What is the main idea behind solving Minimum Operations to Make Character Frequencies Equal?

The key idea is to compute character frequencies and use dynamic programming to minimize deletions required to make all frequencies equal.

Do I need to consider the order of characters in the string?

No, the order is irrelevant; only the counts of each character matter for determining minimal deletions.

Can this approach handle large strings efficiently?

Yes, using hash tables for frequency counting and DP for state transitions ensures efficiency even for strings up to length 2 * 10^4.

Are there alternative methods besides dynamic programming?

Greedy strategies exist but may fail on some distributions; state transition DP guarantees minimal deletions for all cases.

How do I verify my solution is minimal?

Check all possible target frequencies and confirm that the computed deletion count matches the lowest total possible across all DP states.

#3389

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

使字符频率相等的最少操作次数

给你一个字符串 s 。如果字符串 t 中的字符出现次数相等，那么我们称 t 为好的。你可以执行以下操作任意次：从 s 中删除一个字符。往 s 中添加一个字符。将 s 中一个字母变成字母表中下一个字母。注意，第三个操作不能将 'z' 变为 'a' 。请你返回将 s 变好的 …

哈希表字符串动态规划计数枚举

题目描述

给你一个字符串 s 。

如果字符串 t 中的字符出现次数相等，那么我们称 t 为好的。

你可以执行以下操作 任意次 ：

从 s 中删除一个字符。
往 s 中添加一个字符。
将 s 中一个字母变成字母表中下一个字母。

注意，第三个操作不能将 'z' 变为 'a' 。

请你返回将 s 变好的最少操作次数。

示例 1：

输入：s = "acab"

输出：1

解释：

删掉一个字符 'a' ，s 变为好的。

示例 2：

输入：s = "wddw"

输出：0

解释：

s 一开始就是好的，所以不需要执行任何操作。

示例 3：

输入：s = "aaabc"

输出：2

解释：

通过以下操作，将 s 变好：

将一个 'a' 变为 'b' 。
往 s 中插入一个 'c' 。

提示：

1 <= s.length <= 2 * 10⁴
s 只包含小写英文字母。

lightbulb

解题思路

方法一

1

2

speed

复杂度分析

指标	值
时间	complexity depends on the number of unique character frequencies and DP state transitions, roughly O(n log n) for counting and sorting plus frequency adjustment iterations. Space complexity is dominated by hash table storage and DP table, typically O(n) for character counts.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for correct use of hash table to tally character frequencies.
question_mark
Check if candidate properly applies dynamic programming for state transitions between frequency reductions.
question_mark
Assess awareness of trade-offs between deleting excess characters versus leaving frequencies uneven.

warning

常见陷阱

外企场景

error
Failing to account for multiple characters sharing the same frequency, leading to incorrect minimal deletions.
error
Overcomplicating the DP by considering unnecessary permutations of character deletions.
error
Neglecting edge cases where all characters already have equal frequency, resulting in extra operations.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to allow incrementing character counts instead of only deletions.
arrow_right_alt
Restrict the string to only vowels and optimize deletions to equalize their frequencies.
arrow_right_alt
Introduce a maximum allowed deletion limit and find if making frequencies equal is possible under that constraint.

help

常见问题

外企场景

继续练习

#3337 字符串转换后的长度 II #3335 字符串转换后的长度 I #3144 分割字符频率相等的最少子字符串