We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Mars Exploration
  2. Discussions

Mars Exploration

Sort by

recency

|

207 Discussions

|

  • nikhilsaji2k
     + 0 comments
    def marsExploration(s):
    sos_count = len(s)//3
    original_sos = sos_count * "SOS"
    changed_chars = sum(1 for i in range(len(s)) if s[i] != original_sos[i])
    return changed_chars

  • asif2000nm
     + 0 comments
    def marsExploration(s):
    # Write your code here
    testStr = "SOS"
    lengthOfTestStr = len(testStr)
    index = 0
    result = 0
    for char in s:
        if char != testStr[index % lengthOfTestStr]:
            result += 1
        index += 1
    return result

  • inq_daniel_eliz1
     + 0 comments
    def marsExploration(s):
    count = 0
    segments = re.findall(r'...', s)
    print(segments)
    
    for i in segments:
        for j, l in enumerate('SOS'):
            if l != i[j]:
                count += 1
                
    return count

  • alfredocapasso73
     + 0 comments

    Js solution:

    return s.split('').reduce((accumulator ,letter, currentIndex: number) => {
    if(
        ((currentIndex%3 === 0 || currentIndex%3 === 2) && letter !== 'S') ||
        (currentIndex%3 === 1 && letter !== 'O')
    ){
        return accumulator += 1;
    }
    return accumulator;
}, 0);

    }

  • annafedoseeva201
     + 0 comments

    counter = 0 for i in range(0, len(s), 3): if s[i] !='S': counter+=1 if s[i+1] != 'O': counter+=1 if s[i+2] !='S': counter+=1 return counter