Printing Longest Common Subsequence
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.
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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] = '\0'; // 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;} |
Java
// Dynamic Programming implementation of LCS problem in Javaimport 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] = '\u0000'; // Set the terminating character // Start from the right-most-bottom-most corner and // one by one store characters in lcs[] int i = m; int 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 |
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 programX = "AGGTAB"Y = "GXTXAYB"m = len(X)n = len(Y)lcs(X, Y, m, n) # This code is contributed by BHAVYA JAIN |
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 |
PHP
<?php // Dynamic Programming implementation of LCS problem // Returns length of LCS for X[0..m-1], Y[0..n-1] function lcs( $X, $Y, $m, $n ){ $L = array_fill(0, $m + 1, array_fill(0, $n + 1, NULL)); /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for ($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]); } } // Following code is used to print LCS $index = $L[$m][$n]; $temp = $index; // Create a character array to store the lcs string $lcs = array_fill(0, $index + 1, NULL); $lcs[$index] = ''; // Set the terminating character // Start from the right-most-bottom-most corner // and one by one store characters in lcs[] $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]) { // Put current character in result $lcs[$index - 1] = $X[$i - 1]; $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 echo "LCS of " . $X . " and " . $Y . " is "; for($k = 0; $k < $temp; $k++) echo $lcs[$k];} // Driver Code$X = "AGGTAB";$Y = "GXTXAYB";$m = strlen($X);$n = strlen($Y);lcs($X, $Y, $m, $n); // This code is contributed by ita_c?> |
Output:
LCS of AGGTAB and GXTXAYB is GTAB
References:
http://en.wikipedia.org/wiki/Longest_common_subsequence_problem
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