What is the main idea behind Redistribute Characters to Make All Strings Equal?

The main idea is counting each character's total frequency and checking if it is divisible by the number of strings to allow even redistribution.

Can I ignore character positions when solving this problem?

Yes, only the frequency of characters matters; positions within strings are irrelevant for equality.

What is the time complexity for this solution?

The time complexity is O(n * k) because each character in every string must be counted.

How does GhostInterview handle impossible cases?

It checks character count divisibility first, quickly returning false if any character cannot be evenly distributed.

Does this problem rely on a specific algorithm pattern?

Yes, it follows a Hash Table plus String pattern, where counting frequencies determines feasibility of redistribution.

#1897

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

重新分配字符使所有字符串都相等

给你一个字符串数组 words （下标从 0 开始计数）。在一步操作中，需先选出两个不同下标 i 和 j ，其中 words[i] 是一个非空字符串，接着将 words[i] 中的任一字符移动到 words[j] 中的任一位置上。如果执行任意步操作可以使 words 中的每个字符…

哈希表字符串计数

题目描述

给你一个字符串数组 words（下标 从 0 开始 计数）。

在一步操作中，需先选出两个不同下标 i 和 j，其中 words[i] 是一个非空字符串，接着将 words[i] 中的任一字符移动到 words[j] 中的任一位置上。

如果执行任意步操作可以使 words 中的每个字符串都相等，返回 true ；否则，返回 false 。

示例 1：

输入：words = ["abc","aabc","bc"]
输出：true
解释：将 words[1] 中的第一个 'a' 移动到 words[2] 的最前面。
使 words[1] = "abc" 且 words[2] = "abc" 。
所有字符串都等于 "abc" ，所以返回 true 。

示例 2：

输入：words = ["ab","a"]
输出：false
解释：执行操作无法使所有字符串都相等。

提示：

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] 由小写英文字母组成

lightbulb

解题思路

方法一：计数

根据题目描述，只要每个字符的出现次数能被字符串数组的长度整除，就可以通过移动字符使所有字符串相等。

因此，我们用哈希表或者一个长度为 $26$ 的整数数组 $\textit{cnt}$ 统计每个字符出现的次数，最后判断是否每个字符的出现次数能被字符串数组的长度整除即可。

时间复杂度 $O(L)$ ，空间复杂度 $O(|\Sigma|)$ 。其中 $L$ 为数组 $\textit{words}$ 中所有字符串的长度之和，而 $\Sigma$ 为字符集，这里为小写字母集合，所以 $|\Sigma|=26$ 。

1

2

3

4

5

6

7

8

9

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        cnt = Counter()
        for w in words:
            for c in w:
                cnt[c] += 1
        n = len(words)
        return all(v % n == 0 for v in cnt.values())

speed

复杂度分析

指标	值
时间	complexity is O(n * k) where n is the number of strings and k is the average string length, due to counting each character. Space complexity is O(1) because only 26 lowercase letters are counted regardless of input size.
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Notice that the order of characters in strings does not matter, only total counts matter.
question_mark
Check for divisibility by the number of strings as a quick feasibility check.
question_mark
Consider edge cases with strings of different lengths or empty strings.

warning

常见陷阱

外企场景

error
Ignoring the character frequency divisibility check and trying to simulate movements directly.
error
Assuming character positions matter rather than just counts.
error
Overlooking edge cases where a character total count is not divisible by the number of strings.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to include uppercase letters or digits, changing the hash table size.
arrow_right_alt
Limit the number of allowed moves and determine if equality is still achievable.
arrow_right_alt
Allow only adjacent character swaps and analyze feasibility differently.

help

常见问题

外企场景

继续练习

#1876 长度为三且各字符不同的子字符串 #1941 检查是否所有字符出现次数相同 #1790 仅执行一次字符串交换能否使两个字符串相等