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Maximize count of Substrings containing at least 1 vowel and 1 consonant

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  • Last Updated : 30 Aug, 2022
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Given a string S consisting of only lowercase English letters of length N, the task is to find the count of substrings such that each substring contains at least 1 vowel and 1 consonant.

Examples:

Input: S = “happybirthday”
Output: 3
Explanation: S can be divided as “ha”, “ppybi”, “rthday”

Input: S = “geeksforgeeks”
Output 5
Explanation: S can be divided as ge“, “ek“, “sfo“, “rge”, “eks”

Approach: To solve the problem follow the below idea:

The idea is to apply the greedy approach, traverse the string, and every time we encounter 1 vowel and 1 consonant increase the count by 1 and reset the number of vowels and consonants to 0.

Follow the steps below to solve the problem:

  • Initialize variable ans = 0, haveVowels = false (current string have vowels or not), haveConsonants = false (current string have consonants or not)
  • Travel over the string by each character
    • If current character is vowels make the haveVowels = true else haveConsonants = true
    • Whenever both are true then consider current subsegment is a valid substring, increase ans by 1, make haveVowels = false and haveConsonants = false.

Below is the implementation of the above approach.

C++




#include <iostream>
using namespace std;
int strengthOfString(string S)
{
    int ans = 0;
    int N = S.size();
 
    // For checking at least 1 vowels
    // and 1 consonants
    bool haveVowels = false;
    bool haveConsonants = false;
 
    // Travel over the loop
    for(int i=0;i<N;i++)
    {
        // Check current letter have
        // vowels or not
        if(S[i]=='a'||S[i]=='e'||S[i]=='i'||S[i]=='o'||S[i]=='u')
        haveVowels = true;
        else
        haveConsonants = true;
         
        // Both the type of letter are present
        // then increase the ans count
        if (haveVowels && haveConsonants)
        {
            ans += 1;
            haveVowels = false;
            haveConsonants = false;
        }
    }
     
    // Return the main ans
    return ans;
}
int main()
{
    string S = "happybirthday";
 
    // Function Call
    int ans = strengthOfString(S);
    cout << ans;
 
    return 0;
}
//this code is contributed by aditya942003patil


Java




// Java program to implement the approach
import java.io.*;
 
class GFG {
    public static int strengthOfString(String S)
    {
        int ans = 0;
        int N = S.length();
 
        // For checking at least 1 vowels
        // and 1 consonants
        boolean haveVowels = false;
        boolean haveConsonants = false;
 
        // Travel over the loop
        for (int i = 0; i < N; i++) {
            // Check current letter have
            // vowels or not
            if (S.charAt(i) == 'a' || S.charAt(i) == 'e'
                || S.charAt(i) == 'i' || S.charAt(i) == 'o'
                || S.charAt(i) == 'u')
                haveVowels = true;
            else
                haveConsonants = true;
 
            // Both the type of letter are present
            // then increase the ans count
            if (haveVowels == true
                && haveConsonants == true) {
                ans += 1;
                haveVowels = false;
                haveConsonants = false;
            }
        }
 
        // Return the main ans
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "happybirthday";
 
        // Function Call
        int ans = strengthOfString(S);
        System.out.print(ans);
    }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Python program to implement the approach
 
# Function to calculate maximum
# strength of a string
def strengthOfString(S):
    ans = 0
    N = len(S)
 
    # For checking at least 1 vowels
    # and 1 consonants
    haveVowels = False
    haveConsonants = False
 
    # Travel over the loop
    for s in S:
 
        # Check current letter have
        # vowels or not
        if s in "aeiou":
            haveVowels = True
        else:
            haveConsonants = True
 
        # Both the type of letter are present
        # then increase the ans count
        if haveVowels and haveConsonants:
            ans += 1
            haveVowels = False
            haveConsonants = False
 
    # Return the main ans
    return ans
 
# Driver Code
if __name__ == "__main__":
    S = "happybirthday"
 
    # Function Call
    ans = strengthOfString(S)
    print(ans)


C#




// C# program to implement the approach
 
using System;
 
public class GFG {
 
    public static int strengthOfString(String S)
    {
        int ans = 0;
        int N = S.Length;
 
        // For checking at least 1 vowels
        // and 1 consonants
        bool haveVowels = false;
        bool haveConsonants = false;
 
        // Travel over the loop
        for (int i = 0; i < N; i++) {
            // Check current letter have
            // vowels or not
            if (S[i] == 'a' || S[i] == 'e' || S[i] == 'i'
                || S[i] == 'o' || S[i] == 'u')
                haveVowels = true;
            else
                haveConsonants = true;
 
            // Both the type of letter are present
            // then increase the ans count
            if (haveVowels == true
                && haveConsonants == true) {
                ans += 1;
                haveVowels = false;
                haveConsonants = false;
            }
        }
 
        // Return the main ans
        return ans;
    }
 
    static public void Main()
    {
 
        // Code
        string S = "happybirthday";
 
        // Function Call
        int ans = strengthOfString(S);
        Console.Write(ans);
    }
}
 
// This code is contributed by lokeshmvs21.


Javascript




<script>
     
   // JavaScript code for the above approach
 
    function strengthOfString(S)
    {
        let ans = 0;
        let N = S.length;
  
        // For checking at least 1 vowels
        // and 1 consonants
        let haveVowels = false;
        let haveConsonants = false;
  
        // Travel over the loop
        for (let i = 0; i < N; i++) {
            // Check current letter have
            // vowels or not
            if (S.charAt(i) == 'a' || S.charAt(i) == 'e'
                || S.charAt(i) == 'i' || S.charAt(i) == 'o'
                || S.charAt(i) == 'u')
                haveVowels = true;
            else
                haveConsonants = true;
  
            // Both the type of letter are present
            // then increase the ans count
            if (haveVowels == true
                && haveConsonants == true) {
                ans += 1;
                haveVowels = false;
                haveConsonants = false;
            }
        }
  
        // Return the main ans
        return ans;
    }
 
    // Driver code
     
        let S = "happybirthday";
  
        // Function Call
        let ans = strengthOfString(S);
        document.write(ans);
     
    // This code is contributed by code_hunt.
</script>


Output

3

Time Complexity: O(N)
Auxiliary Space: O(1)


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