Why is the code is not removing “u” from the input string?

Question

I am learning C++. Today I have written a code to remove vowels form a string. It works fine in some tests. But this test is failing to remove "u" from a string. M

Answer 1

Here's how you can do it more easily:

#include <algorithm>
#include <array>
#include <cctype>
#include <iostream>
#include <string>

constexpr auto isVowel(unsigned char const ch) noexcept {
  constexpr std::array<unsigned char, 10> vowels{
      'a', 'e', 'i', 'o', 'u',
  };
  return std::find(vowels.begin(), vowels.end(), std::tolower(ch)) !=
         vowels.end();
}

int main() {
  std::string str = "Tour";
  str.erase(std::remove_if(str.begin(), str.end(), isVowel), str.end());
  std::cout << str << '\n';
}

Try online

Answer 2

After erase, i++ is performed. That means one element is bypassed by the check.

Change the for loop to

for (int i = 0; i < word.length(); )
{
    if (word[i] == 'a' || word[i] == 'e' || word[i] == 'i' || word[i] == 'o' || word[i] == 'u')
    {
        word.erase(word.begin() + i);  // Removing specific character
    } 
    else 
    {
        i++; // increment when no erasion is performed
    }
}

Answer 3

When you erase a character, your index i is still being incremented, which means you are skipping checking characters that appear right after a vowel.

You need to decrement i when you erase a character:

if (word[i] == 'a' || word[i] == 'e' || word[i] == 'i' || word[i] == 'o' || word[i] == 'u')
{
    word.erase(word.begin() + i);  // Removing specific character
    --i;  // make sure to check the next character
}

Here's a demo.

Also, please avoid the following 2 lines:

#include <bits/stdc++.h>
using namespace std;

They are dangerous and should not be used.