# 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
Samual Sam

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