Check if frequency of each character is equal to its position in English Alphabet

An integral consideration when examining character frequencies and their place within the English alphabet is determining whether each character's frequency within a string aligns with its corresponding location within the alphabet. Each letter of this 26-letter system holds an assigned position ranging from one through twenty-six. Consequently, we must explore how one can recognize if a character's count in each string correlates with its alphabetical place. The following topic will delve into this issue while exploring whether identifying these frequencies can facilitate validating and investigating connections between character frequency and placement within our beloved language's alphabetical order.

Methods

Here are two methods to check if the frequency of each character in a string is equal to its position in the English alphabet −

Method 1: Using a Frequency Array

Method 2: Using a Dictionary

Method 1: Using a Frequency Array

The Frequency Array method is a technique for determining if each character's frequency in a string corresponds to its location in the English alphabet. This method entails building an array to hold each character's frequency and comparing it to that character's corresponding location in the alphabet.

Syntax

def check_frequency_position(string):

• Initialize an array to store the frequency of each character

frequency = [0] * 26

• Convert the input string to lowercase

string = string.lower()

• Iterate through each character in the string

 for char in string:

• Check if the character is an English alphabet letter

if 'a' <= char <= 'z':

• Increment the frequency of the character

frequency[ord(char) - ord('a')] += 1

• Iterate through the frequency array

for i in range(26):

• Check if the frequency of the character is equal to its position in the English alphabet

if frequency[i] != i + 1:
return False

• If all characters have the correct frequency, return True

return True


Algorithm

Verifying whether a given string adheres to an English alphabet's character frequency requires using a step-by-step approach known as the Frequency Array method; follow these steps −

Step 1 − First off, create an array titled "Frequency"that comprises 26 entries pertaining to each letter within the English alphabet while setting every component at zero values initially.

Step 2 − Next up, convert all strings' characters into lowercase format for consistent treatment of capital or small letters.

Step 3 − Loop through each character (c) in the input string iteratively −

• Use ASCII value of c to increase the corresponding element in the frequency field if c is a letter of the alphabet.

• Subtract the ASCII value of 'a' from 'c' to find the index.

• For example, if c is the character "d", the index will be 3 (d minus an in ASCII).

Step 4 − Repeat each character (c) in the input string −

• If c is letter of the alphabet, check that its frequency corresponds to its place in the English alphabet.

• Subtract the ASCII value of 'a' from ‘c' and then multiply the result by 1. This gives you the expected frequency.

• For example, if c is "d", the predicted frequency will be 4 (the ASCII value of "d" minus the ASCII value of "a" plus 1).

• Returns false if the frequency c (specified by the frequency field) is different from the expected value.

Step 5 − Return true if all characters pass the frequency check of step 4.

Example 1

Here is a C++ example that shows how to use the Frequency Array method to determine whether each character in each string occurs at the same frequency as its location in the English alphabet −

The check Frequency Array function takes the string as the input and finds whether the frequency of each character in the string resembles its position in the English alphabet or not. If the frequencies for each character resembles, the method returns true otherwise, it returns false.

The main function uses the sample string "HelloWorld" to show how to use the check Frequency Array function. The program then prints whether the frequency of each character in the string corresponds to its position in the English alphabet.

#include <iostream>
#include <string>

bool checkFrequencyArray(const std::string& str) {
int freq[26] = {0};  // Frequency array for 26 characters

// Count the frequency of each character
for (char ch : str) {
if (isalpha(ch)) {
ch = tolower(ch);  // Convert to lowercase
freq[ch - 'a']++;  // Increment frequency
}
}

// Check if frequency matches the position in the English alphabet
for (int i = 0; i < 26; i++) {
if (freq[i] != (i + 1)) {
return false;  // Frequency doesn't match
}
}

return true;  // Frequency matches for all characters
}

int main() {
std::string str = "HelloWorld";

if (checkFrequencyArray(str)) {
std::cout << "The frequency of each character matches its position in the English alphabet." << std::endl;
} else {
std::cout << "The frequency of each character does not match its position in the English alphabet." << std::endl;
}

return 0;
}


Output

The frequency of each character does not match its position in the English alphabet.


Method 2: Using a Dictionary

The dictionary method compares the frequency of each character in word to its place in the English alphabet by storing the character frequencies in a dictionary. The method includes going over the word repeatedly, updating the dictionary's frequency tables, and then comparing the frequency of each character to its corresponding location in the alphabet. The method returns False if any character do not follow this requirement and True if it follows.

Syntax

def check_frequency_position(string):

• Create a dictionary to store the character frequencies

frequencies = {}

• Count the frequency of each character in the string

for char in string:

• Convert the character to lowercase

char = char.lower()

• Check if the character is a letter

if char.isalpha():
if char in frequencies:
frequencies[char] += 1
else:
frequencies[char] = 1

• Check if frequency equals position for each character

for char, freq in frequencies.items():

• Calculate the position in the alphabet

position = ord(char) - ord('a') + 1
if freq != position:
return False

return True


Algorithm

Sure! Using a dictionary, see if each character's frequency matches its location in the English alphabet by following this step-by-step algorithm −

Step 1 − To hold the frequency of each character, create an empty dictionary called char_freq.

Step 2 − Continue the input string iteration character by character.

• If the current character already appears in the dictionary char_freq, increase its frequency by 1.

• If the current character does not already exist in the dictionary char_freq, add it as a key with a frequency of 1.

Step 3 − To keep track of whether the requirement is met for each character, initialise the variable valid as True.

Step 4 − Go through the char_freq dictionary iteratively.

• Verify that the frequency matches the character's position in the English alphabet for each key (character) and value (frequency) pair. By changing the character to lowercase and deducting the letter 'a's' ASCII value (97) from its ASCII value, you may determine the position.Set valid to False and terminate the loop if the frequency is not the same as the position.

Step 5 − If valid is True at the end of the loop, the condition has been met for every character. Print a message stating that each character appears as often as it does in the English alphabet. Print a message stating that the condition is not met if valid is False.

Example 2

In this example, the function check Character Frequency accepts a string as input and returns true if each character's frequency matches its location in the English alphabet. If not, false is displayed.

The frequency of each character in the string is stored in the function using an unordered map called frequency. Each character in the string is iterated over, and the appropriate frequency on the map is increased.

It then cycles through each character once again to determine its position in the alphabet by changing it to lowercase, taking away the 'a' character's ASCII value, and adding 1. The function returns false if the frequency of the current character doesn't match its place. The function returns true if each character passes the test.

The sample string "abbcccddddeeee" is used in the main function to show how to utilise the check Character Frequency method. Depending on the returned result, it prints the required message.

#include <iostream>
#include <string>
#include <unordered_map>

bool checkCharacterFrequency(const std::string& str) {
std::unordered_map<char, int> frequency;

// Count the frequency of each character in the string
for (char c : str) {
frequency[c]++;
}

// Check if the frequency of each character matches its position in the alphabet
for (char c : str) {
int charPosition = std::tolower(c) - 'a' + 1;  // Get the position of the character

if (frequency[c] != charPosition) {
return false;
}
}
return true;
}
int main() {
std::string str = "abbcccddddeeeee";

if (checkCharacterFrequency(str)) {
std::cout << "The frequency of each character is equal to its position in the English alphabet." << std::endl;
} else {
std::cout << "The frequency of at least one character is not equal to its position in the English alphabet." << std::endl;
}
return 0;
}


Output

The frequency of each character is equal to its position in the English alphabet.


Conclusion

In essence, determining whether each character's frequency in a sentence aligns with its placement in the English alphabet poses a captivating yet challenging pursuit. Analysis of character frequency distribution vis-à-vis positional reference holds promise for unraveling patterned or structured facets of text. That said, undertaking this activity demands meticulous data processing as well as precision in accounting for variations stemming from capitalization or punctuation. Should we apply programming languages or statistical approaches to accomplish this goal; we are left with an appreciation for language's elaborate intricacy and systemic nature.

Updated on: 31-Jul-2023

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