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<bits/stdc++.h>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[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 program to test the above functionint 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 ofshortest 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 codepublic 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 subsequencedef 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 =- 1print("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 ofshortest 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 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 codepublic 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
<script>// JavaScript code to implement the approachconst MAX = 1005// A recursive function to// find the length of shortest// uncommon subsequencefunction shortestSeq(S, T, m, n){ // S String is empty if(m == 0) return MAX // T String is empty if(n <= 0) return 1 let k; // Loop to search for // current character 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(S.substring(1,), T, m - 1, n), 1 + shortestSeq((S.substring(1,)), T.substring(k + 1,), m - 1, n - k - 1))}// Driver codelet S = "babab"let T = "babba"let m = S.lengthlet n = T.lengthlet ans = shortestSeq(S, T, m, n)if(ans >= MAX) ans =- 1document.write("Length of shortest subsequence is: "+ ans,"</br>")// This code is contributed by shinjanpatra</script> |
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<bits/stdc++.h>using namespace std;#define MAX 1005// Returns length of shortest common subsequenceint 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][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] = 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 functionint 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 subsequencedef 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 Codeif __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
<?php// A dynamic programming based PHP program // to find shortest uncommon subsequence. $GLOBALS['MAX'] = 1005;// Returns length of shortest// common subsequence function shortestSeq($S, $T) { $m = strlen($S); $n = strlen($T); // declaring 2D array of m + 1 rows // and n + 1 columns dynamically $dp = array(array()); // T string is empty for ($i = 0; $i <= $m; $i++) $dp[$i][0] = 1; // S string is empty for ($i = 0; $i <= $n; $i++) $dp[0][$i] = $GLOBALS['MAX']; for ($i = 1; $i <= $m; $i++) { for ($j = 1; $j <= $n; $j++) { $ch = $S[$i - 1]; 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); } } $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
<script>// A dynamic programming based JavaScript program// to find shortest uncommon subsequence.const MAX = 1005// Returns length of shortest common subsequencefunction shortestSeq(S, T){ let m = S.length, n = T.length; // declaring 2D array of m + 1 rows and // n + 1 columns dynamically let dp = new Array(m+1).fill(0).map(()=>new Array(n+1).fill(0)); // T string is empty for (let i = 0; i <= m; i++) dp[i][0] = 1; // S string is empty for (let i = 0; i <= n; i++) dp[0][i] = MAX; for (let i = 1; i <= m; i++) { for (let j = 1; j <= n; j++) { let ch = S[i-1]; let 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); } } let ans = dp[m][n]; if (ans >= MAX) ans = -1; return ans;}// Driver program to test the above functionlet S = "babab";let T = "babba";let m = S.length, n = T.length;document.write("Length of shortest subsequence is : "+shortestSeq(S, T),"</br>");// This code is contributed by shinjanpatra</script> |
Output
Length of shortest subsequence is : 3
Time Complexity: O(m*n)
Auxiliary Space: O(m*n)
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