. Which of the following is NOT an option for increasinghard drive storage space?
1 answer:
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Answer:
Check the explanation
Explanation:
Code to copy:
// ConsoleApplication7.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <cstring>
#include <string>
using namespace std;
int * letterFrequency(char s[]);
int main()
{
int *freq_letters;
char *s = new char[100];
freq_letters = new int[26];
//To read the input string in order to find the frequency of each letter
cout << "Enter a string: ";
cin.getline(s, ' ');
//call the function to find the occurrence of each alphabet
freq_letters = letterFrequency(s);
//Display the count
cout << "Letter Frequency " << endl;
for (int i = 0; i < 26; i++)
{
//Constriant to check if the letter appeared at least once in the string and so printing the frequency of its occurence
if (freq_letters[i] != 0)
cout << " " << static_cast<char>(i + 'a') << " " << freq_letters[i] << endl;
}
system("pause");
return 0;
}
//Define the function to find occurrence of each letter in the input string
int * letterFrequency(char s[])
{
int *occurrence_array;
//to store the output of occurrences for each alphabet
occurrence_array = new int[26];
// to store the count of occurrence for each letter temporarily
int letter_count;
// for loop to check the occurrence for all 26 alphabets
for (int i = 0; i < 26; i++)
{
letter_count = 0;
for (int j = 0; j < strlen(s); j++)
{
/*comparing the ascii values of each alphabet with every character from the string by converting it to lower case i.e. case insensitive*/
if (int('a') + i == int(tolower(s[j])))
letter_count++;
}
occurrence_array[i] = letter_count;//To store the count calculated for each alphabet
}
return occurrence_array;
}
The following below code screenshot and output shows that when the string is entered, the output will shows each lettercount irrespective of whether the letter is in capitals or small and does not count non-letter characters (i.e spaces, punctuations etc.)
