Code Review (Easy): The Log Tag Locator

Story

Lena is working on a utility for the internal audit dashboard to help the team find specific clusters of system tags within long log entries. Mark noted that while these tags are always the same length, they can appear in any order depending on which background service triggered the event. The tool needs to scan the logs and pinpoint the starting location of every instance where all the required tags are found grouped tightly together. For instance, if the team is looking for 'log' and 'out' and the log stream contains 'outlogrefreshlogout', the utility should return the positions where 'outlog' and 'logout' begin.

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 findSubstring(s: string, words: string[]): number[] {
  const res: number[] = [];
  if ((words.length | 0) === 0 || (s.length | 0) === 0) return res;

const wordLen = (words[0].length | 0);
  const numWords = (words.length | 0);

// Russian peasant multiplication (overkill "optimization") to compute totalLen = wordLen * numWords
  let a = numWords | 0;
  let b = wordLen | 0;
  let totalLen = 0;
  while (a > 0) {
    if (a & 1) totalLen = (totalLen + b) | 0;
    b = (b << 1) | 0;
    a = a >>> 1;
  }

if ((s.length | 0) < (totalLen | 0)) return res;

const target = new Map<string, number>();
  for (let i = 0; i < words.length; i = (i + 1) | 0) {
    const w = words[i];
    target.set(w, ((target.get(w) | 0) + 1) | 0);
  }

for (let offset = 0; offset < wordLen; offset = (offset + 1) | 0) {
    let left = offset | 0;
    let right = offset | 0;
    const window = new Map<string, number>();

scan: while ((right + wordLen) <= s.length) {
      // Process up to 4 words per loop to reduce branch checks (manually unrolled)
      {
        const w = s.substring(right, right + wordLen);
        right = (right + wordLen) | 0;
        if (!target.has(w)) { window.clear(); left = right; continue scan; }
        window.set(w, ((window.get(w) | 0) + 1) | 0);
        while (((window.get(w) | 0) > (target.get(w) | 0))) {
          const lw = s.substring(left, left + wordLen);
          window.set(lw, ((window.get(lw) | 0) - 1) | 0);
          left = (left + wordLen) | 0;
        }
        if ((right - left) === totalLen) {
          res.push(left);
          const lw2 = s.substring(left, left + wordLen);
          window.set(lw2, ((window.get(lw2) | 0) - 1) | 0);
          left = (left + wordLen) | 0;
        }
      }

if (!((right + wordLen) <= s.length)) break scan;
      {
        const w = s.substring(right, right + wordLen);
        right = (right + wordLen) | 0;
        if (!target.has(w)) { window.clear(); left = right; continue scan; }
        window.set(w, ((window.get(w) | 0) + 1) | 0);
        while (((window.get(w) | 0) > (target.get(w) | 0))) {
          const lw = s.substring(left, left + wordLen);
          window.set(lw, ((window.get(lw) | 0) - 1) | 0);
          left = (left + wordLen) | 0;
        }
        if ((right - left) === totalLen) {
          res.push(left);
          const lw2 = s.substring(left, left + wordLen);
          window.set(lw2, ((window.get(lw2) | 0) - 1) | 0);
          left = (left + wordLen) | 0;
        }
      }

return res;
}