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  1. Prepare
  2. Algorithms
  3. Strings
  4. Alternating Characters
  5. Discussions

Alternating Characters

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1955 Discussions

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  • busragokmen671
     + 0 comments
        count = 0
    i = 0
    for i in range(i, len(s)-1):
       if s[i] == s[i+1]:
          count = count + 1         
    return count

  • alban_tyrex
     + 0 comments

    Here are my c++ approaches of solving this, video explanation here : https://youtu.be/KoDlxS38_ig

    Solution 1 : Using loop

    int alternatingCharacters(string s) {
    int result = 0;
    for(int i = 1; i < s.size(); i++) if(s[i] == s[i-1]) result++;
    return result;
}

    Solution 2 : Using regex

    int alternatingCharacters(string s) {
    regex re("A{2,}|B{2,}");
    return s.size() - regex_replace(s, re, "*").size();
}

  • mukeshprajapati6
     + 0 comments

    def alternatingCharacters(s): n = len(s) count = 0 for i in range(n-1): if s[i] == s[i+1]: count += 1 return count

  • misterjesusingl1
     + 0 comments

    php

    function alternatingCharacters($s) {
    // Write your code here
    $stack = [];
    $deletions = 0;
    for ($i = 0; $i < strlen($s); $i++) {
        if (count($stack) === 0) {
            $stack[] = $s[$i];
        } else {
            $last = $stack[count($stack) - 1];
            if ($last === $s[$i]) {
                $deletions++;
            } else {
                $stack[] = $s[$i];
            }
        }
    }
    return $deletions;
}

  • maneth_21
     + 0 comments
    public static int alternatingCharacters(String s) {
    // Write your code here
    int count = 0 ;
    for (int i = 1 ; i < s.length(); i++){
        if(s.charAt(i) == s.charAt(i-1)){
            count +=1;
        }
    }
    return count;

    }