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  1. Misère Nim
  2. Discussions

Misère Nim

  • AmCharlie
     + 0 comments

    If you're familiar with Nim-Sum, this problem will be very easy; otherwise, it's incredibly difficult. This type of problem should be used to test well-read individuals, not programmers.

        let allOnes = true;
    
    for (let stone of stones) {
        xorSum ^= stone;
        if (stone > 1) {
            allOnes = false;
        }
    }
    
    if (allOnes) {
        return stones.length % 2 === 0 ? "First" : "Second";
    } else {
        return xorSum === 0 ? "Second" : "First";
    }