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  2. Algorithms
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  4. Mars Exploration
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Mars Exploration

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  • geovanni_luna
     + 0 comments

    Python solution with O(n/3) time and O(1) time. I am not sure if there is a more efficient solution. If anyone can find something more efficient please comment:

    def marsExploration(s):
    i = 2
    res = 0
    while i < len(s):
        if s[i - 2] != 'S':
            res += 1
        if s[i - 1] != 'O':
            res += 1
        if s[i] != 'S':
            res += 1
            
        i += 3
    
    return res

  • narias_as
     + 0 comments

    For C#

    return string.Concat(Enumerable.Range(0, s.Length / 3) .Select(s => "SOS")) .Select((c, i) => c != s[i] ? 1 : 0) .Sum();

  • yashdeora98294
     + 0 comments

    Here is problem solution in python java c++ c and Javascript - https://programmingoneonone.com/hackerrank-mars-exploration-problem-solution.html

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can find the explanation here : https://youtu.be/Gf0fpGE83r4

    int marsExploration(string s) {
   int result = 0;
    string base = "SOS";
    for(int i = 0; i < s.size(); i++) if(s[i] != base[i%3]) result++;
    return result;
}

  • robertbielicki
     + 0 comments
    fn marsExploration(s: &str) -> i32 {
    let temp: Vec<char> = s.chars().collect();
    let mut valid = 0;

    for (idx, c) in temp.iter().enumerate() {
        if (idx % 3 == 0 || idx % 3 == 2) && *c == 'S' {
            valid += 1;
        } else if idx % 3 == 1 && *c == 'O' {
            valid += 1;
        }
    }
    s.len() as i32 - valid
}