Find longest occurrence of same number in array

Question

Using JavaScript, I\'m trying to find a way to find the longest occurrence of the same number (in this case, 1) in an array.

For instance, here\'s a sample array:

Answer 1

Input array:

const seq = [
0, 0, 0,
1, 1, 1,
1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 1, 1,
];

Shortest solutions:

console.log(Math.max(...Array.from(seq.join("").matchAll(/(.)\1+/g), m=>m[0].length)))

Alternative with regexp (spoiler: it's ~25%, slower than solution with reduce(). See "Modern approach with reduce()" below):

const longestSeq = (seq) => {
    let max = 0;
    seq.join("").replace(/(.)\1+/g, m=> max = Math.max(max, m.length));
    return max;
};

Straightforward, old-school style, human readable and fastest solution:

let longestSeq = () => {
    let maxCount = 0,
        curCount = 0,
        curItem, prevItem,
        l = seq.length+2, // +1+1 to finish last sequence and compare 'undefined' with previous
        i = 0;
    for (; i < l; ++i) {
      curItem = seq[i];
      if (curItem === prevItem) ++curCount;
      else {
        if (curCount > maxCount) maxCount = curCount;
        curCount = 1;
        prevItem = curItem;
      }
    }
    return maxCount;
}

Modern approach with reduce() (just very little slower than old-school code above):

const longestSeq = (seq) => seq
    .reduce(
        ({count, max}, item) => item === 0
        ? { count: ++count, max: Math.max(count, max) }
        : { count: 0, max: max },
      { count: 0, max: 0} )
    .max;

Performance test, Reduce() vs old-school for(): https://jsbench.me/ifkgsin56z/1