Given a string S and Q queries, each query contains a string T. The task is to print “Yes” if T is a subsequence of S, else print “No”.
Examples:
Input : S = "neveropen" Query 1: "gg" Query 2: "gro" Query 3: "gfg" Query 4: "orf" Output : Yes No Yes No
For each query, using the brute force, start iterating over S looking for the first character of T. As soon as, the first character is found, continue to iterate S now looking for the second character of T and so on (Refer this for details). If manage to find all the character of T, print “Yes”, else “No”. Time complexity is O(Q*N), N is the length of S.
The efficient approach can be if we know the position of the next character of T in S. Then simply skip all the character between current and position of the next character and jump to that position. This can be done by making |S| x 26 size matrix and storing the next position of each character from every position of S.
Below is the implementation of the above idea :
C++
// C++ program to answer subsequence queries for a // given string. #include <bits/stdc++.h> #define MAX 10000 #define CHAR_SIZE 26 using namespace std; // Precompute the position of each character from // each position of String S void precompute(int mat[MAX][CHAR_SIZE], char str[], int len) { for (int i = 0; i < CHAR_SIZE; ++i) mat[len][i] = len; // Computing position of each character from // each position of String S for (int i = len-1; i >= 0; --i) { for (int j = 0; j < CHAR_SIZE; ++j) mat[i][j] = mat[i+1][j]; mat[i][str[i]-'a'] = i; } } // Print "Yes" if T is subsequence of S, else "No" bool query(int mat[MAX][CHAR_SIZE], const char *str, int len) { int pos = 0; // Traversing the string T for (int i = 0; i < strlen(str); ++i) { // If next position is greater than // length of S set flag to false. if (mat[pos][str[i] - 'a'] >= len) return false; // Setting position of next character else pos = mat[pos][str[i] - 'a'] + 1; } return true; } // Driven Program int main() { char S[]= "neveropen"; int len = strlen(S); int mat[MAX][CHAR_SIZE]; precompute(mat, S, len); query(mat, "gg", len)? cout << "Yes\n" : cout << "No\n"; query(mat, "gro", len)? cout << "Yes\n" : cout << "No\n"; query(mat, "gfg", len)? cout << "Yes\n" : cout << "No\n"; query(mat, "orf", len)? cout << "Yes\n" : cout << "No\n"; return 0; } |
Java
// Java program to answer subsequence queries for // a given string. public class Query_Subsequence { static final int MAX = 10000; static final int CHAR_SIZE = 26; // Precompute the position of each character from // each position of String S static void precompute(int mat[][], String str, int len) { for (int i = 0; i < CHAR_SIZE; ++i) mat[len][i] = len; // Computing position of each character from // each position of String S for (int i = len-1; i >= 0; --i) { for (int j = 0; j < CHAR_SIZE; ++j) mat[i][j] = mat[i+1][j]; mat[i][str.charAt(i)-'a'] = i; } } // Print "Yes" if T is subsequence of S, else "No" static boolean query(int mat[][], String str, int len) { int pos = 0; // Traversing the string T for (int i = 0; i < str.length(); ++i) { // If next position is greater than // length of S set flag to false. if (mat[pos][str.charAt(i) - 'a'] >= len) return false; // Setting position of next character else pos = mat[pos][str.charAt(i) - 'a'] + 1; } return true; } // Driven Program public static void main(String args[]) { String S= "neveropen"; int len = S.length(); int[][] mat = new int[MAX][CHAR_SIZE]; precompute(mat, S, len); String get = query(mat, "gg", len)? "Yes" :"No"; System.out.println(get); get = query(mat, "gro", len)? "Yes" :"No"; System.out.println(get); get = query(mat, "gfg", len)? "Yes" :"No"; System.out.println(get); get = query(mat, "orf", len)? "Yes" :"No"; System.out.println(get); } } // This code is contributed by Sumit Ghosh |
Python3
# Python3 program to answer # subsequence queries for # a given string. MAX = 10000CHAR_SIZE = 26 # Precompute the position of # each character from # each position of String S def precompute(mat, str, Len): for i in range(CHAR_SIZE): mat[Len][i] = Len # Computing position of each # character from each position # of String S for i in range(Len - 1, -1, -1): for j in range(CHAR_SIZE): mat[i][j] = mat[i + 1][j] mat[i][ord(str[i]) - ord('a')] = i # Print "Yes" if T is # subsequence of S, else "No" def query(mat, str, Len): pos = 0 # Traversing the string T for i in range(len(str)): # If next position is greater than # length of S set flag to false. if(mat[pos][ord(str[i]) - ord('a')] >= Len): return False # Setting position of next character else: pos = mat[pos][ord(str[i]) - ord('a')] + 1 return True # Driven code S = "neveropen"Len = len(S) mat = [[0 for i in range(CHAR_SIZE)] for j in range(MAX)] precompute(mat, S, Len) get = "No"if(query(mat, "gg", Len)): get = "Yes"print(get) get = "No"if(query(mat, "gro", Len)): get = "Yes"print(get) get = "No"if(query(mat, "gfg", Len)): get = "Yes"print(get) get = "No"if(query(mat, "orf", Len)): get = "Yes"print(get) # This code is contributed by avanitrachhadiya2155 |
C#
// C# program to answer subsequence // queries for a given string using System; public class Query_Subsequence { static int MAX = 10000; static int CHAR_SIZE = 26; // Precompute the position of each // character from each position // of String S static void precompute(int [,]mat, string str, int len) { for (int i = 0; i < CHAR_SIZE; ++i) mat[len, i] = len; // Computing position of each // character from each position // of String S for (int i = len - 1; i >= 0; --i) { for (int j = 0; j < CHAR_SIZE; ++j) mat[i, j] = mat[i + 1, j]; mat[i, str[i] - 'a'] = i; } } // Print "Yes" if T is subsequence // of S, else "No" static bool query(int [,]mat, string str, int len) { int pos = 0; // Traversing the string T for (int i = 0; i < str.Length; ++i) { // If next position is greater than // length of S set flag to false. if (mat[pos,str[i] - 'a'] >= len) return false; // Setting position of next character else pos = mat[pos,str[i] - 'a'] + 1; } return true; } // Driver Code public static void Main() { string S= "neveropen"; int len = S.Length; int[,] mat = new int[MAX,CHAR_SIZE]; precompute(mat, S, len); string get = query(mat, "gg", len)? "Yes" :"No"; Console.WriteLine(get); get = query(mat, "gro", len)? "Yes" :"No"; Console.WriteLine(get); get = query(mat, "gfg", len)? "Yes" :"No"; Console.WriteLine(get); get = query(mat, "orf", len)? "Yes" :"No"; Console.WriteLine(get); } } // This code is contributed by vt_m. |
Javascript
<script> // Javascript program to answer subsequence queries for // a given string. let MAX = 10000; let CHAR_SIZE = 26; // Precompute the position of each character from // each position of String S function precompute(mat, str, len) { for (let i = 0; i < CHAR_SIZE; ++i) mat[len][i] = len; // Computing position of each character from // each position of String S for (let i = len-1; i >= 0; --i) { for (let j = 0; j < CHAR_SIZE; ++j) mat[i][j] = mat[i+1][j]; mat[i][str[i].charCodeAt()-'a'.charCodeAt()] = i; } } // Print "Yes" if T is subsequence of S, else "No" function query(mat, str, len) { let pos = 0; // Traversing the string T for (let i = 0; i < str.length; ++i) { // If next position is greater than // length of S set flag to false. if (mat[pos][str[i].charCodeAt() - 'a'.charCodeAt()] >= len) return false; // Setting position of next character else pos = mat[pos][str[i].charCodeAt() - 'a'.charCodeAt()] + 1; } return true; } let S= "neveropen"; let len = S.length; let mat = new Array(MAX); for(let i = 0; i < MAX; i++) { mat[i] = new Array(CHAR_SIZE); for(let j = 0; j < CHAR_SIZE; j++) { mat[i][j] = 0; } } precompute(mat, S, len); let get = query(mat, "gg", len)? "Yes" :"No"; document.write(get + "</br>"); get = query(mat, "gro", len)? "Yes" :"No"; document.write(get + "</br>"); get = query(mat, "gfg", len)? "Yes" :"No"; document.write(get + "</br>"); get = query(mat, "orf", len)? "Yes" :"No"; document.write(get + "</br>"); </script> |
Output
Yes No Yes No
This article is contributed by Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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