Question - Given a string, Write a program that prints the number of vowels in it.
function numVowels (str) {
let count = 0;
// Iterate over each character in the string
for (let i=0; i<str.length; i++) {
// check whether this character (in lower case) is a, e, i, o, or u (converting in lower case would reduce the amount of conditons by half since there would be no need to check A, E, I, O or U)
if (
str[i].toLowerCase() === 'a' ||
str[i].toLowerCase() === 'e' ||
str[i].toLowerCase() === 'i' ||
str[i].toLowerCase() === 'o' ||
str[i].toLowerCase() === 'u'
) {
count++;
}
}
// Print and return the number of vowels
console.log(`The number of vowels in ${str} is = ${count}`);
return count;
}
numVowels ('hello');
numVowels ('Greetings');function numVowels (str) {
let count = 0;
// Define a vowels array
let vowels = ['a', 'e', 'i', 'o', 'u'];
// Iterate over each character in the string
for (let i=0; i<str.length; i++) {
// check whether this character (in lower case) is a, e, i, o, or u (converting in lower case would reduce the amount of conditons by half since there would be no need to check A, E, I, O or U)
if (vowels.indexOf(str[i].toLowerCase()) >= 0) {
count++;
}
}
// Print and return the number of vowels
console.log(`The number of vowels in ${str} is = ${count}`);
return count;
}
numVowels ('hEllo');
numVowels ('Greetings');// Store vowels in an array
// check each character of the string with each element in vowels array
function numVowels (str) {
// Initialize count to be zero
let count = 0;
// Define an array of vowels
let vowels = ['a', 'e', 'i', 'o', 'u'];
// Check each character of string
for (let char of str) {
for (let vowel of vowels) {
if (char.toLowerCase() === vowel) count++;
}
}
// Print the result
console.log(`The number of vowels in "${str}" is = ${count}`);
return count;
}
numVowels ('hello');
numVowels ('Greetings');// Similar to previous method
// Instead of storing vowels in an array, store them in an object as keys and values as the number of occurances
function numVowels (str) {
console.log('\nString: ', str);
// Initialize count to be zero
let count = 0;
// Store the vowels in an object
let vowels = {
'a': 0,
'b': 0,
'c': 0,
'd': 0,
'e': 0
};
// Iterate through each character of string to check vowels
for (let char of str) {
for (vowel in vowels) {
if (char.toLowerCase() === vowel) {
count++;
vowels[vowel]++;
}
}
}
// Print the result
console.log('Vowel Count: ');
for (let vowel in vowels) {
console.log(`Vowel: ${vowel} appears ${vowels[vowel]} number of times in the string "${str}"`);
}
console.log("Total number of vowels: " + count);
return count;
}
numVowels ('hello');
numVowels ('Greetings');import java.util.Scanner;
public class NumVowels {
public static void main (String[] args) {
// set count = 0
int count = 0;
// Input the string
Scanner input = new Scanner(System.in);
System.out.println("/* ===== Number of Vowels ===== */");
System.out.print("\nEnter the string: ");
String str = input.next();
// Convert input string to lower case
str = str.toLowerCase();
// Run a loop from 0 to string length
for (int i=0; i<str.length(); i++) {
if (
str.charAt(i) == 'a' ||
str.charAt(i) == 'e' ||
str.charAt(i) == 'i' ||
str.charAt(i) == 'o' ||
str.charAt(i) == 'u'
) {
count++;
}
}
// Print the result
System.out.println("Number of vowels in \"" + str + "\" = " + count);
}
}import java.util.Scanner;
public class NumVowels2 {
public static void main (String[] args) {
// set count = 0
int count = 0;
// Input the string
Scanner input = new Scanner(System.in);
System.out.println("/* ===== Number of Vowels ===== */");
System.out.print("\nEnter the string: ");
String str = input.next();
// Convert input string to lower case
str = str.toLowerCase();
// Create an array of vowels
char[] vowels = new char[]{ 'a', 'e', 'i', 'o', 'u' };
// Run a loop from 0 to string length
for (int i=0; i<str.length(); i++) {
// Check whether the current character exists in the vowels array
for (int j=0; j<5; j++) {
if (str.charAt(i) == vowels[j]) {
count++;
break;
}
}
}
// Print the result
System.out.println("Number of vowels in \"" + str + "\" = " + count);
}
}import re
"""
This solution makes use of python's re module i.e regular expressions
Here findall method takes two parameters 1st - regualr expression, 2nd - input string
The regular expression simply checks for all vowels present in the input string
and returns them as a list.
"""
print(len(re.findall(r'[a,e,i,o,u,A,E,I,O,U]', input())))# Input the String.
string=input("Enter the String : ")
# Create a list of vowels in lowercase.
vowels=['a','e','i','o','u']
# Initialize the count variable to zero.
count=0
# Now iterate every character in the string.
for char in string:
# And if the current character we are iterating is
# present in the vowels list then increment count.
if char.lower() in vowels:
count+=1
# Print the result
print("Number of vowels in the string are : ",count)a=input("Enter the string to count no. of vowels?")
b=list(a.replace(" ","").lower())
c=['a','e','i','o','u']
count=0
for i in b:
for j in c:
if (j==i):
count=count+1
print(count)/**
* @author: Rajdeep Roy Chowdhury<[email protected]>
* @github: https://github.com/razdeep
* @date: 25/12/2018
**/
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
// set count = 0
int count = 0;
string str;
cout << "/* ===== Number of Vowels ===== */" << endl;
cout << "\nEnter the string: ";
cin >> str;
// Convert input string to lower case
// using transform() function and ::tolower in STL
transform(str.begin(), str.end(), str.begin(), ::tolower);
// Run a loop from 0 to string length
for (int i = 0; i < str.length(); i++)
{
if (
str[i] == 'a' ||
str[i] == 'e' ||
str[i] == 'i' ||
str[i] == 'o' ||
str[i] == 'u')
{
count++;
}
}
// Print the result
cout<<"Number of vowels in \""<<str<<"\" = "<<count<<endl;
return 0;
}/*
* @author: imkaka
* @date: 25/12/2018
*/
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int count = 0;
string str;
cout << "/* ===== Number of Vowels ===== */" << endl;
cout << "\nEnter the string: ";
cin >> str;
// transform(str.begin(), str.end(), str.begin(), ::tolower);
for (int i = 0; i < str.size(); i++)
{
if (
tolower(str[i]) == 'a' ||
tolower(str[i]) == 'e' ||
tolower(str[i]) == 'i' ||
tolower(str[i]) == 'o' ||
tolower(str[i]) == 'u')
{
count++;
}
}
cout<<"Number of vowels in \""<<str<<"\" = "<<count<<endl;
return 0;
}
/**
* @author:divyakhetan
* @date: 30/12/2018
*/
#include<bits/stdc++.h>
using namespace std;
int main(){
string s1;
cin >> s1;
int count = 0;
for(int i = 0; i < s1.length(); i++){
if(s1[i] == 'a' || s1[i] == 'e' ||s1[i] == 'i' || s1[i] == 'o' ||s1[i] == 'u') count++;
}
cout << "Vowels are:" << count;
return 0;
}
# Input the String.
string=input("Enter the String : ")
# Create a list of vowels in lowercase.
vowels=['a','e','i','o','u']
# Initialize the count variable to zero.
count=0
# Now iterate every character in the string.
for char in string:
# And if the current character we are iterating is
# present in the vowels list then increment count.
if char.lower() in vowels:
count+=1
# Print the result
print("Number of vowels in the string are : ",count)"""
* @author: Shashank Jain
* @date: 25/12/2018
"""
a=input("Enter the string to count no. of vowels?")
b=list(a.replace(" ","").lower())
c=['a','e','i','o','u']
count=0
for i in b:
for j in c:
if (j==i):
count=count+1
print(count)/**
* @author : ashwek
* @date : 25/12/2018
*/
#include<stdio.h>
#include<ctype.h>
void main(){
char Str[50];
int i, count=0;
printf("Enter a string = ");
scanf("%s", Str);
for(i=0; Str[i]!='\0'; i++){
switch(tolower(Str[i])){
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
count++;
}
}
printf("number of vowels = %d", count);
}=begin
@author: aaditkamat
@date: 25/12/2018
=end
def count_vowels(str)
if str === nil or not str.class === String
-1
end
ctr = 0
str.downcase!
vowels = ['a', 'e', 'i', 'o', 'u']
vowels.each do |vowel|
ctr += str.count(vowel)
end
ctr
end
print "Enter a string: "
str = gets
str.chomp!
puts "The number of vowels in #{str} is : #{count_vowels(str)}"Question - Given a string, write a program to return the character that appears most frequently in that string
function maxChars (str) {
let letterCount = {};
// Complete the letter count object
for(let currentChar of str) {
let flag = 0;
for (letter in letterCount) {
if (currentChar.toLowerCase() === letter) {
flag = 1;
letterCount[letter]++;
}
}
if (flag === 0) {
letterCount[currentChar] = 1;
}
}
// Find the maximum value key from the letterCount
let max = 0, maxChar = '';
for (letter in letterCount) {
if (letterCount[letter] >= max) {
max = letterCount[letter];
maxChar = letter;
}
}
// Print the result
console.log(`The maximum frequency letter is: "${maxChar}" and it appears ${letterCount[maxChar]} number of times!`);
return letterCount[maxChar];
}
maxChars('helllllo worlld');function maxChars (sentence) {
let characters = {};
// Update characters object with count of each character in sentence
for (let letter of sentence) {
// increment the count of existing letter OR add the new character with value = 1
characters[letter] = characters[letter] + 1 || 1;
}
let max = 0,
maxChar = '';
for (let character in characters) {
if (max < characters[character]) {
max = characters[character];
maxChar = character;
}
}
// Print the result
console.log(`The maximum frequency letter is: "${maxChar}" and it appears ${characters[maxChar]} number of times!`);
return characters[maxChar];
}
maxChars('helllllo worlld');input_str = input()
dictnry = {}
maxCount, maxChar = 0, ''
for char in input_str:
"""
In this solution we are simply maintaining a dictionary of characters
in the input string as key and their count as value
"""
if char in dictnry.keys():
dictnry[char] += 1
else:
dictnry[char] = 1
"""
We check for maxCount of each character in a single loop
by comparing it with present maxCount, hence not iterating over the dictionary again
to find character with maximum count.
"""
if dictnry[char] > maxCount:
maxCount = dictnry[char]
maxChar = char
print(maxChar)# Input the string
string=input("Enter the string : ")
# Create an empty dictionary to store the frequency of characters
characters={}
# Iterate every character in the string
for char in string:
# And if the character already exists in the dictionary then
# increment its frequency by 1.
if char.lower() in characters:
characters[char.lower()]+=1
# Else initialize its frequency by 1
else:
characters[char.lower()]=1
# Print the character which has the maximum frequency
print("The most occouring character in the string is : ", max(characters,key=characters.get))###[Solution 3] (./Python/Shashankchar.py)
a=input("Enter the string to count frequent occuring characters?")
b=list(a.replace(" ","").lower())
c=[]
for i in b:
d=(i,b.count(i))
c.append(d)
e=dict(list(set(c)))
f=max(e)
g=max(e.values())
print("maximum occurence is of {0}:{1}".format(f,g))/**
* @author: Rajdeep Roy Chowdhury<[email protected]>
* @github: https://github.com/razdeep
* @date: 25/12/2018
*
* Max Char Problem Solution
*
**/
#include <iostream>
#include <algorithm>
#include <vector>
#include <climits>
using namespace std;
int main()
{
int count = 0;
// Input the string
string str;
cout << "/* ===== Number of Vowels ===== */" << endl;
cout << "\nEnter the string: ";
cin >> str;
// Convert input string to lower case
// using transform() function and ::tolower in STL
transform(str.begin(), str.end(), str.begin(), ::tolower);
int max_count = INT_MIN;
char max_label;
for (int i = 0; i < str.size(); i++)
{
int this_count = std::count(str.begin(), str.end(), str[i]);
if (this_count > max_count)
{
max_count = this_count;
max_label = str[i];
}
}
cout << "'" << (char)max_label << "' has " << max_count << " occurences." << endl;
return 0;
}/*
* @author : imkaka
* @date : 25/12.2018
*/
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
string str;
cout << "Enter the string: ";
cin >> str;
int max_count = -1;
char label;
for (int i = 0; i < str.size(); ++i)
{
int current_count = count(str.begin(), str.end(), str[i]);
if (current_count > max_count)
{
max_count = current_count;
label = str[i];
}
}
cout << "'" << (char)label << "' has MAX " << max_count << " occurences in " << str << endl;
return 0;
}
// Time Compllexity = O(size(str) ^2)
// We can Improve that by using hashing to which will increse space complexity to O(n)
// Its called time-space trade off, which we generally do most times./**
* @author:divyakhetan
* @date: 30/12/2018
*/
#include<bits/stdc++.h>
using namespace std;
int main(){
string s1;
cin >> s1;
int freqlo[26];
memset(freqlo, 0, sizeof(freqlo));
int freqhi[26];
memset(freqhi, 0, sizeof(freqhi));
int max = 0;
char maxc;
//count for lowercase
for(int i = 0; i < s1.length(); i++){
if(s1[i] >= 'a' && s1[i] <= 'z'){
freqlo[s1[i] - 'a']++;
if(freqlo[s1[i] - 'a']> max){
max = freqlo[s1[i] - 'a'];
maxc = s1[i];
}
}
else if(s1[i] >= 'A' && s1[i] <= 'Z'){
freqhi[s1[i] - 'A']++;
if(freqhi[s1[i] - 'A']> max){
max = freqhi[s1[i] - 'A'];
maxc = s1[i];
}
}
}
cout << "Max freq is:" << max << " for " << maxc;
return 0;
}# Input the string
string=input("Enter the string : ")
# Create an empty dictionary to store the frequency of characters
characters={}
# Iterate every character in the string
for char in string:
# And if the character already exists in the dictionary then
# increment its frequency by 1.
if char.lower() in characters:
characters[char.lower()]+=1
# Else initialize its frequency by 1
else:
characters[char.lower()]=1
# Print the character which has the maximum frequency
print("The most occouring character in the string is : ", max(characters,key=characters.get))"""
* @author: Shashank Jain
* @date: 25/12/2018
"""
a=input("Enter the string to count frequent occuring characters?")
b=list(a.replace(" ","").lower())
c=[]
for i in b:
d=(i,b.count(i))
c.append(d)
e=dict(list(set(c)))
f=max(e)
g=max(e.values())
print("maximum occurence is of {0}:{1}".format(f,g))/**
* @author : ashwek
* @date : 25/12/2018
*/
#include<stdio.h>
void main(){
char Str[50];
int i, Max = 0;
int possibleChar[94] = {0}; //ASCII 32-126
printf("Enter a string = ");
scanf("%[a-zA-Z-0-9 ]s", Str); //Modify control string to accept blank spaces in input
for(i=0; Str[i]!='\0'; i++){
possibleChar[ (int)Str[i] - 32]++;
if( possibleChar[Max] < possibleChar[ (int)Str[i] - 32] )
Max = (int)Str[i] - 32;
}
printf("Most frequent character = \'%c\'\n", (Max+32));
}=begin
@author: aaditkamat
@date: 25/12/2018
=end
def most_frequent_character(str)
if str === nil or not str.class === String
nil
end
counts = {}
str.downcase!
str.each_char do |ch|
if not counts.key?(ch)
counts[ch] = 1
else
counts[ch] = counts[ch] + 1
end
end
counts.key(counts.values.max)
end
print "Enter a string: "
str = gets
str.chomp!
puts "The most frequent character in #{str} is : #{most_frequent_character(str)}"The beauty of programming lies in the fact that there is never a single solution to any problem.
In case you have an alternative way to solve this problem, do contribute to this repository (https://github.com/CodeToExpress/dailycodebase) :)