Calcular la dificultad de una oración dada. Aquí una palabra se considera difícil si tiene 4 consonantes consecutivas o el número de consonantes es mayor que el número de vocales. Otra palabra es fácil. La dificultad de la oración se define como 5*(número de palabras difíciles) + 3*(número de palabras fáciles).
Ejemplos:
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
Preguntado en: Microsoft
Implementación:
Comience a atravesar la string y realice los siguientes pasos:
- Incrementa el conteo de vocales si el carácter actual es una vocal y establece el consonante consecutivo = 0.
- De lo contrario, incremente el recuento de consonantes, también incremente el recuento de consonantes consecutivas.
- Compruebe si las consonantes consecutivas se convierten en 4, entonces la palabra actual es difícil, así que incremente su conteo y pase a la siguiente palabra. Restablezca todos los conteos a 0.
- De lo contrario, verifique si una palabra está completa y el número de consonantes es mayor que el número de vocales,
entonces es una palabra difícil o una palabra fácil. Restablezca todos los conteos a 0.
Implementación:
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 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
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 maintain 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.
Javascript
<script>
// JavaScript program to find
// difficulty of a sentence
// Utility method to check character
// is vowel or not
function isVowel(ch)
{
return (ch == 'a' || ch == 'e' ||
ch == 'i' || ch == 'o' ||
ch == 'u');
}
// Method to calculate difficulty
function calcDiff(str)
{
let count_vowels = 0, count_conso = 0;
let hard_words = 0, easy_words = 0;
let consec_conso = 0;
// Start traversing the string
for (let i = 0; i < str.length; i++)
{
// Check if current character
// is vowel or consonant
if (str[i] != ' ' &&
isVowel(str[i].toLowerCase()))
{
// 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;
}
let str = "I am a geek";
let str2 = "We are geeks";
document.write(calcDiff(str) + "</br>");
document.write(calcDiff(str2));
</script>
Producción
12 11
Complejidad temporal : O(n)
Espacio auxiliar : O(1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA