Calculate the difficulty of a sentence

Calculate difficulty of a given sentence. Here a Word is considered hard if it has 4 consecutive consonants or number of consonants are more than number of vowels. Else word is easy. Difficulty of sentence is defined as 5*(number of hard words) + 3*(number of easy words).

Examples:

Input : str = "Difficulty of sentence"
Output : 13
Hard words = 2(Difficulty and sentence)
Easy words = 1(of)
So, answer is 5*2+3*1 = 13

Asked in : Microsoft

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Implementation:
Start traversing the string and perform following steps:-

Increment vowels count, if current character is vowel and set conecutive consonants count=0.
Else increment consonants count, also increment consecutive consonants count.
Check if consecutive consonants becomes 4, then current word is hard, so increment its count
and move to the next word.Reset all counts to 0.
Else check if a word is completed and count of consonants is greater than count of vowels,
then it is a hard word else easy word.Reset all counts to 0.

C++

// C++ program to find difficulty of a sentence 
#include <iostream> 
using namespace std; 
  
// Utility function to check character is vowel 
// or not 
bool isVowel(char ch) 
{ 
    return ( ch == 'a' || ch == 'e' || 
             ch == 'i' || ch == 'o' || 
             ch == 'u'); 
} 
  
// Function to calculate difficulty 
int calcDiff(string str) 
{ 
  
    int count_vowels = 0, count_conso = 0; 
    int hard_words = 0, easy_words = 0; 
    int consec_conso = 0; 
  
    // Start traversing the string 
    for (int i = 0; i < str.length(); i++) 
    { 
        // Check if current character is vowel 
        // or consonant 
        if (str[i] != ' ' && isVowel(tolower(str[i]))) 
        { 
            // Increment if vowel 
            count_vowels++; 
            consec_conso = 0; 
        } 
  
        // Increment counter for consonant 
        // also mainatin a separate counter for 
        // counting consecutive consonants 
        else if (str[i]!= ' ') 
        { 
            count_conso++; 
            consec_conso++; 
        } 
  
        // If we get 4 consecutive consonants 
        // then it is a hard word 
        if (consec_conso == 4) 
        { 
            hard_words++; 
  
            // Move to the next word 
            while (i < str.length() && str[i]!= ' ') 
                i++; 
  
            // Reset all counts 
            count_conso = 0; 
            count_vowels = 0; 
            consec_conso = 0; 
        } 
  
        else if ( i < str.length() && 
                  (str[i] == ' ' || i == str.length()-1)) 
        { 
            // Increment hard_words, if no. of consonants are 
            // higher than no. of vowels, otherwise increment 
            // count_vowels 
            count_conso > count_vowels ? hard_words++ 
                                       : easy_words++; 
  
            // Reset all counts 
            count_conso = 0; 
            count_vowels = 0; 
            consec_conso = 0; 
        } 
    } 
  
    // Return difficulty of sentence 
    return 5 * hard_words + 3 * easy_words; 
} 
  
// Drivers code 
int main() 
{ 
    string str = "I am a geek"; 
    string str2 = "We are geeks"; 
    cout << calcDiff(str) << endl; 
    cout << calcDiff(str2) << endl; 
  
    return 0; 
} 

Java

// Java program to find difficulty of a sentence 
  
class GFG  
{ 
    // Utility method to check character is vowel 
    // or not 
    static boolean isVowel(char ch) 
    { 
        return ( ch == 'a' || ch == 'e' || 
                 ch == 'i' || ch == 'o' || 
                 ch == 'u'); 
    } 
       
    // Method to calculate difficulty 
    static int calcDiff(String str) 
    { 
       
        int count_vowels = 0, count_conso = 0; 
        int hard_words = 0, easy_words = 0; 
        int consec_conso = 0; 
       
        // Start traversing the string 
        for (int i = 0; i < str.length(); i++) 
        { 
            // Check if current character is vowel 
            // or consonant 
            if (str.charAt(i) != ' ' && isVowel(Character.toLowerCase(str.charAt(i)))) 
            { 
                // Increment if vowel 
                count_vowels++; 
                consec_conso = 0; 
            } 
       
            // Increment counter for consonant 
            // also maintain a separate counter for 
            // counting consecutive consonants 
            else if (str.charAt(i)!= ' ') 
            { 
                count_conso++; 
                consec_conso++; 
            } 
       
            // If we get 4 consecutive consonants 
            // then it is a hard word 
            if (consec_conso == 4) 
            { 
                hard_words++; 
       
                // Move to the next word 
                while (i < str.length() && str.charAt(i)!= ' ') 
                    i++; 
       
                // Reset all counts 
                count_conso = 0; 
                count_vowels = 0; 
                consec_conso = 0; 
            } 
       
            else if ( i < str.length() && 
                      (str.charAt(i) == ' ' || i == str.length()-1)) 
            { 
                // Increment hard_words, if no. of consonants are 
                // higher than no. of vowels, otherwise increment 
                // count_vowels 
                if(count_conso > count_vowels) 
                    hard_words++; 
                else
                    easy_words++; 
       
                // Reset all counts 
                count_conso = 0; 
                count_vowels = 0; 
                consec_conso = 0; 
            } 
        } 
       
        // Return difficulty of sentence 
        return 5 * hard_words + 3 * easy_words; 
    } 
      
    // Driver method 
    public static void main (String[] args) 
    { 
        String str = "I am a geek"; 
        String str2 = "We are geeks"; 
        System.out.println(calcDiff(str)); 
        System.out.println(calcDiff(str2)); 
    } 
} 

Python 3

# Python3 program to find difficulty  
# of a sentence 
  
# Utility function to check character  
# is vowel or not 
def isVowel(ch): 
    return (ch == 'a' or ch == 'e' or 
            ch == 'i' or ch == 'o' or 
            ch == 'u') 
  
# Function to calculate difficulty 
def calcDiff(str): 
    str = str.lower() 
    count_vowels = 0
    count_conso = 0
    consec_conso = 0
    hard_words = 0
    easy_words = 0
  
    # Start traversing the string 
    for i in range(0, len(str)): 
          
        # Check if current character is  
        # vowel or consonant 
        if(str[i]!= " " and isVowel(str[i])): 
              
            # Increment if vowel 
            count_vowels += 1
            consec_conso = 0
              
        # Increment counter for consonant 
        # also mainatin a separate counter for 
        # counting consecutive consonants  
        elif(str[i] != " "): 
            count_conso += 1
            consec_conso += 1
  
        # If we get 4 consecutive consonants 
        # then it is a hard word  
        if(consec_conso == 4): 
            hrad_words += 1
  
            # Move to the next word 
            while(i < len(str) and str[i] != " "): 
                i += 1
                  
            # Reset all counts  
            count_conso = 0
            count_vowels = 0
            consec_conso = 0
        elif(i < len(str) and (str[i] == ' ' or
                          i == len(str) - 1)): 
                                
            # Increment hard_words, if no. of  
            # consonants are higher than no. of  
            # vowels, otherwise increment count_vowels 
            if(count_conso > count_vowels): 
                hard_words += 1
            else: 
                easy_words += 1
  
            # Reset all counts  
            count_conso = 0
            count_vowels = 0
            consec_conso = 0
              
    # Return difficulty of sentence      
    return (5 * hard_words + 3 * easy_words) 
      
# Driver Code 
if __name__=="__main__": 
    str = "I am a geek"
    str2 = "We are geeks"
    print(calcDiff(str)) 
    print(calcDiff(str2)) 
  
# This code is contributed 
# by Sairahul Jella 

C#

// C# program to find difficulty 
// of a sentence 
using System; 
      
public class GFG { 
      
    // Utility method to check character 
    // is vowel or not 
    static bool isVowel(char ch) 
    { 
        return (ch == 'a' || ch == 'e' || 
                ch == 'i' || ch == 'o' || 
                ch == 'u'); 
    } 
      
    // Method to calculate difficulty 
    static int calcDiff(string str) 
    { 
        int count_vowels = 0, count_conso = 0; 
        int hard_words = 0, easy_words = 0; 
        int consec_conso = 0; 
      
        // Start traversing the string 
        for (int i = 0; i < str.Length; i++) 
        { 
              
            // Check if current character  
            // is vowel or consonant 
            if (str[i] != ' ' &&  
                isVowel(char.ToLower( str[i]))) 
            { 
                // Increment if vowel 
                count_vowels++; 
                consec_conso = 0; 
            } 
      
            // Increment counter for consonant 
            // also maintain a separate counter for 
            // counting consecutive consonants 
            else if (str[i]!= ' ') 
            { 
                count_conso++; 
                consec_conso++; 
            } 
      
            // If we get 4 consecutive consonants 
            // then it is a hard word 
            if (consec_conso == 4) 
            { 
                hard_words++; 
      
                // Move to the next word 
                while (i < str.Length && str[i]!= ' ') 
                    i++; 
      
                // Reset all counts 
                count_conso = 0; 
                count_vowels = 0; 
                consec_conso = 0; 
            } 
      
            else if ( i < str.Length && 
                    (str[i] == ' ' || i == str.Length-1)) 
            { 
                  
                // Increment hard_words, if no. of 
                // consonants are higher than no. 
                // of vowels, otherwise increment 
                // count_vowels 
                if(count_conso > count_vowels) 
                    hard_words++; 
                else
                    easy_words++; 
      
                // Reset all counts 
                count_conso = 0; 
                count_vowels = 0; 
                consec_conso = 0; 
            } 
        } 
      
        // Return difficulty of sentence 
        return 5 * hard_words + 3 * easy_words; 
    } 
      
    // Driver code 
    public static void Main () 
    { 
        string str = "I am a geek"; 
        string str2 = "We are geeks"; 
        Console.WriteLine(calcDiff(str)); 
        Console.WriteLine(calcDiff(str2)); 
    } 
} 
  
// This code is contributed by Sam007. 

Output:

12
11

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python 3

C#

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