我正在嘗試使用打字稿解決這個版本的通配符問題

Question

我正在嘗試使用打字稿解決這個版本的通配符問題。 但是，我提供的一些測試用例沒有通過。 這是代碼如何工作的總結。

平衡字符串是字符串中每個字符與其他字符出現次數相等的字符串

例如，“ab”、“aaabbb”和“ababaabb”是平衡的，但“abb”和“abbaa”不是。

function balanced(s: string): boolean {
  const MAX = 1000;
  let wildcards: number = 0;
  const map: { [key: string]: number } = {};
  let characterCount = 0;
  for (const char of s) {
    if (char === '*') {
      wildcards++;
    } else {
      if (!map[char]) {
        map[char] = 0;
      }
      map[char]++;
      if (map[char] > characterCount) {
        characterCount = map[char];
      }
    }
  }
  const mapSize = Object.keys(map).length;
  if (mapSize === MAX && characterCount === 1) {
    return wildcards % MAX === 0;
  }
  if (mapSize < MAX && characterCount === 1 && wildcards <= (MAX - mapSize)) {
    return true;
  }
  if (wildcards === s.length) {
    return true;
  }

  for (const char in map) {
    while (map[char] + 1 <= characterCount) {
      map[char]++;
      wildcards--;
    }
    if (wildcards < 0) return false;
  }
  if (wildcards % characterCount === 0) {
    return true;
  }
  return false;
}

我正在針對以下輸入測試代碼

1: "a" should return true
2: "ab" should return true
3: "abc" should return true
4: "abcb" should return false
5: "Aaa" should return false
6: "abcb*" should return false
7: "abcb**" shoud return true
8: "***********" should return true
9: "" should return true
10: "abd*xdx*yba*" should return true
11: "aabb***" should return false
12: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" should return false
13: "JB**JTIT*****EY" should return false

Answer 1

12: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" should return false

不，它應該返回true所有其他測試通過

https://stackblitz.com/edit/vitest-dev-vitest-afgmgj?file=balanced.test.ts&initialPath=__vitest_ _

Answer 2

我發現了類似的問題。 我只是稍微扭曲了解決方案。

const factorPairs = (n: number) =>
  [...Array(Math.floor(Math.sqrt(n)))].map((_, i) => i + 1)
    .flatMap(f => n % f == 0 ? [[f, n / f], [n / f, f]] : []);

const balanced = (s: string): boolean => {
  if (!s) return true;
  const ss = s.split('');
  const chars = [...new Set(ss.filter(s => s !== '*'))];
  const counts = ss.reduce(
    (counts, s) => s === '*' ? counts : ((counts[s] += 1), counts),
    Object.fromEntries(chars.map(l => [l, 0]))
  );
  const maxCount = Math.max(...Object.values(counts));
  return factorPairs(ss.length).some(
    ([count, letters]) =>
      letters <= 52 &&
      letters >= chars.length &&
      count >= maxCount
  );
};

鏈接到原始解決方案