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Longest Common Subsequence
The longest common subsequence is a type of subsequence which is present in both of the given sequences or arrays.
We can see that there are many subproblems, which are computed again and again to solve this problem. By using the Overlapping Substructure Property of Dynamic programming, we can overcome the computational efforts. Once the result of subproblems is calculated and stored in the table, we can simply calculate the next results by use of the previous results.
Input and Output
Input: Two strings with different letters or symbols. string 1: AGGTAB string 2: GXTXAYB Output: The length of the longest common subsequence. Here it is 4. AGGTAB and GXTXAYB. The underlined letters are present in both strings and in correct sequence.
Algorithm
longestComSubSeq(str1, str2)
Input − Two strings to find the length of Longest Common Subsequence.
Output − The length of LCS.
Begin m := length of str1 n := length of str2 define longSubSeq matrix of order (m+1) x (n+1) for i := 0 to m, do for j := 0 to n, do if i = 0 or j = 0, then longSubSeq[i,j] := 0 else if str1[i-1] = str2[j-1], then longSubSeq[i,j] := longSubSeq[i-1,j-1] + 1 else longSubSeq[i,j] := maximum of longSubSeq[i-1, j] and longSubSeq[i, j-1] done done longSubSeq[m, n] End
Example
#include<iostream>
using namespace std;
int max(int a, int b) {
return (a > b)? a : b;
}
int longestComSs( string str1, string str2) {
int m = str1.size();
int n = str2.size();
int longSubSeq[m+1][n+1];
//longSubSeq[i,j] will hold the LCS of str1 (0 to i-1) and str2 (0 to j-1)
for (int i=0; i<=m; i++) {
for (int j=0; j<=n; j++) {
if (i == 0 || j == 0)
longSubSeq[i][j] = 0;
else if (str1[i-1] == str2[j-1])
longSubSeq[i][j] = longSubSeq[i-1][j-1] + 1;
else
longSubSeq[i][j] = max(longSubSeq[i-1][j], longSubSeq[i][j-1]);
}
}
return longSubSeq[m][n];
}
int main() {
string str1 = "AGGTAB";
string str2 = "GXTXAYB";
cout << "Length of Longest Common Subsequence is: " << longestComSs( str1, str2);
}Output
Length of Longest Common Subsequence is: 4
Updated on: 16-Jun-2020
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