# Count pair of strings whose concatenation has every vowel

• Difficulty Level : Expert
• Last Updated : 13 Dec, 2022

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:
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:
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 using namespace std; // Function to return the count of all// concatenated string with each vowel// at least onceint 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 Codeint 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 approachimport java.io.*;class GFG{ // Function to return the count of all// concatenated String with each vowel// at least oncestatic 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 Codepublic 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 oncedef 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 Codeif __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 approachusing System; class GFG{ // Function to return the count of all// concatenated String with each vowel// at least oncestatic 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 Codepublic 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



Output:

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 using namespace std; // Function to return the count of all// concatenated string with each vowel// at least onceint 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 Codeint 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 approachimport java.io.*;public class GFG{     // Function to return the count of all// concatenated string with each vowel// at least oncepublic 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 codepublic 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 oncedef 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 stringsStr = [ "aaweiolkju", "oxdfgujkmi" ]N = len(Str) print(good_pairs(Str, N)) # This code is contributed by avanitrachhadiya2155

## C#

 // C# program for the above approachusing System;class GFG{      // Function to return the count of all// concatenated string with each vowel// at least oncepublic 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 codepublic 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



Output:

1

Time Complexity: O(N2)
Auxiliary Space: O(32)

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