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Solve LCS through recursion

Exuberant Elephant answered on February 16, 2022 Popularity 4/10 Helpfulness 1/10

answer Solve LCS through recursion

Solve LCS through recursion

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#include <iostream>
#include <cstring>
using namespace std;
int f(string s1,string s2)
{
	if(s1.length()==0||s2.length()==0)
		return 0;
	if(s1[0]==s2[0])
		return 1+f(s1.substr(1),s2.substr(1)); 
	else
		return max(f(s1.substr(1),s2),f(s1,s2.substr(1)));
}
int main()
{
	cout<<f("ABCD","ABD")<<endl;
	return 0;
}

Popularity 4/10 Helpfulness 1/10 Language whatever

Source: blog.csdn.net

Tags: lcs recursion whatever

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Contributed on Apr 29 2022

Exuberant Elephant

0 Answers Avg Quality 2/10

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LCS Problem C Recursion

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/* Dynamic Programming C implementation of LCS problem */
#include <bits/stdc++.h>
 
int max(int a, int b);
 
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
int L[m+1][n+1];
int i, j;
 
/* Following steps build L[m+1][n+1] in bottom up fashion. Note
    that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
for (i=0; i<=m; i++)
{
    for (j=0; j<=n; j++)
    {
    if (i == 0 || j == 0)
        L[i][j] = 0;
 
    else if (X[i-1] == Y[j-1])
        L[i][j] = L[i-1][j-1] + 1;
 
    else
        L[i][j] = max(L[i-1][j], L[i][j-1]);
    }
}
     
/* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
return L[m][n];
}
 
/* Utility function to get max of 2 integers */
int max(int a, int b)
{
    return (a > b)? a : b;
}
 
/* Driver program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
 
int m = strlen(X);
int n = strlen(Y);
 
printf("Length of LCS is %d", lcs( X, Y, m, n ) );
 
return 0;
}

Popularity 6/10 Helpfulness 1/10 Language whatever

Source: www.geeksforgeeks.org

Tags: whatever lcs recursion c

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Contributed on Feb 16 2022

Zealous Zebra

0 Answers Avg Quality 2/10

Solve LCS through recursion

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Solve LCS through recursion

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