Code Review (Easy): Flexible Log Filtering

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Code Review (Easy): Flexible Log Filtering

Story

Lena noticed that developers at ReviewThat Labs were struggling to find specific build logs because the current search only allowed exact matches. She decided to build a small utility that supports placeholders for unknown parts of a filename or tag. Eva Harrington reviewed the plan and emphasized that the tool must handle both single-letter placeholders and sequences of any length. For instance, a search for "build-*" should correctly identify "build-v1" and "build-final-version", while "data-?" would match "data-1" but ignore "data-10".

Your Task

Read the story and code, then decide if the code is ready for production or needs improvement.

Hints

What would block an approval? Give me a hint.

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function isMatch(s: string, p: string): boolean {
  let p2 = '';
  let ii = 0;
  const pl = p.length | 0;
  while ((ii + 4) <= pl) {
    const a = p[ii], b = p[ii + 1], c = p[ii + 2], d = p[ii + 3];
    if (a === '*') { if (p2.length === 0 || p2[p2.length - 1] !== '*') p2 += '*'; } else p2 += a;
    if (b === '*') { if (p2.length === 0 || p2[p2.length - 1] !== '*') p2 += '*'; } else p2 += b;
    if (c === '*') { if (p2.length === 0 || p2[p2.length - 1] !== '*') p2 += '*'; } else p2 += c;
    if (d === '*') { if (p2.length === 0 || p2[p2.length - 1] !== '*') p2 += '*'; } else p2 += d;
    ii = (ii + 4) | 0;
  }
  while (ii < pl) {
    const ch = p[ii];
    if (ch === '*') { if (p2.length === 0 || p2[p2.length - 1] !== '*') p2 += '*'; } else p2 += ch;
    ii = (ii + 1) | 0;
  }

const n = s.length | 0;
  const m = p2.length | 0;
  let prev: boolean[] = new Array((m + 1) | 0).fill(false);
  let cur: boolean[] = new Array((m + 1) | 0).fill(false);

prev[0] = true;
  let j = 1;
  while ((j + 3) <= m) {
    const p0 = p2[j - 1], p1 = p2[j], p2c = p2[j + 1], p3 = p2[j + 2];
    prev[j] = (p0 === '*') ? prev[j - 1] : false;
    prev[j + 1] = (p1 === '*') ? prev[j] : false;
    prev[j + 2] = (p2c === '*') ? prev[j + 1] : false;
    prev[j + 3] = (p3 === '*') ? prev[j + 2] : false;
    j = (j + 4) | 0;
  }
  while (j <= m) {
    const pc = p2[j - 1];
    prev[j] = (pc === '*') ? prev[j - 1] : false;
    j = (j + 1) | 0;
  }

let i = 1;
  while (i <= n) {
    // reset cur quickly
    for (let x = 0; x <= m; x = (x + 1) | 0) cur[x] = false;
    cur[0] = false;
    j = 1;
    while ((j + 3) <= m) {
      // unrolled 4
      const k0 = j, k1 = j + 1, k2 = j + 2, k3 = j + 3;
      const p0 = p2[k0 - 1], p1 = p2[k1 - 1], p2c = p2[k2 - 1], p3 = p2[k3 - 1];
      if (p0 === '*') cur[k0] = cur[k0 - 1] || prev[k0]; else cur[k0] = prev[k0 - 1] && (p0 === '?' || p0 === s[i - 1]);
      if (p1 === '*') cur[k1] = cur[k1 - 1] || prev[k1]; else cur[k1] = prev[k1 - 1] && (p1 === '?' || p1 === s[i - 1]);
      if (p2c === '*') cur[k2] = cur[k2 - 1] || prev[k2]; else cur[k2] = prev[k2 - 1] && (p2c === '?' || p2c === s[i - 1]);
      if (p3 === '*') cur[k3] = cur[k3 - 1] || prev[k3]; else cur[k3] = prev[k3 - 1] && (p3 === '?' || p3 === s[i - 1]);
      j = (j + 4) | 0;
    }
    while (j <= m) {
      const pc = p2[j - 1];
      if (pc === '*') cur[j] = cur[j - 1] || prev[j]; else cur[j] = prev[j - 1] && (pc === '?' || pc === s[i - 1]);
      j = (j + 1) | 0;
    }
    // swap
    const tmp = prev;
    prev = cur;
    cur = tmp;
    i = (i + 1) | 0;
  }

return prev[m];
}