Given two sequences, print the longest subsequence present in both of them.

**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.

We have discussed Longest Common Subsequence (LCS) problem in a previous post. The function discussed there was mainly to find the length of LCS. To find length of LCS, a 2D table L[][] was constructed. In this post, the function to construct and print LCS is discussed.

Following is detailed algorithm to print the LCS. It uses the same 2D table L[][].

**1)** Construct L[m+1][n+1] using the steps discussed in previous post.

**2)** The value L[m][n] contains length of LCS. Create a character array lcs[] of length equal to the length of lcs plus 1 (one extra to store \0).

**2)** Traverse the 2D array starting from L[m][n]. Do following for every cell L[i][j]

…..**a)** If characters (in X and Y) corresponding to L[i][j] are same (Or X[i-1] == Y[j-1]), then include this character as part of LCS.

…..**b)** Else compare values of L[i-1][j] and L[i][j-1] and go in direction of greater value.

The following table (taken from Wiki) shows steps (highlighted) followed by the above algorithm.

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | ||
---|---|---|---|---|---|---|---|---|---|

Ø | M | Z | J | A | W | X | U | ||

0 | Ø | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | X | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 |

2 | M | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |

3 | J | 0 | 1 | 1 | 2 |
2 | 2 | 2 | 2 |

4 | Y | 0 | 1 | 1 | 2 | 2 | 2 | 2 | 2 |

5 | A | 0 | 1 | 1 | 2 | 3 |
3 | 3 | 3 |

6 | U | 0 | 1 | 1 | 2 | 3 | 3 | 3 | 4 |

7 | Z | 0 | 1 | 2 | 2 | 3 | 3 | 3 | 4 |

Following is the implementation of above approach.

## C/C++

`/* Dynamic Programming implementation of LCS problem */` `#include<iostream> ` `#include<cstring> ` `#include<cstdlib> ` `using` `namespace` `std; ` ` ` `/* Returns length of LCS for X[0..m-1], Y[0..n-1] */` `void` `lcs( ` `char` `*X, ` `char` `*Y, ` `int` `m, ` `int` `n ) ` `{ ` ` ` `int` `L[m+1][n+1]; ` ` ` ` ` `/* 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` `(` `int` `i=0; i<=m; i++) ` ` ` `{ ` ` ` `for` `(` `int` `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]); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Following code is used to print LCS ` ` ` `int` `index = L[m][n]; ` ` ` ` ` `// Create a character array to store the lcs string ` ` ` `char` `lcs[index+1]; ` ` ` `lcs[index] = ` `''` `; ` `// Set the terminating character ` ` ` ` ` `// Start from the right-most-bottom-most corner and ` ` ` `// one by one store characters in lcs[] ` ` ` `int` `i = m, j = n; ` ` ` `while` `(i > 0 && j > 0) ` ` ` `{ ` ` ` `// If current character in X[] and Y are same, then ` ` ` `// current character is part of LCS ` ` ` `if` `(X[i-1] == Y[j-1]) ` ` ` `{ ` ` ` `lcs[index-1] = X[i-1]; ` `// Put current character in result ` ` ` `i--; j--; index--; ` `// reduce values of i, j and index ` ` ` `} ` ` ` ` ` `// If not same, then find the larger of two and ` ` ` `// go in the direction of larger value ` ` ` `else` `if` `(L[i-1][j] > L[i][j-1]) ` ` ` `i--; ` ` ` `else` ` ` `j--; ` ` ` `} ` ` ` ` ` `// Print the lcs ` ` ` `cout << ` `"LCS of "` `<< X << ` `" and "` `<< Y << ` `" is "` `<< lcs; ` `} ` ` ` `/* Driver program to test above function */` `int` `main() ` `{ ` ` ` `char` `X[] = ` `"AGGTAB"` `; ` ` ` `char` `Y[] = ` `"GXTXAYB"` `; ` ` ` `int` `m = ` `strlen` `(X); ` ` ` `int` `n = ` `strlen` `(Y); ` ` ` `lcs(X, Y, m, n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Dynamic Programming implementation of LCS problem in Java ` `import` `java.io.*; ` ` ` `class` `LongestCommonSubsequence ` `{ ` ` ` `// Returns length of LCS for X[0..m-1], Y[0..n-1] ` ` ` `static` `void` `lcs(String X, String Y, ` `int` `m, ` `int` `n) ` ` ` `{ ` ` ` `int` `[][] L = ` `new` `int` `[m+` `1` `][n+` `1` `]; ` ` ` ` ` `// 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` `(` `int` `i=` `0` `; i<=m; i++) ` ` ` `{ ` ` ` `for` `(` `int` `j=` `0` `; j<=n; j++) ` ` ` `{ ` ` ` `if` `(i == ` `0` `|| j == ` `0` `) ` ` ` `L[i][j] = ` `0` `; ` ` ` `else` `if` `(X.charAt(i-` `1` `) == Y.charAt(j-` `1` `)) ` ` ` `L[i][j] = L[i-` `1` `][j-` `1` `] + ` `1` `; ` ` ` `else` ` ` `L[i][j] = Math.max(L[i-` `1` `][j], L[i][j-` `1` `]); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Following code is used to print LCS ` ` ` `int` `index = L[m][n]; ` ` ` `int` `temp = index; ` ` ` ` ` `// Create a character array to store the lcs string ` ` ` `char` `[] lcs = ` `new` `char` `[index+` `1` `]; ` ` ` `lcs[index] = ` `''` `; ` `// Set the terminating character ` ` ` ` ` `// Start from the right-most-bottom-most corner and ` ` ` `// one by one store characters in lcs[] ` ` ` `int` `i = m, j = n; ` ` ` `while` `(i > ` `0` `&& j > ` `0` `) ` ` ` `{ ` ` ` `// If current character in X[] and Y are same, then ` ` ` `// current character is part of LCS ` ` ` `if` `(X.charAt(i-` `1` `) == Y.charAt(j-` `1` `)) ` ` ` `{ ` ` ` `// Put current character in result ` ` ` `lcs[index-` `1` `] = X.charAt(i-` `1` `); ` ` ` ` ` `// reduce values of i, j and index ` ` ` `i--; ` ` ` `j--; ` ` ` `index--; ` ` ` `} ` ` ` ` ` `// If not same, then find the larger of two and ` ` ` `// go in the direction of larger value ` ` ` `else` `if` `(L[i-` `1` `][j] > L[i][j-` `1` `]) ` ` ` `i--; ` ` ` `else` ` ` `j--; ` ` ` `} ` ` ` ` ` `// Print the lcs ` ` ` `System.out.print(` `"LCS of "` `+X+` `" and "` `+Y+` `" is "` `); ` ` ` `for` `(` `int` `k=` `0` `;k<=temp;k++) ` ` ` `System.out.print(lcs[k]); ` ` ` `} ` ` ` ` ` `// driver program ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `String X = ` `"AGGTAB"` `; ` ` ` `String Y = ` `"GXTXAYB"` `; ` ` ` `int` `m = X.length(); ` ` ` `int` `n = Y.length(); ` ` ` `lcs(X, Y, m, n); ` ` ` `} ` `} ` ` ` `// Contributed by Pramod Kumar ` |

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## Python

`# Dynamic programming implementation of LCS problem ` ` ` `# Returns length of LCS for X[0..m-1], Y[0..n-1] ` `def` `lcs(X, Y, m, n): ` ` ` `L ` `=` `[[` `0` `for` `x ` `in` `xrange` `(n` `+` `1` `)] ` `for` `x ` `in` `xrange` `(m` `+` `1` `)] ` ` ` ` ` `# 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 ` `in` `xrange` `(m` `+` `1` `): ` ` ` `for` `j ` `in` `xrange` `(n` `+` `1` `): ` ` ` `if` `i ` `=` `=` `0` `or` `j ` `=` `=` `0` `: ` ` ` `L[i][j] ` `=` `0` ` ` `elif` `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` `]) ` ` ` ` ` `# Following code is used to print LCS ` ` ` `index ` `=` `L[m][n] ` ` ` ` ` `# Create a character array to store the lcs string ` ` ` `lcs ` `=` `[""] ` `*` `(index` `+` `1` `) ` ` ` `lcs[index] ` `=` `"" ` ` ` ` ` `# Start from the right-most-bottom-most corner and ` ` ` `# one by one store characters in lcs[] ` ` ` `i ` `=` `m ` ` ` `j ` `=` `n ` ` ` `while` `i > ` `0` `and` `j > ` `0` `: ` ` ` ` ` `# If current character in X[] and Y are same, then ` ` ` `# current character is part of LCS ` ` ` `if` `X[i` `-` `1` `] ` `=` `=` `Y[j` `-` `1` `]: ` ` ` `lcs[index` `-` `1` `] ` `=` `X[i` `-` `1` `] ` ` ` `i` `-` `=` `1` ` ` `j` `-` `=` `1` ` ` `index` `-` `=` `1` ` ` ` ` `# If not same, then find the larger of two and ` ` ` `# go in the direction of larger value ` ` ` `elif` `L[i` `-` `1` `][j] > L[i][j` `-` `1` `]: ` ` ` `i` `-` `=` `1` ` ` `else` `: ` ` ` `j` `-` `=` `1` ` ` ` ` `print` `"LCS of "` `+` `X ` `+` `" and "` `+` `Y ` `+` `" is "` `+` `"".join(lcs) ` ` ` `# Driver program ` `X ` `=` `"AGGTAB"` `Y ` `=` `"GXTXAYB"` `m ` `=` `len` `(X) ` `n ` `=` `len` `(Y) ` `lcs(X, Y, m, n) ` ` ` `# This code is contributed by BHAVYA JAIN ` |

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## C#

`// Dynamic Programming implementation ` `// of LCS problem in C# ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Returns length of LCS for X[0..m-1], Y[0..n-1] ` ` ` `static` `void` `lcs(String X, String Y, ` `int` `m, ` `int` `n) ` ` ` `{ ` ` ` `int` `[,] L = ` `new` `int` `[m+1, n+1]; ` ` ` ` ` `// 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` `(` `int` `i = 0; i <= m; i++) ` ` ` `{ ` ` ` `for` `(` `int` `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] = Math.Max(L[i-1, j], L[i, j-1]); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Following code is used to print LCS ` ` ` `int` `index = L[m, n]; ` ` ` `int` `temp = index; ` ` ` ` ` `// Create a character array ` ` ` `// to store the lcs string ` ` ` `char` `[] lcs = ` `new` `char` `[index+1]; ` ` ` ` ` `// Set the terminating character ` ` ` `lcs[index] = ` `'\0'` `; ` ` ` ` ` `// Start from the right-most-bottom-most corner ` ` ` `// and one by one store characters in lcs[] ` ` ` `int` `k = m, l = n; ` ` ` `while` `(k > 0 && l > 0) ` ` ` `{ ` ` ` `// If current character in X[] and Y ` ` ` `// are same, then current character ` ` ` `// is part of LCS ` ` ` `if` `(X[k-1] == Y[l-1]) ` ` ` `{ ` ` ` `// Put current character in result ` ` ` `lcs[index-1] = X[k-1]; ` ` ` ` ` `// reduce values of i, j and index ` ` ` `k--; ` ` ` `l--; ` ` ` `index--; ` ` ` `} ` ` ` ` ` `// If not same, then find the larger of two and ` ` ` `// go in the direction of larger value ` ` ` `else` `if` `(L[k-1, l] > L[k, l-1]) ` ` ` `k--; ` ` ` `else` ` ` `l--; ` ` ` `} ` ` ` ` ` `// Print the lcs ` ` ` `Console.Write(` `"LCS of "` `+ X + ` `" and "` `+ Y + ` `" is "` `); ` ` ` `for` `(` `int` `q = 0; q <= temp; q++) ` ` ` `Console.Write(lcs[q]); ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `String X = ` `"AGGTAB"` `; ` ` ` `String Y = ` `"GXTXAYB"` `; ` ` ` `int` `m = X.Length; ` ` ` `int` `n = Y.Length; ` ` ` `lcs(X, Y, m, n); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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Output:

LCS of AGGTAB and GXTXAYB is GTAB

**References:**

http://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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