给定两个字符串str1 和 str2，任务是找到同时包含 str1 和 str2 作为子序列的最短字符串的长度。
例子 ：
Input: str1 = "geek", str2 = "eke"
Output: 5
Explanation:
String "geeke" has both string "geek"
and "eke" as subsequences.
Input: str1 = "AGGTAB", str2 = "GXTXAYB"
Output: 9
Explanation:
String "AGXGTXAYB" has both string
"AGGTAB" and "GXTXAYB" as subsequences.这个问题与最长公共子序列问题密切相关。下面是步骤。
1)找出两个给定字符串的最长公共子序列 (lcs)。例如，“geek”和“eke”的lcs就是“ek”。
2) 将非 lcs字符(按字符串中的原始顺序)插入到上面找到的 lcs 中，并返回结果。所以“ek”变成了“geeke”，这是最短的公共超序列。
让我们考虑另一个例子，str1 = “AGGTAB”和 str2 = “GXTXAYB”。 str1 和 str2 的 LCS 是“GTAB”。一旦找到 LCS，我们按顺序插入两个字符串的字符，我们得到“AGXGTXAYB”
这是如何运作的?
我们需要找到一个字符串，它具有两个字符串作为子序列，并且是最短的字符串。如果两个字符串的所有字符不同，则结果是两个给定字符串的长度之和。如果有公共字符，那么我们不希望它们多次出现，因为任务是最小化长度。因此，我们首先找到最长的公共子序列，取该子序列出现的一次并添加额外的字符。
Length of the shortest supersequence
= (Sum of lengths of given two strings)
- (Length of LCS of two given strings) 下面是上述想法的实现。下面的实现只找到最短超序列的长度。
C++
// C++ program to find length of the
// shortest supersequence
#include
using namespace std;
// Utility function to get max
// of 2 integers
int max(int a, int b) { return (a > b) ? a : b; }
// Returns length of LCS for
// X[0..m - 1], Y[0..n - 1]
int lcs(char* X, char* Y, int m, int n);
// Function to find length of the
// shortest supersequence of X and Y.
int shortestSuperSequence(char* X, char* Y)
{
int m = strlen(X), n = strlen(Y);
// find lcs
int l = lcs(X, Y, m, n);
// Result is sum of input string
// lengths - length of lcs
return (m + n - l);
}
// 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];
}
// Driver code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
cout << "Length of the shortest supersequence is "
<< shortestSuperSequence(X, Y) << endl;
return 0;
}
// This code is contributed by Akanksha Rai C
// C program to find length of
// the shortest supersequence
#include
#include
// Utility function to get
// max of 2 integers
int max(int a, int b) { return (a > b) ? a : b; }
// Returns length of LCS for
// X[0..m - 1], Y[0..n - 1]
int lcs(char* X, char* Y, int m, int n);
// Function to find length of the
// shortest supersequence of X and Y.
int shortestSuperSequence(char* X, char* Y)
{
int m = strlen(X), n = strlen(Y);
// find lcs
int l = lcs(X, Y, m, n);
// Result is sum of input string
// lengths - length of lcs
return (m + n - l);
}
// 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];
}
// Driver code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
printf("Length of the shortest supersequence is %d\n",
shortestSuperSequence(X, Y));
return 0;
} Java
// Java program to find length of
// the shortest supersequence
class GFG {
// Function to find length of the
// shortest supersequence of X and Y.
static int shortestSuperSequence(String X, String Y)
{
int m = X.length();
int n = Y.length();
// find lcs
int l = lcs(X, Y, m, n);
// Result is sum of input string
// lengths - length of lcs
return (m + n - l);
}
// Returns length of LCS
// for X[0..m - 1], Y[0..n - 1]
static int lcs(String X, String Y, int m, int n)
{
int[][] L = new int[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.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]);
}
}
// L[m][n] contains length of LCS
// for X[0..n - 1] and Y[0..m - 1]
return L[m][n];
}
// Driver code
public static void main(String args[])
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println("Length of the shortest "
+ "supersequence is "
+ shortestSuperSequence(X, Y));
}
}
// This article is contributed by Sumit GhoshPython
# Python program to find length
# of the shortest supersequence
# Function to find length of the
# shortest supersequence of X and Y.
def shortestSuperSequence(X, Y):
m = len(X)
n = len(Y)
l = lcs(X, Y, m, n)
# Result is sum of input string
# lengths - length of lcs
return (m + n - l)
# Returns length of LCS for
# X[0..m - 1], Y[0..n - 1]
def lcs(X, Y, m, n):
L = [[0] * (n + 2) for i in
range(m + 2)]
# 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 range(m + 1):
for j in range(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])
# L[m][n] contains length of
# LCS for X[0..n - 1] and Y[0..m - 1]
return L[m][n]
# Driver code
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of the shortest supersequence is %d"
% shortestSuperSequence(X, Y))
# This code is contributed by Ansu KumariC#
// C# program to find length of
// the shortest supersequence
using System;
class GFG {
// Function to find length of the
// shortest supersequence of X and Y.
static int shortestSuperSequence(String X, String Y)
{
int m = X.Length;
int n = Y.Length;
// find lcs
int l = lcs(X, Y, m, n);
// Result is sum of input string
// lengths - length of lcs
return (m + n - l);
}
// Returns length of LCS for
// X[0..m - 1], Y[0..n - 1]
static int lcs(String X, String Y, int m, int n)
{
int[, ] L = new int[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] = Math.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];
}
// Driver code
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine("Length of the shortest"
+ "supersequence is "
+ shortestSuperSequence(X, Y));
}
}
// This code is contributed by Sam007PHP
Javascript
C++
// A Naive recursive C++ program to find
// length of the shortest supersequence
#include
using namespace std;
int superSeq(char* X, char* Y, int m, int n)
{
if (!m)
return n;
if (!n)
return m;
if (X[m - 1] == Y[n - 1])
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver Code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
cout << "Length of the shortest supersequence is "
<< superSeq(X, Y, strlen(X), strlen(Y));
return 0;
} Java
// A Naive recursive Java program to find
// length of the shortest supersequence
class GFG {
static int superSeq(String X, String Y, int m, int n)
{
if (m == 0)
return n;
if (n == 0)
return m;
if (X.charAt(m - 1) == Y.charAt(n - 1))
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ Math.min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver code
public static void main(String args[])
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println(
"Length of the shortest"
+ "supersequence is: "
+ superSeq(X, Y, X.length(), Y.length()));
}
}
// This article is contributed by Sumit GhoshPython
# A Naive recursive python program to find
# length of the shortest supersequence
def superSeq(X, Y, m, n):
if (not m):
return n
if (not n):
return m
if (X[m - 1] == Y[n - 1]):
return 1 + superSeq(X, Y, m - 1, n - 1)
return 1 + min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1))
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of the shortest supersequence is %d"
% superSeq(X, Y, len(X), len(Y)))
# This code is contributed by Ansu KumariC#
// A Naive recursive C# program to find
// length of the shortest supersequence
using System;
class GFG {
static int superSeq(String X, String Y, int m, int n)
{
if (m == 0)
return n;
if (n == 0)
return m;
if (X[m - 1] == Y[n - 1])
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ Math.Min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver Code
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine(
"Length of the shortest supersequence is: "
+ superSeq(X, Y, X.Length, Y.Length));
}
}
// This code is contributed by Sam007PHP
Javascript
C++
// A dynamic programming based C program to
// find length of the shortest supersequence
#include
using namespace std;
// Returns length of the shortest
// supersequence of X and Y
int superSeq(char* X, char* Y, int m, int n)
{
int dp[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (!i)
dp[i][j] = j;
else if (!j)
dp[i][j] = i;
else if (X[i - 1] == Y[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j]
= 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m][n];
}
// Driver Code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
cout << "Length of the shortest supersequence is "
<< superSeq(X, Y, strlen(X), strlen(Y));
return 0;
} Java
// A dynamic programming based Java program to
// find length of the shortest supersequence
class GFG {
// Returns length of the shortest
// supersequence of X and Y
static int superSeq(String X, String Y, int m, int n)
{
int[][] dp = new int[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (i == 0)
dp[i][j] = j;
else if (j == 0)
dp[i][j] = i;
else if (X.charAt(i - 1) == Y.charAt(j - 1))
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j] = 1
+ Math.min(dp[i - 1][j],
dp[i][j - 1]);
}
}
return dp[m][n];
}
// Driver Code
public static void main(String args[])
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println(
"Length of the shortest supersequence is "
+ superSeq(X, Y, X.length(), Y.length()));
}
}
// This article is contributed by Sumit GhoshPython
# A dynamic programming based python program
# to find length of the shortest supersequence
# Returns length of the shortest supersequence of X and Y
def superSeq(X, Y, m, n):
dp = [[0] * (n + 2) for i in range(m + 2)]
# Fill table in bottom up manner
for i in range(m + 1):
for j in range(n + 1):
# Below steps follow above recurrence
if (not i):
dp[i][j] = j
elif (not j):
dp[i][j] = i
elif (X[i - 1] == Y[j - 1]):
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j],
dp[i][j - 1])
return dp[m][n]
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of the shortest supersequence is %d"
% superSeq(X, Y, len(X), len(Y)))
# This code is contributed by Ansu KumariC#
// A dynamic programming based C# program to
// find length of the shortest supersequence
using System;
class GFG {
// Returns length of the shortest
// supersequence of X and Y
static int superSeq(String X, String Y, int m, int n)
{
int[, ] dp = new int[m + 1, n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (i == 0)
dp[i, j] = j;
else if (j == 0)
dp[i, j] = i;
else if (X[i - 1] == Y[j - 1])
dp[i, j] = 1 + dp[i - 1, j - 1];
else
dp[i, j] = 1
+ Math.Min(dp[i - 1, j],
dp[i, j - 1]);
}
}
return dp[m, n];
}
// Driver code
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine(
"Length of the shortest supersequence is "
+ superSeq(X, Y, X.Length, Y.Length));
}
}
// This code is contributed by Sam007PHP
Javascript
Python3
# A dynamic programming based python program
# to find length of the shortest supersequence
# Returns length of the
# shortest supersequence of X and Y
import numpy as np
def superSeq(X,Y,n,m,lookup):
if m==0 or n==0:
lookup[n][m] = n+m
if (lookup[n][m] == 0):
if X[n-1]==Y[m-1]:
lookup[n][m] = superSeq(X,Y,n-1,m-1,lookup)+1
else:
lookup[n][m] = min(superSeq(X,Y,n-1,m,lookup)+1,
superSeq(X,Y,n,m-1,lookup)+1)
return lookup[n][m]
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
lookup = np.zeros([len(X)+1,len(Y)+1])
print("Length of the shortest supersequence is {}"
.format(superSeq(X,Y,len(X),len(Y),lookup)))
# This code is contributed by Tanmay Ambadkar输出：
Length of the shortest supersequence is 9下面是解决上述问题的另一种方法。
一个简单的分析产生以下简单的递归解决方案。
Let X[0..m - 1] and Y[0..n - 1] be two
strings and m and n be respective
lengths.
if (m == 0) return n;
if (n == 0) return m;
// If last characters are same, then
// add 1 to result and
// recur for X[]
if (X[m - 1] == Y[n - 1])
return 1 + SCS(X, Y, m - 1, n - 1);
// Else find shortest of following two
// a) Remove last character from X and recur
// b) Remove last character from Y and recur
else
return 1 + min( SCS(X, Y, m - 1, n), SCS(X, Y, m, n - 1) );下面是基于上述递归公式的简单朴素递归解决方案。
C++
// A Naive recursive C++ program to find
// length of the shortest supersequence
#include
using namespace std;
int superSeq(char* X, char* Y, int m, int n)
{
if (!m)
return n;
if (!n)
return m;
if (X[m - 1] == Y[n - 1])
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver Code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
cout << "Length of the shortest supersequence is "
<< superSeq(X, Y, strlen(X), strlen(Y));
return 0;
}
Java
// A Naive recursive Java program to find
// length of the shortest supersequence
class GFG {
static int superSeq(String X, String Y, int m, int n)
{
if (m == 0)
return n;
if (n == 0)
return m;
if (X.charAt(m - 1) == Y.charAt(n - 1))
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ Math.min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver code
public static void main(String args[])
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println(
"Length of the shortest"
+ "supersequence is: "
+ superSeq(X, Y, X.length(), Y.length()));
}
}
// This article is contributed by Sumit Ghosh
Python
# A Naive recursive python program to find
# length of the shortest supersequence
def superSeq(X, Y, m, n):
if (not m):
return n
if (not n):
return m
if (X[m - 1] == Y[n - 1]):
return 1 + superSeq(X, Y, m - 1, n - 1)
return 1 + min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1))
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of the shortest supersequence is %d"
% superSeq(X, Y, len(X), len(Y)))
# This code is contributed by Ansu Kumari
C＃
// A Naive recursive C# program to find
// length of the shortest supersequence
using System;
class GFG {
static int superSeq(String X, String Y, int m, int n)
{
if (m == 0)
return n;
if (n == 0)
return m;
if (X[m - 1] == Y[n - 1])
return 1 + superSeq(X, Y, m - 1, n - 1);
return 1
+ Math.Min(superSeq(X, Y, m - 1, n),
superSeq(X, Y, m, n - 1));
}
// Driver Code
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine(
"Length of the shortest supersequence is: "
+ superSeq(X, Y, X.Length, Y.Length));
}
}
// This code is contributed by Sam007
PHP
Javascript
输出：
Length of the shortest supersequence is 9上述解的时间复杂度指数为 O(2 min(m, n) )。由于存在重叠子问题，我们可以使用动态规划有效地解决这个递归问题。下面是基于动态规划的实现。该解决方案的时间复杂度为 O(mn)。
C++
// A dynamic programming based C program to
// find length of the shortest supersequence
#include
using namespace std;
// Returns length of the shortest
// supersequence of X and Y
int superSeq(char* X, char* Y, int m, int n)
{
int dp[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (!i)
dp[i][j] = j;
else if (!j)
dp[i][j] = i;
else if (X[i - 1] == Y[j - 1])
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j]
= 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m][n];
}
// Driver Code
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
cout << "Length of the shortest supersequence is "
<< superSeq(X, Y, strlen(X), strlen(Y));
return 0;
}
Java
// A dynamic programming based Java program to
// find length of the shortest supersequence
class GFG {
// Returns length of the shortest
// supersequence of X and Y
static int superSeq(String X, String Y, int m, int n)
{
int[][] dp = new int[m + 1][n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (i == 0)
dp[i][j] = j;
else if (j == 0)
dp[i][j] = i;
else if (X.charAt(i - 1) == Y.charAt(j - 1))
dp[i][j] = 1 + dp[i - 1][j - 1];
else
dp[i][j] = 1
+ Math.min(dp[i - 1][j],
dp[i][j - 1]);
}
}
return dp[m][n];
}
// Driver Code
public static void main(String args[])
{
String X = "AGGTAB";
String Y = "GXTXAYB";
System.out.println(
"Length of the shortest supersequence is "
+ superSeq(X, Y, X.length(), Y.length()));
}
}
// This article is contributed by Sumit Ghosh
Python
# A dynamic programming based python program
# to find length of the shortest supersequence
# Returns length of the shortest supersequence of X and Y
def superSeq(X, Y, m, n):
dp = [[0] * (n + 2) for i in range(m + 2)]
# Fill table in bottom up manner
for i in range(m + 1):
for j in range(n + 1):
# Below steps follow above recurrence
if (not i):
dp[i][j] = j
elif (not j):
dp[i][j] = i
elif (X[i - 1] == Y[j - 1]):
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j],
dp[i][j - 1])
return dp[m][n]
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
print("Length of the shortest supersequence is %d"
% superSeq(X, Y, len(X), len(Y)))
# This code is contributed by Ansu Kumari
C＃
// A dynamic programming based C# program to
// find length of the shortest supersequence
using System;
class GFG {
// Returns length of the shortest
// supersequence of X and Y
static int superSeq(String X, String Y, int m, int n)
{
int[, ] dp = new int[m + 1, n + 1];
// Fill table in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// Below steps follow above recurrence
if (i == 0)
dp[i, j] = j;
else if (j == 0)
dp[i, j] = i;
else if (X[i - 1] == Y[j - 1])
dp[i, j] = 1 + dp[i - 1, j - 1];
else
dp[i, j] = 1
+ Math.Min(dp[i - 1, j],
dp[i, j - 1]);
}
}
return dp[m, n];
}
// Driver code
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
Console.WriteLine(
"Length of the shortest supersequence is "
+ superSeq(X, Y, X.Length, Y.Length));
}
}
// This code is contributed by Sam007
PHP
Javascript
输出：
Length of the shortest supersequence is 9感谢 Gaurav Ahirwar 提出这个解决方案。
自上而下的记忆方法：
这个想法是遵循简单的递归解决方案，使用查找表来避免重新计算。在计算输入的结果之前，我们检查结果是否已经计算出来。如果已经计算，我们返回该结果。
蟒蛇3
# A dynamic programming based python program
# to find length of the shortest supersequence
# Returns length of the
# shortest supersequence of X and Y
import numpy as np
def superSeq(X,Y,n,m,lookup):
if m==0 or n==0:
lookup[n][m] = n+m
if (lookup[n][m] == 0):
if X[n-1]==Y[m-1]:
lookup[n][m] = superSeq(X,Y,n-1,m-1,lookup)+1
else:
lookup[n][m] = min(superSeq(X,Y,n-1,m,lookup)+1,
superSeq(X,Y,n,m-1,lookup)+1)
return lookup[n][m]
# Driver Code
X = "AGGTAB"
Y = "GXTXAYB"
lookup = np.zeros([len(X)+1,len(Y)+1])
print("Length of the shortest supersequence is {}"
.format(superSeq(X,Y,len(X),len(Y),lookup)))
# This code is contributed by Tanmay Ambadkar
输出：
Length of the shortest supersequence is 9.0如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。