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# Shortest Uncommon Subsequence

• Difficulty Level : Hard
• Last Updated : 27 Jan, 2023

Given two strings S and T, find the length of the shortest subsequence in S which is not a subsequence in T. If no such subsequence is possible, return -1. A subsequence is a sequence that appears in the same relative order, but is not necessarily contiguous. A string of length n has 2^n different possible subsequences.

String S of length m (1 <= m <= 1000)
String T of length n (1 <= n <= 1000)

Examples:

Input : S = “babab” T = “babba”
Output : 3
Explanation: The subsequence “aab” of length 3 is present in S but not in T.

Input :  S = “abb” T = “abab”
Output : -1
Explanation: There is no subsequence that is present in S but not in T.

### Brute Force Method

A simple approach will be to generate all the subsequences of string S and for each subsequence check if it is present in string T or not. Consider example 2 in which S = “abb”,
its subsequences are “”, ”a”, ”b”, ”ab”, ”bb”, ”abb” each of which is present in “abab”. The time complexity of this solution will be exponential.

### Efficient (Dynamic Programming):

1. Optimal substructure: Consider two strings S and T of length m and n respectively & let the function to find the shortest uncommon subsequence be shortestSeq (char *S, char *T). For each character in S, if it is not present in T then that character is the answer itself.

Otherwise, if it is found at index k then we have the choice of either including it in the shortest uncommon subsequence or not.

```If it is included answer = 1 + ShortestSeq( S + 1, T + k + 1)
If not included answer =  ShortestSeq( S + 1, T)
The minimum of the two is the answer.```

Thus, we can see that this problem has optimal substructure properties as it can be solved by using solutions to subproblems.

2. Overlapping Subproblems: Following is a simple recursive implementation of the above problem.

## C++

 `// A simple recursive C++ program to find shortest` `// uncommon subsequence.` `#include` `using` `namespace` `std;`   `#define MAX 1005`   `/* A recursive function to find the length of` `   ``shortest uncommon subsequence*/` `int` `shortestSeq(``char` `*S, ``char` `*T, ``int` `m, ``int` `n)` `{` `    ``// S string is empty` `    ``if` `(m == 0)` `        ``return` `MAX;`   `    ``// T string is empty` `    ``if` `(n <= 0)` `        ``return` `1;`   `    ``// Loop to search for current character` `    ``int` `k;` `    ``for` `(k=0; k < n; k++)` `        ``if` `(T[k] == S)` `            ``break``;`   `    ``// char not found in T` `    ``if` `(k == n)` `        ``return` `1;`   `    ``// Return minimum of following two` `    ``// Not including current char in answer` `    ``// Including current char` `    ``return` `min(shortestSeq(S+1, T, m-1, n),` `            ``1 + shortestSeq(S+1, T+k+1, m-1, n-k-1));` `}`   `// Driver program to test the above function` `int` `main()` `{` `    ``char` `S[] = ``"babab"``;` `    ``char` `T[] = ``"babba"``;` `    ``int` `m = ``strlen``(S), n = ``strlen``(T);` `    ``int` `ans = shortestSeq(S, T, m, n);` `    ``if` `(ans >= MAX)` `       ``ans = -1;` `    ``cout << ``"Length of shortest subsequence is: "` `         ``<< ans << endl;` `    ``return` `0;` `}`

## Java

 `// A simple recursive Java program to find shortest` `// uncommon subsequence.` `import` `java.util.*;`   `class` `GFG` `{`   `static` `final` `int` `MAX = ``1005``;`   `/* A recursive function to find the length of` `shortest uncommon subsequence*/` `static` `int` `shortestSeq(``char` `[]S, ``char` `[]T, ``int` `m, ``int` `n)` `{` `    ``// S String is empty` `    ``if` `(m == ``0``)` `        ``return` `MAX;`   `    ``// T String is empty` `    ``if` `(n <= ``0``)` `        ``return` `1``;`   `    ``// Loop to search for current character` `    ``int` `k;` `    ``for` `(k = ``0``; k < n; k++)` `        ``if` `(T[k] == S[``0``])` `            ``break``;`   `    ``// char not found in T` `    ``if` `(k == n)` `        ``return` `1``;`   `    ``// Return minimum of following two` `    ``// Not including current char in answer` `    ``// Including current char` `    ``return` `Math.min(shortestSeq(Arrays.copyOfRange(S, ``1``, S.length), T, m - ``1``, n),` `                    ``1` `+ shortestSeq(Arrays.copyOfRange(S, ``1``, S.length), ` `                    ``Arrays.copyOfRange(T, k + ``1``, T.length), m - ``1``, n - k - ``1``));` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``char` `S[] = ``"babab"``.toCharArray();` `    ``char` `T[] = ``"babba"``.toCharArray();` `    ``int` `m = S.length, n = T.length;` `    ``int` `ans = shortestSeq(S, T, m, n);` `    ``if` `(ans >= MAX)` `    ``ans = -``1``;` `    ``System.out.print(``"Length of shortest subsequence is: "` `        ``+ ans +``"\n"``);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# A simple recursive Python ` `# program to find shortest ` `# uncommon subsequence. ` `MAX` `=` `1005`   `# A recursive function to ` `# find the length of shortest ` `# uncommon subsequence` `def` `shortestSeq(S, T, m, n):`   `    ``# S String is empty ` `    ``if` `m ``=``=` `0``:` `        ``return` `MAX`   `    ``# T String is empty ` `    ``if``(n <``=` `0``):` `        ``return` `1`   `    ``# Loop to search for ` `    ``# current character ` `    ``for` `k ``in` `range``(n):` `        ``if``(T[k] ``=``=` `S[``0``]):` `            ``break`   `    ``# char not found in T ` `    ``if``(k ``=``=` `n):` `        ``return` `1`   `    ``# Return minimum of following ` `    ``# two Not including current ` `    ``# char in answer Including ` `    ``# current char ` `    ``return` `min``(shortestSeq(S[``1` `: ], T, m ``-` `1``, n), ` `               ``1` `+` `shortestSeq((S[``1` `: ]), T[k ``+` `1` `: ], ` `                                ``m ``-` `1``, n ``-` `k ``-` `1``))`   `# Driver code ` `S ``=` `"babab"` `T ``=` `"babba"`   `m ``=` `len``(S)` `n ``=` `len``(T)` `ans ``=` `shortestSeq(S, T, m, n)` `if``(ans >``=` `MAX``):` `    ``ans ``=``-` `1` `print``(``"Length of shortest subsequence is:"``, ans)`   `# This code is contributed by avanitrachhadiya2155`

## C#

 `// A simple recursive C# program to find shortest` `// uncommon subsequence.` `using` `System;`   `class` `GFG` `{`   `static` `readonly` `int` `MAX = 1005;`   `/* A recursive function to find the length of` `shortest uncommon subsequence*/` `static` `int` `shortestSeq(``char` `[]S, ``char` `[]T, ``int` `m, ``int` `n)` `{` `    ``// S String is empty` `    ``if` `(m == 0)` `        ``return` `MAX;`   `    ``// T String is empty` `    ``if` `(n <= 0)` `        ``return` `1;`   `    ``// Loop to search for current character` `    ``int` `k;` `    ``for` `(k = 0; k < n; k++)` `        ``if` `(T[k] == S)` `            ``break``;`   `    ``// char not found in T` `    ``if` `(k == n)` `        ``return` `1;`   `    ``// Return minimum of following two` `    ``// Not including current char in answer` `    ``// Including current char` `    ``char` `[]St1 = ``new` `Char[S.Length - 1];` `    ``Array.Copy(S, 1, St1, 0, S.Length - 1);` `    ``char` `[]St2 = ``new` `Char[S.Length - 1];` `    ``Array.Copy(S, 1, St2, 0, S.Length - 1);` `    ``char` `[]Tt1 = ``new` `Char[T.Length - (k + 1)];` `    ``Array.Copy(T, k + 1, Tt1, 0, T.Length - (k + 1));` `    ``return` `Math.Min(shortestSeq(St1, T, m - 1, n),` `                    ``1 + shortestSeq(St2, Tt1, m - 1, n - k - 1));` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``char` `[]S = ``"babab"``.ToCharArray();` `    ``char` `[]T = ``"babba"``.ToCharArray();` `    ``int` `m = S.Length, n = T.Length;` `    ``int` `ans = shortestSeq(S, T, m, n);` `    ``if` `(ans >= MAX)` `    ``ans = -1;` `    ``Console.Write(``"Length of shortest subsequence is: "` `        ``+ ans +``"\n"``);` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`Length of shortest subsequence is: 3`

Time complexity: O(m*n),The time complexity of this algorithm is O(m*n), where m and n are the lengths of the two strings. This is because the algorithm uses a recursive approach to find the shortest uncommon subsequence which involves comparing each character of the two strings.

Auxiliary Space: O(m*n),The space complexity of this algorithm is also O(m*n) because we are using a two-dimensional array to store the result of the recursive call.

If we draw the complete recursion tree, then we can see that there are many subproblems that are solved again and again. So this problem has Overlapping Substructure property and re-computation of the same subproblems can be avoided by either using Memoization or Tabulation. The following is a tabulated implementation of the problem. ## C++

 `// A dynamic programming based C++ program` `// to find shortest uncommon subsequence.` `#include` `using` `namespace` `std;`   `#define MAX 1005`   `// Returns length of shortest common subsequence` `int` `shortestSeq(``char` `*S, ``char` `*T)` `{` `    ``int` `m = ``strlen``(S), n = ``strlen``(T);`   `    ``// declaring 2D array of m + 1 rows and` `    ``// n + 1 columns dynamically` `    ``int` `dp[m+1][n+1];`   `    ``// T string is empty` `    ``for` `(``int` `i = 0; i <= m; i++)` `        ``dp[i] = 1;`   `    ``// S string is empty` `    ``for` `(``int` `i = 0; i <= n; i++)` `        ``dp[i] = MAX;`   `    ``for` `(``int` `i = 1; i <= m; i++)` `    ``{` `        ``for` `(``int` `j = 1; j <= n; j++)` `        ``{` `            ``char` `ch = S[i-1];` `            ``int` `k;` `            ``for` `(k = j-1; k >= 0; k--)` `                ``if` `(T[k] == ch)` `                    ``break``;`   `            ``// char not present in T` `            ``if` `(k == -1)` `                ``dp[i][j] = 1;` `            ``else` `               ``dp[i][j] = min(dp[i-1][j], dp[i-1][k] + 1);` `        ``}` `    ``}`   `    ``int` `ans = dp[m][n];` `    ``if` `(ans >= MAX)` `        ``ans = -1;`   `    ``return` `ans;` `}`   `// Driver program to test the above function` `int` `main()` `{` `    ``char` `S[] = ``"babab"``;` `    ``char` `T[] = ``"babba"``;` `    ``int` `m = ``strlen``(S), n = ``strlen``(T);` `    ``cout << ``"Length of shortest subsequence is : "` `         ``<< shortestSeq(S, T) << endl;` `}`

## Java

 `// A dynamic programming based Java program ` `// to find shortest uncommon subsequence. ` `import` `java.io.*;` `class` `GFG ` `{`   `    ``static` `final` `int` `MAX = ``1005``;`   `    ``// Returns length of shortest common subsequence ` `    ``static` `int` `shortestSeq(``char``[] S, ``char``[] T) ` `    ``{` `        ``int` `m = S.length, n = T.length;`   `        ``// declaring 2D array of m + 1 rows and ` `        ``// n + 1 columns dynamically ` `        ``int` `dp[][] = ``new` `int``[m + ``1``][n + ``1``];`   `        ``// T string is empty ` `        ``for` `(``int` `i = ``0``; i <= m; i++) ` `        ``{` `            ``dp[i][``0``] = ``1``;` `        ``}`   `        ``// S string is empty ` `        ``for` `(``int` `i = ``0``; i <= n; i++)` `        ``{` `            ``dp[``0``][i] = MAX;` `        ``}`   `        ``for` `(``int` `i = ``1``; i <= m; i++) ` `        ``{` `            ``for` `(``int` `j = ``1``; j <= n; j++) ` `            ``{` `                ``char` `ch = S[i - ``1``];` `                ``int` `k;` `                ``for` `(k = j - ``1``; k >= ``0``; k--)` `                ``{` `                    ``if` `(T[k] == ch)` `                    ``{` `                        ``break``;` `                    ``}` `                ``}`   `                ``// char not present in T ` `                ``if` `(k == -``1``) ` `                ``{` `                    ``dp[i][j] = ``1``;` `                ``} ` `                ``else` `                ``{` `                    ``dp[i][j] = Math.min(dp[i - ``1``][j],` `                                    ``dp[i - ``1``][k] + ``1``);` `                ``}` `            ``}` `        ``}`   `        ``int` `ans = dp[m][n];` `        ``if` `(ans >= MAX) ` `        ``{` `            ``ans = -``1``;` `        ``}` `        ``return` `ans;` `    ``}`   `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``char` `S[] = ``"babab"``.toCharArray();` `        ``char` `T[] = ``"babba"``.toCharArray();` `        ``int` `m = S.length, n = T.length;` `        ``System.out.println(``"Length of shortest"` `+` `                            ``"subsequence is : "` `+ ` `                            ``shortestSeq(S, T));` `    ``}` `} `   `// This code is contributed by 29AjayKumar`

## Python3

 `# A dynamic programming based Python program` `# to find shortest uncommon subsequence.` `MAX` `=` `1005`   `# Returns length of shortest common subsequence` `def` `shortestSeq(S: ``list``, T: ``list``):` `    ``m ``=` `len``(S)` `    ``n ``=` `len``(T)`   `    ``# declaring 2D array of m + 1 rows and` `    ``# n + 1 columns dynamically` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n ``+` `1``)] ` `             ``for` `j ``in` `range``(m ``+` `1``)]`   `    ``# T string is empty` `    ``for` `i ``in` `range``(m ``+` `1``):` `        ``dp[i][``0``] ``=` `1`   `    ``# S string is empty` `    ``for` `i ``in` `range``(n ``+` `1``):` `        ``dp[``0``][i] ``=` `MAX`   `    ``for` `i ``in` `range``(``1``, m ``+` `1``):` `        ``for` `j ``in` `range``(``1``, n ``+` `1``):` `            ``ch ``=` `S[i ``-` `1``]` `            ``k ``=` `j ``-` `1` `            ``while` `k >``=` `0``:` `                ``if` `T[k] ``=``=` `ch:` `                    ``break` `                ``k ``-``=` `1`   `            ``# char not present in T` `            ``if` `k ``=``=` `-``1``:` `                ``dp[i][j] ``=` `1` `            ``else``:` `                ``dp[i][j] ``=` `min``(dp[i ``-` `1``][j], ` `                               ``dp[i ``-` `1``][k] ``+` `1``)`   `    ``ans ``=` `dp[m][n]` `    ``if` `ans >``=` `MAX``:` `        ``ans ``=` `-``1`   `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``S ``=` `"babab"` `    ``T ``=` `"babba"`   `    ``print``(``"Length of shortest subsequence is:"``, ` `                             ``shortestSeq(S, T))`   `# This code is contributed by` `# sanjeev2552`

## C#

 `// A dynamic programming based C# program ` `// to find shortest uncommon subsequence.` `using` `System;`   `class` `GFG ` `{`   `    ``static` `readonly` `int` `MAX = 1005;`   `    ``// Returns length of shortest common subsequence ` `    ``static` `int` `shortestSeq(``char``[] S, ``char``[] T) ` `    ``{` `        ``int` `m = S.Length, n = T.Length;`   `        ``// declaring 2D array of m + 1 rows and ` `        ``// n + 1 columns dynamically ` `        ``int` `[,]dp = ``new` `int``[m + 1, n + 1];`   `        ``// T string is empty ` `        ``for` `(``int` `i = 0; i <= m; i++) ` `        ``{` `            ``dp[i, 0] = 1;` `        ``}`   `        ``// S string is empty ` `        ``for` `(``int` `i = 0; i <= n; i++)` `        ``{` `            ``dp[0, i] = MAX;` `        ``}`   `        ``for` `(``int` `i = 1; i <= m; i++) ` `        ``{` `            ``for` `(``int` `j = 1; j <= n; j++) ` `            ``{` `                ``char` `ch = S[i - 1];` `                ``int` `k;` `                ``for` `(k = j - 1; k >= 0; k--)` `                ``{` `                    ``if` `(T[k] == ch)` `                    ``{` `                        ``break``;` `                    ``}` `                ``}`   `                ``// char not present in T ` `                ``if` `(k == -1) ` `                ``{` `                    ``dp[i, j] = 1;` `                ``} ` `                ``else` `                ``{` `                    ``dp[i, j] = Math.Min(dp[i - 1, j],` `                                    ``dp[i - 1, k] + 1);` `                ``}` `            ``}` `        ``}`   `        ``int` `ans = dp[m, n];` `        ``if` `(ans >= MAX) ` `        ``{` `            ``ans = -1;` `        ``}` `        ``return` `ans;` `    ``}`   `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{` `        ``char` `[]S = ``"babab"``.ToCharArray();` `        ``char` `[]T = ``"babba"``.ToCharArray();` `        ``int` `m = S.Length, n = T.Length;` `        ``Console.WriteLine(``"Length of shortest"` `+` `                            ``"subsequence is : "` `+ ` `                            ``shortestSeq(S, T));` `    ``}` `}`   `// This code contributed by Rajput-Ji`

## PHP

 `= 0; ``\$k``--) ` `                ``if` `(``\$T``[``\$k``] == ``\$ch``) ` `                    ``break``; `   `            ``// char not present in T ` `            ``if` `(``\$k` `== -1) ` `                ``\$dp``[``\$i``][``\$j``] = 1; ` `            ``else` `            ``\$dp``[``\$i``][``\$j``] = min(``\$dp``[``\$i` `- 1][``\$j``],  ` `                              ``\$dp``[``\$i` `- 1][``\$k``] + 1); ` `        ``} ` `    ``} `   `    ``\$ans` `= ``\$dp``[``\$m``][``\$n``]; ` `    ``if` `(``\$ans` `>= ``\$GLOBALS``[``'MAX'``]) ` `        ``\$ans` `= -1; `   `    ``return` `\$ans``; ` `} `   `// Driver Code` `\$S` `= ``"babab"``; ` `\$T` `= ``"babba"``; ` `\$m` `= ``strlen``(``\$S``);` `\$n` `= ``strlen``(``\$T``); ` `echo` `"Length of shortest subsequence is : "``, ` `                        ``shortestSeq(``\$S``, ``\$T``); `   `// This code is contributed by Ryuga` `?>`

## Javascript

 ``

Output

`Length of shortest subsequence is : 3`

Time Complexity: O(m*n)
Auxiliary Space: O(m*n)