as input: s1 and s2.
- 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.
- Define a function lcs(s1, s2, n, m) that takes s1, s2, n, and m as input.
- If n or m is equal to 0, return 0 as the base case.
- If dp[n][m] is not equal to -1, return dp[n][m].
- If the last characters of s1 and s2 match, then return 1 + lcs(s1, s2, n-1, m-1).
- Otherwise, return the maximum of lcs(s1, s2, n-1, m) and lcs(s1, s2, n, m-1).
- In the main function, call lcs(s1, s2, n, m) and print the result.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int dp[1000][1000];
int lcs(string s, string k, int n, int m)
{
if (n == 0 or m == 0) {
return 0;
}
if (dp[n][m] != -1)
return dp[n][m];
if (s[n - 1] == k[m - 1]) {
return dp[n][m] = 1 + lcs(s, k, n - 1, m - 1);
}
return dp[n][m] = max(lcs(s, k, n - 1, m),
lcs(s, k, n, m - 1));
}
int main()
{
string s1 = "1110" ;
string s2 = "1101" ;
memset (dp, -1, sizeof (dp));
cout << lcs(s1, s2,
s1.size(), s2.size());
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
static int [][] dp;
static int lcs(String s, String k, int n, int m) {
if (n == 0 || m == 0 ) {
return 0 ;
}
if (dp[n][m] != - 1 ) {
return dp[n][m];
}
if (s.charAt(n - 1 ) == k.charAt(m - 1 )) {
return dp[n][m] = 1 + lcs(s, k, n - 1 , m - 1 );
}
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" ;
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()));
}
}
|
Python3
def lcs(s, k, n, m):
if n = = 0 or m = = 0 :
return 0
if dp[n][m] ! = - 1 :
return dp[n][m]
if s[n - 1 ] = = k[m - 1 ]:
dp[n][m] = 1 + lcs(s, k, n - 1 , m - 1 )
return dp[n][m]
dp[n][m] = max (lcs(s, k, n - 1 , m), lcs(s, k, n, m - 1 ))
return dp[n][m]
s1 = "1110"
s2 = "1101"
dp = [[ - 1 for _ in range ( len (s2) + 1 )] for _ in range ( len (s1) + 1 )]
print (lcs(s1, s2, len (s1), len (s2)))
|
C#
using System;
public class GFG
{
static int [,] dp;
public static int LCS( string s, string k, int n, int m)
{
if (n == 0 || m == 0)
{
return 0;
}
if (dp[n, m] != -1)
return dp[n, m];
if (s[n - 1] == k[m - 1])
{
return dp[n, m] = 1 + LCS(s, k, n - 1, m - 1);
}
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" ;
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));
}
}
|
Javascript
function lcs(s, k, n, m) {
const dp = new Array(n + 1).fill().map(() => new Array(m + 1).fill(-1));
if (n === 0 || m === 0) {
return 0;
}
if (dp[n][m] !== -1) {
return dp[n][m];
}
if (s[n - 1] === k[m - 1]) {
return dp[n][m] = 1 + lcs(s, k, n - 1, m - 1);
}
return dp[n][m] = Math.max(lcs(s, k, n - 1, m),
lcs(s, k, n, m - 1));
}
function main() {
const s1 = "1110" ;
const s2 = "1101" ;
const n = s1.length;
const m = s2.length;
console.log(lcs(s1, s2, n, m));
}
main();
|
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.
Last Updated :
06 Nov, 2023
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...