Count pair of strings whose concatenation has every vowel
Given an array arr[] of N strings. The task is to find the count of all possible pairs of strings such that their concatenations has every vowel at least once.
Examples:
Input: str[] = {“oakitie”, “aemiounau”, “aazuaxvbga”, “eltdgo”}
Output: 4
Explanation:
Concatenation of pair of string are:
1. (“oakitie”, “aazuaxvbga”),
2. (“oakitie”, “aemiounau”),
3. (“aazuaxvbga”, “aemiounau”),
4. (“aemiounau”, “eltdgo”)
They all have the vowel at least once.Input: str[] = {“aaweiolkju”, “oxdfgujkmi”}
Output: 1
Explanation:
Concatenation of pair of string (“aaweiolkju”, “oxdfgujkmi”) has all the vowel character at least once.
Naive Approach: The idea is to generate all possible pairs from the given array and check if the concatenation of every possible pair of two strings has all the vowels at least once or not. If yes then include this in the count. Print the count after all the operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to return the count of all // concatenated string with each vowel // at least once int good_pair(string str[], int N) { int countStr = 0; // Concatenating all possible // pairs of string for ( int i = 0; i < N; i++) { for ( int j = i + 1; j < N; j++) { string res = str[i] + str[j]; // Creating an array which checks, // the presence of each vowel int vowel[5] = { 0 }; // Checking for each vowel by // traversing the concatenated // string for ( int k = 0; k < res.length(); k++) { if (res[k] == 'a' ) vowel[0] = 1; else if (res[k] == 'e' ) vowel[1] = 1; else if (res[k] == 'i' ) vowel[2] = 1; else if (res[k] == 'o' ) vowel[3] = 1; else if (res[k] == 'u' ) vowel[4] = 1; } // Checking if all the elements // are set in vowel[] int temp = 0; for ( int ind = 0; ind < 5; ind++) { if (vowel[ind] == 1) temp++; } // Check if all vowels are // present or not if (temp == 5) countStr++; } } // Return the final count return countStr; } // Driver Code int main() { // Given array of strings string arr[] = { "aaweiolkju" , "oxdfgujkmi" }; int N = sizeof (arr) / sizeof (arr[0]); // Function Call cout << good_pair(arr, N); } |
Java
// Java program for the above approach import java.io.*; class GFG{ // Function to return the count of all // concatenated String with each vowel // at least once static int good_pair(String str[], int N) { int countStr = 0 ; // Concatenating all possible // pairs of String for ( int i = 0 ; i < N; i++) { for ( int j = i + 1 ; j < N; j++) { String res = str[i] + str[j]; // Creating an array which checks, // the presence of each vowel int vowel[] = new int [ 5 ]; // Checking for each vowel by // traversing the concatenated // String for ( int k = 0 ; k < res.length(); k++) { if (res.charAt(k) == 'a' ) vowel[ 0 ] = 1 ; else if (res.charAt(k) == 'e' ) vowel[ 1 ] = 1 ; else if (res.charAt(k) == 'i' ) vowel[ 2 ] = 1 ; else if (res.charAt(k) == 'o' ) vowel[ 3 ] = 1 ; else if (res.charAt(k) == 'u' ) vowel[ 4 ] = 1 ; } // Checking if all the elements // are set in vowel[] int temp = 0 ; for ( int ind = 0 ; ind < 5 ; ind++) { if (vowel[ind] == 1 ) temp++; } // Check if all vowels are // present or not if (temp == 5 ) countStr++; } } // Return the final count return countStr; } // Driver Code public static void main(String[] args) { // Given array of Strings String arr[] = { "aaweiolkju" , "oxdfgujkmi" }; int N = arr.length; // Function Call System.out.print(good_pair(arr, N)); } } // This code is contributed by Rohit_ranjan |
Python3
# Python3 program for the above approach # Function to return the count of all # concatenated string with each vowel # at least once def good_pair(st, N): countStr = 0 # Concatenating all possible # pairs of string for i in range (N): for j in range (i + 1 , N): res = st[i] + st[j] # Creating an array which checks, # the presence of each vowel vowel = [ 0 ] * 5 # Checking for each vowel by # traversing the concatenated # string for k in range ( len (res)): if (res[k] = = 'a' ): vowel[ 0 ] = 1 elif (res[k] = = 'e' ): vowel[ 1 ] = 1 elif (res[k] = = 'i' ): vowel[ 2 ] = 1 elif (res[k] = = 'o' ): vowel[ 3 ] = 1 elif (res[k] = = 'u' ): vowel[ 4 ] = 1 # Checking if all the elements # are set in vowel[] temp = 0 for ind in range ( 5 ): if (vowel[ind] = = 1 ): temp + = 1 # Check if all vowels are # present or not if (temp = = 5 ): countStr + = 1 # Return the final count return countStr # Driver Code if __name__ = = "__main__" : # Given array of strings arr = [ "aaweiolkju" , "oxdfgujkmi" ] N = len (arr) # Function call print (good_pair(arr, N)) # This code is contributed by jana_sayantan |
C#
// C# program for the above approach using System; class GFG{ // Function to return the count of all // concatenated String with each vowel // at least once static int good_pair(String []str, int N) { int countStr = 0; // Concatenating all possible // pairs of String for ( int i = 0; i < N; i++) { for ( int j = i + 1; j < N; j++) { String res = str[i] + str[j]; // Creating an array which checks, // the presence of each vowel int []vowel = new int [5]; // Checking for each vowel by // traversing the concatenated // String for ( int k = 0; k < res.Length; k++) { if (res[k] == 'a' ) vowel[0] = 1; else if (res[k] == 'e' ) vowel[1] = 1; else if (res[k] == 'i' ) vowel[2] = 1; else if (res[k] == 'o' ) vowel[3] = 1; else if (res[k] == 'u' ) vowel[4] = 1; } // Checking if all the elements // are set in vowel[] int temp = 0; for ( int ind = 0; ind < 5; ind++) { if (vowel[ind] == 1) temp++; } // Check if all vowels are // present or not if (temp == 5) countStr++; } } // Return the readonly count return countStr; } // Driver Code public static void Main(String[] args) { // Given array of Strings String []arr = { "aaweiolkju" , "oxdfgujkmi" }; int N = arr.Length; // Function call Console.Write(good_pair(arr, N)); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript program for the above approach // Function to return the count of all // concatenated string with each vowel // at least once function good_pair(str, N) { let countStr = 0; // Concatenating all possible // pairs of string for (let i = 0; i < N; i++) { for (let j = i + 1; j < N; j++) { let res = str[i] + str[j]; // Creating an array which checks, // the presence of each vowel let vowel = new Array(5); vowel.fill(0); // Checking for each vowel by // traversing the concatenated // string for (let k = 0; k < res.length; k++) { if (res[k] == 'a' ) vowel[0] = 1; else if (res[k] == 'e' ) vowel[1] = 1; else if (res[k] == 'i' ) vowel[2] = 1; else if (res[k] == 'o' ) vowel[3] = 1; else if (res[k] == 'u' ) vowel[4] = 1; } // Checking if all the elements // are set in vowel[] let temp = 0; for (let ind = 0; ind < 5; ind++) { if (vowel[ind] == 1) temp++; } // Check if all vowels are // present or not if (temp == 5) countStr++; } } // Return the final count return countStr; } // Driver code // Given array of strings let arr = [ "aaweiolkju" , "oxdfgujkmi" ]; let N = arr.length; // Function Call document.write(good_pair(arr, N)); // This code is contributed by divyesh072019 </script> |
1
Time Complexity: O(N4)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above method the idea is to use Hashing. Below are the steps:
- Create a hash hash[] of length 32 with all elements as 0.
- Use hashing by taking a variable weight and performing Bitwise OR operation on this variable with powers of 2 while traversing each string as:
for 'a': Weight | 1, for 'e': Weight | 2, for 'i': Weight | 4, for 'o': Weight | 8, for 'u': Weight | 16
- The value of weight is a hash value which will indicate the combination of vowels the string contains.
For Example:
"aeiouau": Weight = 31, as it has all the vowels and "oaiie": Weight = 15, as it has all the vowels except 'u'
- Increase index denoted by weight in the hash[] by 1 and the elements are the count of the string with indices as their hash value.
- So, all pairs of the hash value that is indices (i, j) which give Bitwise OR as 31 are the pairs that have all the vowels at least once in them. Let the count of such string be countStr. So, the count is given by:
countStr = hash[i] * hash[j] - Add the count of each pairs using the formula in the above step.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to return the count of all // concatenated string with each vowel // at least once int good_pairs(string str[], int N) { // Creating a hash array with // initial value as 0 int arr[32] = { 0 }, strCount = 0; // Traversing through each string // and getting hash value for each // of them for ( int i = 0; i < N; i++) { // Initializing the weight // of each string int Weight = 0; // Find the hash value for // each string for ( int j = 0; j < str[i].size(); j++) { switch (str[i][j]) { case 'a' : Weight = Weight | 1; break ; case 'e' : Weight = Weight | 2; break ; case 'i' : Weight = Weight | 4; break ; case 'o' : Weight = Weight | 8; break ; case 'u' : Weight = Weight | 16; break ; } } // Increasing the count // of the hash value arr[Weight]++; } // Getting all possible pairs // of indexes in hash array for ( int i = 0; i < 32; i++) { for ( int j = i + 1; j < 32; j++) { // Check if the pair which has // hash value 31 and // multiplying the count of // string and add it strCount if ((i | j) == 31) strCount += arr[i] * arr[j]; } } // Corner case, for strings which // independently has all the vowels strCount += (arr[31] * (arr[31] - 1)) / 2; // Return there final count return strCount; } // Driver Code int main() { // Given array of strings string str[] = { "aaweiolkju" , "oxdfgujkmi" }; int N = sizeof (str) / sizeof (str[0]); // Function Call cout << good_pairs(str, N); return 0; } |
Java
// Java program for the above approach import java.io.*; public class GFG{ // Function to return the count of all // concatenated string with each vowel // at least once public static int good_pairs(String[] str, int N) { // Creating a hash array with // initial value as 0 int arr[] = new int [ 32 ]; int strCount = 0 ; // Traversing through each string // and getting hash value for each // of them for ( int i = 0 ; i < N; i++) { // Initializing the weight // of each string int Weight = 0 ; // Find the hash value for // each string for ( int j = 0 ; j < str[i].length(); j++) { switch (str[i].charAt(j)) { case 'a' : Weight = Weight | 1 ; break ; case 'e' : Weight = Weight | 2 ; break ; case 'i' : Weight = Weight | 4 ; break ; case 'o' : Weight = Weight | 8 ; break ; case 'u' : Weight = Weight | 16 ; break ; } } // Increasing the count // of the hash value arr[Weight]++; } // Getting all possible pairs // of indexes in hash array for ( int i = 0 ; i < 32 ; i++) { for ( int j = i + 1 ; j < 32 ; j++) { // Check if the pair which has // hash value 31 and // multiplying the count of // string and add it strCount if ((i | j) == 31 ) strCount += arr[i] * arr[j]; } } // Corner case, for strings which // independently has all the vowels strCount += (arr[ 31 ] * (arr[ 31 ] - 1 )) / 2 ; // Return there final count return strCount; } // Driver code public static void main(String[] args) { // Given array of strings String str[] = { "aaweiolkju" , "oxdfgujkmi" }; int N = str.length; // Function call System.out.println(good_pairs(str, N)); } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for the above approach # Function to return the count of all # concatenated string with each vowel # at least once def good_pairs( Str , N): # Creating a hash array with # initial value as 0 arr = [ 0 for i in range ( 32 )] strCount = 0 # Traversing through each string # and getting hash value for each # of them for i in range (N): # Initializing the weight # of each string Weight = 0 # Find the hash value for # each string for j in range ( len ( Str [i])): switcher = { 'a' : 1 , 'e' : 2 , 'i' : 4 , 'o' : 8 , 'u' : 16 , } Weight = Weight | switcher.get( Str [i][j], 0 ) # Increasing the count # of the hash value arr[Weight] + = 1 # Getting all possible pairs # of indexes in hash array for i in range ( 32 ): for j in range (i + 1 , 32 ): # Check if the pair which has # hash value 31 and # multiplying the count of # string and add it strCount if ((i | j) = = 31 ): strCount + = arr[i] * arr[j] # Corner case, for strings which # independently has all the vowels strCount + = int ((arr[ 31 ] * (arr[ 31 ] - 1 )) / 2 ) # Return there final count return strCount # Driver Code # Given array of strings Str = [ "aaweiolkju" , "oxdfgujkmi" ] N = len ( Str ) print (good_pairs( Str , N)) # This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the above approach using System; class GFG{ // Function to return the count of all // concatenated string with each vowel // at least once public static int good_pairs( string [] str, int N) { // Creating a hash array with // initial value as 0 int [] arr = new int [32]; int strCount = 0; // Traversing through each string // and getting hash value for each // of them for ( int i = 0; i < N; i++) { // Initializing the weight // of each string int Weight = 0; // Find the hash value for // each string for ( int j = 0; j < str[i].Length; j++) { switch (str[i][j]) { case 'a' : Weight = Weight | 1; break ; case 'e' : Weight = Weight | 2; break ; case 'i' : Weight = Weight | 4; break ; case 'o' : Weight = Weight | 8; break ; case 'u' : Weight = Weight | 16; break ; } } // Increasing the count // of the hash value arr[Weight]++; } // Getting all possible pairs // of indexes in hash array for ( int i = 0; i < 32; i++) { for ( int j = i + 1; j < 32; j++) { // Check if the pair which has // hash value 31 and // multiplying the count of // string and add it strCount if ((i | j) == 31) strCount += arr[i] * arr[j]; } } // Corner case, for strings which // independently has all the vowels strCount += (arr[31] * (arr[31] - 1)) / 2; // Return there final count return strCount; } // Driver code public static void Main( string [] args) { // Given array of strings string [] str = { "aaweiolkju" , "oxdfgujkmi" }; int N = str.Length; // Function call Console.Write(good_pairs(str, N)); } } // This code is contributed by rock_cool |
Javascript
<script> // Javascript program for the above approach // Function to return the count of all // concatenated string with each vowel // at least once function good_pairs(str, N) { // Creating a hash array with // initial value as 0 let arr = new Array(32); arr.fill(0); let strCount = 0; // Traversing through each string // and getting hash value for each // of them for (let i = 0; i < N; i++) { // Initializing the weight // of each string let Weight = 0; // Find the hash value for // each string for (let j = 0; j < str[i].length; j++) { switch (str[i][j]) { case 'a' : Weight = Weight | 1; break ; case 'e' : Weight = Weight | 2; break ; case 'i' : Weight = Weight | 4; break ; case 'o' : Weight = Weight | 8; break ; case 'u' : Weight = Weight | 16; break ; } } // Increasing the count // of the hash value arr[Weight]++; } // Getting all possible pairs // of indexes in hash array for (let i = 0; i < 32; i++) { for (let j = i + 1; j < 32; j++) { // Check if the pair which has // hash value 31 and // multiplying the count of // string and add it strCount if ((i | j) == 31) strCount += arr[i] * arr[j]; } } // Corner case, for strings which // independently has all the vowels strCount += parseInt((arr[31] * (arr[31] - 1)) / 2, 10); // Return there final count return strCount; } // Given array of strings let str = [ "aaweiolkju" , "oxdfgujkmi" ]; let N = str.length; // Function call document.write(good_pairs(str, N)); </script> |
1
Time Complexity: O(N2)
Auxiliary Space: O(32)
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