Project Euler Problems #18 and #67

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The new solution for Problem #18 (compared to the old one) is more elegant: I reverse and reduce the triangle in a manner that doesn’t require me to check. The solution for #67 is a bit more longform, so I turned the method I use here into an NPM package: Triangutron.

const eighteen = `
    75
    95 64
    17 47 82
    18 35 87 10
    20 04 82 47 65
    19 01 23 75 03 34
    88 02 77 73 07 63 67
    99 65 04 28 06 16 70 92
    41 41 26 56 83 40 80 70 33
    41 48 72 33 47 32 37 16 94 29
    53 71 44 65 25 43 91 52 97 51 14
    70 11 33 28 77 73 17 78 39 68 17 57
    91 71 52 38 17 14 91 43 58 50 27 29 48
    63 66 04 68 89 53 67 30 73 16 69 87 40 31
    04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
`;


function parse(string) {
    return string.trim().split('\n').map(row => {
        return row.trim().split(' ').map(parseFloat);
    }).filter(row => {
        return row.every(Number.isInteger);
    });
}

function max(lastRow, row) {
    return row.map((cell, index) => {
        return cell + Math.max(lastRow[index], lastRow[index + 1]);
    });
}

const triangle = parse(eighteen).reverse();
console.log(triangle.slice(1).reduce(max, triangle[0])[0]);


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