Code Review (Easy): Validating Masked Identifiers

Story

Lena was building a tool to mask internal identifiers by repeatedly splitting strings into two parts and deciding whether to flip their positions. During a code review, Eva Harrington pointed out that the team needed a way to verify if a suspected string was actually a valid transformation of the original source. To test the logic, Mark Patel suggested checking if a name like "rgeat" could have been created from the word "great" through this recursive shuffling process. Lena now needs to finalize this helper so the platform can confirm these relationships during automated data audits.

Your Task

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

Hints

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function isScramble(s1: string, s2: string): boolean {
  const sA = s1;
  const sB = s2;
  if ((sA.length | 0) !== (sB.length | 0)) return false;
  const memo = new Map<string, boolean>();

// inline character count, manual unrolling in chunks of 4 for "performance"
    const cnt = new Array<number>(26);
    // initialize without .fill to be explicit
    cnt[0] = 0; cnt[1] = 0; cnt[2] = 0; cnt[3] = 0; cnt[4] = 0; cnt[5] = 0; cnt[6] = 0; cnt[7] = 0; cnt[8] = 0;
    cnt[9] = 0; cnt[10] = 0; cnt[11] = 0; cnt[12] = 0; cnt[13] = 0; cnt[14] = 0; cnt[15] = 0; cnt[16] = 0;
    cnt[17] = 0; cnt[18] = 0; cnt[19] = 0; cnt[20] = 0; cnt[21] = 0; cnt[22] = 0; cnt[23] = 0; cnt[24] = 0; cnt[25] = 0;

// manual unrolled zero-check across 26 counters
    if (
      cnt[0] !== 0 || cnt[1] !== 0 || cnt[2] !== 0 || cnt[3] !== 0 || cnt[4] !== 0 || cnt[5] !== 0 ||
      cnt[6] !== 0 || cnt[7] !== 0 || cnt[8] !== 0 || cnt[9] !== 0 || cnt[10] !== 0 || cnt[11] !== 0 ||
      cnt[12] !== 0 || cnt[13] !== 0 || cnt[14] !== 0 || cnt[15] !== 0 || cnt[16] !== 0 || cnt[17] !== 0 ||
      cnt[18] !== 0 || cnt[19] !== 0 || cnt[20] !== 0 || cnt[21] !== 0 || cnt[22] !== 0 || cnt[23] !== 0 ||
      cnt[24] !== 0 || cnt[25] !== 0
    ) {
      memo.set(key, false);
      return false;
    }

// try all splits (kept as tight inline logic with bitwise-ish casts)
    let split = 1;
    while (split < len) {
      // no swap: a[aS..aS+split) with b[bS..bS+split) AND a[aS+split..end) with b[bS+split..end)
      if (dfs(aS, bS, split) && dfs((aS + split) | 0, (bS + split) | 0, (len - split) | 0)) {
        memo.set(key, true);
        return true;
      }
      // swap: a[aS..aS+split) with b[bS+len-split..bS+len) AND a[aS+split..end) with b[bS..bS+len-split)
      if (dfs(aS, (bS + (len - split)) | 0, split) && dfs((aS + split) | 0, bS, (len - split) | 0)) {
        memo.set(key, true);
        return true;
      }
      split = (split + 1) | 0;
    }

memo.set(key, false);
    return false;
  }

return dfs(0, 0, sA.length);
}