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Count pair of strings whose concatenation has every vowel

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  • Difficulty Level : Expert
  • Last Updated : 15 Nov, 2021

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 <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.util.*;
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>


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 <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 
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>


Output: 

1

 

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


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