*LCS Problem Statement:* Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”. So a string of length n has 2^n different possible subsequences.

It is a classic computer science problem, the basis of diff (a file comparison program that outputs the differences between two files), and has applications in bioinformatics.

**Examples:**

LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.

LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.

Let the input sequences be X[0..m-1] and Y[0..n-1] of lengths m and n respectively. And let L(X[0..m-1], Y[0..n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0..m-1], Y[0..n-1]).

If last characters of both sequences match (or X[m-1] == Y[n-1]) then

L(X[0..m-1], Y[0..n-1]) = 1 + L(X[0..m-2], Y[0..n-2])

If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then

L(X[0..m-1], Y[0..n-1]) = MAX ( L(X[0..m-2], Y[0..n-1]), L(X[0..m-1], Y[0..n-2])

`/* A Naive recursive 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) `
`{ ` ` ` `if` `(m == 0 || n == 0) `
` ` `return` `0; `
` ` `if` `(X[m - 1] == Y[n - 1]) `
` ` `return` `1 + lcs(X, Y, m - 1, n - 1); `
` ` `else`
` ` `return` `max(lcs(X, Y, m, n - 1), lcs(X, Y, m - 1, 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\n"` `, lcs(X, Y, m, n)); `
` ` ` ` `return` `0; `
`} ` |

*chevron_right*

*filter_none*

**Output:**

Length of LCS is 4

Following is a tabulated implementation for the LCS problem.

`/* Dynamic Programming C/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\n"` `, lcs(X, Y, m, n)); `
` ` ` ` `return` `0; `
`} ` |

*chevron_right*

*filter_none*

**Output:**

Length of LCS is 4

Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!

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