Discussion on Alternating Characters Challenge

Sort by

recency

|

1971 Discussions

|

2 weeks ago+ 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();
}

1 month ago+ 0 comments

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

2 months ago+ 0 comments

Here is a Python3 implemention. It is intentionally verbose to ensure explainability:

from collections import Counter
def isValid(orig_string):
    # Create a dict with the count of all characters
    string_collect = Counter(list(orig_string))
    str_c = sorted(string_collect.values())
    
    # If the first and last are the same then all characters are equal
    if str_c[0] == str_c[-1]:
        return "YES"
        
    # If the largest count is only one larger than every other value
    elif str_c[-1] == str_c[-2] + 1 and str_c[0] == str_c[-2]:
        return "YES"
        
    # If the smallest count is only one smaller than every other value OR the smallest count is 1 (therefore can be removed completely)
    elif (str_c[1] == str_c[0] + 1 or str_c[0] == 1)and str_c[1] == str_c[-1]:
        return "YES"
        
    else:
        return "NO"

2 months ago+ 0 comments

Java Solution for this with single forloop with O(n) Complexity public static int alternatingCharacters(String s) { // Write your code here StringBuffer buffer = new StringBuffer(); int count = 0; for(int i = 1; i < s.length(); i++){ if(s.charAt(i) == s.charAt(i - 1)){ count++; } } return count; }``

2 months ago+ 0 comments

def alternatingCharacters(s):

    # Write your code here
deletions=0
for i in range(1,len(s)):
    if s[i-1]==s[i]:
        deletions+=1
return deletions

Sort by

|

1971 Discussions

|

Cookie support is required to access HackerRank

Please signup or login in order to view this challenge