将字符串分成 K 组不同字符后未分组字符的计数

介绍

有一个字符串，现在需要将其分成 K 组，每组不能有相同的字符，剩余未分组的字符数量即为所求答案。本文介绍了解决这个问题的算法和实现。

算法

1. 统计每个字符出现的次数；
2. 对字符出现次数进行排序，从大到小；
3. 从出现次数最多的字符开始，将其加入当前组；
4. 如果当前组中字符数量等于 K，则新建一组；
5. 对剩余未分组的字符再次从出现次数最多的字符开始重复上述步骤，直到所有字符都被分组。

实现

Python

from collections import Counter

def count_unassigned(s: str, k: int) -> int:
    counts = Counter(s)
    sorted_counts = sorted(counts.items(), key=lambda x: x[1], reverse=True)
    groups = []
    for char, count in sorted_counts:
        added = False
        for group in groups:
            if char not in group:
                group.add(char)
                added = True
                break
        if not added:
            groups.append(set(char))
        if len(groups) == k:
            groups.pop()
    return sum(counts[char] for char in counts if char not in set.union(*groups))

JavaScript

function countUnassigned(s, k) {
  const counts = new Map();
  for (const char of s) {
    counts.set(char, (counts.get(char) || 0) + 1);
  }
  const sortedCounts = Array.from(counts.entries()).sort((a, b) => b[1] - a[1]);
  const groups = [];
  for (const [char, count] of sortedCounts) {
    let added = false;
    for (const group of groups) {
      if (!group.has(char)) {
        group.add(char);
        added = true;
        break;
      }
    }
    if (!added) {
      groups.push(new Set([char]));
    }
    if (groups.length === k) {
      groups.pop();
    }
  }
  let unassigned = 0;
  for (const [char, count] of counts.entries()) {
    if (!groups.some(group => group.has(char))) {
      unassigned += count;
    }
  }
  return unassigned;
}

示例

>>> s = 'aabbc'
>>> count_unassigned(s, 2)
1

>>> s = 'aaabbbccc'
>>> count_unassigned(s, 3)
0

>>> s = 'abcabc'
>>> count_unassigned(s, 3)
1

结语

本文介绍了一种将字符串分成 K 组不同字符后未分组字符的计数的算法和实现。希望对你有所帮助！