The main problem in counts lies within your conditions.
You're increasing the number of consonants whenever one of the conditions fail [that's what || does, known as the OR operator]. So whenever a character !== "a" OR !== "e", you're increasing the count, which is wrong. [Imagine that a is !== 'e', so you're counting a as a consonant].
Change the || binary operator to && [AND]; this way, you're only increasing the consonant count
whenever the current character str[i] is not among the values you're verifying for [a, e, i, o, u, ' '].
As others pointed out, you're also likely to run into an error as the max value of i should be Length-1.
There also other problems you need to consider:
- What happens when the character is an uppercase letter?
- What happens when the character is a punctuation mark or a number?
For a beginner, this may not be relevant, but it's well worth getting these techniques under your skin: It's more readable to create an array that contains all of the value you're verifying for ["a","e", etc] and then, for each char in the source string, just verify if array.indexOf[str[i]] >= 0 [which means that the character is included in the array].
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Given a string, count total number of consonants in it. A consonant is a English alphabet character that is not vowel [a, e, i, o and u]. Examples of constants are b, c, d, f, g, ..
Examples :
Input : abc de Output : 3 There are three consonants b, c and d. Input : geeksforgeeks portal Output : 12
1. Iterative Method
C++
#include
using namespace std;
bool isConsonant[char ch]
{
ch = toupper[ch];
return ![ch == 'A' || ch == 'E' ||
ch == 'I' || ch == 'O' ||
ch == 'U'] && ch >= 65 && ch
Javascript
function isConsonant[ch]
{
ch = ch.toUpperCase[];
console.log[ch];
return [
![ch == "A"
|| ch == "E" || ch == "I" || ch == "O" || ch == "U"] &&
ch.match[/[A-Z]/i]
];
}
function totalConsonants[str] {
var count = 0;
for [var i = 0; i < str.length; i++]
if [isConsonant[str[i]]] ++count;
return count;
}
var str = "abc de";
document.write[totalConsonants[str]];
Time Complexity: O[n],
where n is the length of the string
Auxiliary Space: O[1]
2. Recursive Method
C++
#include
using namespace std;
bool isConsonant[char ch]
{
ch = toupper[ch];
return ![ch == 'A' || ch == 'E' ||
ch == 'I' || ch == 'O' ||
ch == 'U'] && ch >= 65 && ch