# Longest Common Subsequence

Given two sequences,the task is to find the length of longest sub-sequence present in both of them. A sub-sequence is a sequence that appears in the same relative order, but not necessarily contiguous.For example- “abc”,”fit”,”abfi”,etc are sub-sequences of “abcfit”.A string of length n has 2^n different possible sub-sequences.

Example-

String 1- “BATTING” Length-7
String 2- “BOWLING” Length-7

Longest Common Sub-sequence- “BING” Length-4

Brute Force-

Generate all sub-sequences of both given sequences and find the longest matching sub-sequence.It has an exponential time complexity.

Efficient Algorithm-
lcs(i,j)=
if (X[i]==Y[j])
1+lcs(i-1,j-1)
else
max(lcs(i-1,j),lcs(i,j-1))

where X[],Y[] are strings.

Recursive Solution-

[cpp]

//char X[0,1…m-1],Y[0,1,…n-1] contains the strings
//lcs(m,n) is called initially.

int lcs(int a,int b)
{
if (a == 0 || b == 0)
return 0;

if (X[a-1] == Y[b-1])
return 1 + lcs(a-1,b-1);
else
return max(lcs(a,b-1), lcs(a-1,b));
}

[/cpp]

This problem has overlapping as well as optimal substructure property.Thus,dynamic programming can be applied.A temporary array L[ ][ ] is constructed to store the intermediate results.

DP solution-

[cpp]

int lcs(int m,int n)
{
int L[m+1][n+1];
int i,j;

// 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];
}

[/cpp]
This program has a time complexity of O(m*n) where m,n are lengths of the string which is much better than exponential time complexity. saurabh