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  1. Mars Exploration
  2. Discussions

Mars Exploration

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204 Discussions

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  • 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

  • nwauwa_kelechi_e
     + 0 comments

    My TypeScript solution:

    function marsExploration(s: string): number {
    const length = s.length;
    let changedChars = 0;
    
    for (let i = 0; i < length; i++) {
        const originalChar = (
            i % 3 === 0 ||
            (i + 1) % 3 === 0
        )  ? "S" : "O";
        if (s[i] !== originalChar) changedChars ++;
        
    }
    
    return changedChars;
}

  • outsider973
     + 0 comments

    My rust solution:

     const SOS_LEN: usize = 3;

fn marsExploration(s: &str) -> i32 {
    (0..s.len())
        .step_by(3)
        .map(|i| {
            s[i..i + 3].bytes()
                .zip([b'S', b'O', b'S'])
                .filter(|&zip| zip.0 != zip.1)
                .count() as i32
        })
        .sum()
}

  • vdaons
     + 0 comments

    Python 1 liner.

        return len([c for i, c in enumerate(s) if c != 'SOS'[i % 3]])