The longest common subsequence problem is the problem of finding the longest subsequence common to all sequences in a set of sequences.
If string1 and string2 are the two given sequences, then s is the common subsequece of string1 and string2 if s is a subsequence of both string1 and string2, given that s must be strictly increasing sequence of indices of both string1 and string2.
string1 = {A, C, D, A, B, C, D}
string2 = {A, D, B, C, A}
Then common subsequences are : {A, D}, {A, C}, {A, B, C}, {A, D, B} ...
Among these subsequences, {A, D, B, C} is the longest common subsequence.
Dynamic Programming can be used to find this Longest common subsequence.
In this procedure, table C[m,n] is computed in row major and another table B[m,n] is compute dto construct optimal solution.
Algorithm: LCS-Length-Table-Formulation (X, Y)
m := length(X)
n := length(Y)
for i = 1 to m do
C[i, 0] := 0
for j = 1 to n do
C[0, j] := 0
for i = 1 to m do
for j = 1 to n do
if xi = yj
C[i, j] := C[i - 1, j - 1] + 1
B[i, j] := ‘D’
else
if C[i -1, j] ≥ C[i, j -1]
C[i, j] := C[i - 1, j] + 1
B[i, j] := ‘U’
else
C[i, j] := C[i, j - 1]
B[i, j] := ‘L’
return C and B
Algorithm: Print-LCS (B, X, i, j)
if i = 0 and j = 0
return
if B[i, j] = ‘D’
Print-LCS(B, X, i-1, j-1)
Print(xi)
else if B[i, j] = ‘U’
Print-LCS(B, X, i-1, j)
else
Print-LCS(B, X, i, j-1)
// C program for Longest Common Subsequence
#include <stdio.h>
#include <string.h>
int i, j, m, n, LCS_table[20][20];
char S1[20] = "ACDABCD", S2[20] = "ADBCA", b[20][20];
void lcsAlgo() {
m = strlen(S1);
n = strlen(S2);
// Filling 0's in the matrix
for (i = 0; i <= m; i++)
LCS_table[i][0] = 0;
for (i = 0; i <= n; i++)
LCS_table[0][i] = 0;
// Building the mtrix in bottom-up way
for (i = 1; i <= m; i++)
for (j = 1; j <= n; j++) {
if (S1[i - 1] == S2[j - 1]) {
LCS_table[i][j] = LCS_table[i - 1][j - 1] + 1;
} else if (LCS_table[i - 1][j] >= LCS_table[i][j - 1]) {
LCS_table[i][j] = LCS_table[i - 1][j];
} else {
LCS_table[i][j] = LCS_table[i][j - 1];
}
}
int index = LCS_table[m][n];
char lcsAlgo[index + 1];
lcsAlgo[index] = '\0';
int i = m, j = n;
while (i > 0 && j > 0) {
if (S1[i - 1] == S2[j - 1]) {
lcsAlgo[index - 1] = S1[i - 1];
i--;
j--;
index--;
}
else if (LCS_table[i - 1][j] > LCS_table[i][j - 1])
i--;
else
j--;
}
// Printing the sub sequences
printf("S1 : %s \nS2 : %s \n", S1, S2);
printf("LCS: %s", lcsAlgo);
}
int main() {
lcsAlgo();
printf("\n");
}
#include <iostream>
#include <cstring>
using namespace std;
void lcsAlgo(char *S1, char *S2, int m, int n) {
int LCS_table[m + 1][n + 1];
// Building the mtrix in bottom-up way
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0)
LCS_table[i][j] = 0;
else if (S1[i - 1] == S2[j - 1])
LCS_table[i][j] = LCS_table[i - 1][j - 1] + 1;
else
LCS_table[i][j] = max(LCS_table[i - 1][j], LCS_table[i][j - 1]);
}
}
int index = LCS_table[m][n];
char lcsAlgo[index + 1];
lcsAlgo[index] = '\0';
int i = m, j = n;
while (i > 0 && j > 0) {
if (S1[i - 1] == S2[j - 1]) {
lcsAlgo[index - 1] = S1[i - 1];
i--;
j--;
index--;
}
else if (LCS_table[i - 1][j] > LCS_table[i][j - 1])
i--;
else
j--;
}
// Printing the sub sequences
cout << "S1 : " << S1 << "\nS2 : " << S2 << "\nLCS: " << lcsAlgo << "\n";
}
int main() {
char S1[] = "ACDABCD";
char S2[] = "ADBCA";
int m = strlen(S1);
int n = strlen(S2);
lcsAlgo(S1, S2, m, n);
}
# The longest common subsequence in Python
# Function to find lcs_algo
def lcs_algo(S1, S2, m, n):
L = [[0 for x in range(n+1)] for x in range(m+1)]
# Building the mtrix in bottom-up way
for i in range(m+1):
for j in range(n+1):
if i == 0 or j == 0:
L[i][j] = 0
elif S1[i-1] == S2[j-1]:
L[i][j] = L[i-1][j-1] + 1
else:
L[i][j] = max(L[i-1][j], L[i][j-1])
index = L[m][n]
lcs_algo = [""] * (index+1)
lcs_algo[index] = ""
i = m
j = n
while i > 0 and j > 0:
if S1[i-1] == S2[j-1]:
lcs_algo[index-1] = S1[i-1]
i -= 1
j -= 1
index -= 1
elif L[i-1][j] > L[i][j-1]:
i -= 1
else:
j -= 1
# Printing the sub sequences
print("S1 : " + S1 + "\nS2 : " + S2)
print("LCS: " + "".join(lcs_algo))
S1 = "ACDABCD"
S2 = "ADBCA"
m = len(S1)
n = len(S2)
lcs_algo(S1, S2, m, n)Output:
ADBC
Time complexity: O(m*n)
Auxiliary space: O(m*n)
using System;
class let
{
// 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
for (int q = 0; q <= temp; q++)
Console.Write(lcs[q]);
}
// Driver program
public static void Main()
{
String X = "ACDABCD";
String Y = "ADBCA";
int m = X.Length;
int n = Y.Length;
lcs(X, Y, m, n);
}
}**Output**
ADBC
**Time Complexity: O(m*n)**
**Auxiliary Space: O(m*n)**
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