Make Letter Counts Equal

Hard

You are given a string text of lowercase English letters. Call a string **even** if every letter that appears in it appears the same number of times. You may apply the following actions as many times as you like, each costing 1: - delete any letter; - insert any letter at any position; - replace a letter with the **next** one in the alphabet (for example, c with d). The letter z cannot be replaced this way: the alphabet does not wrap around. Return the smallest number of actions that makes text even.

Example 1

text='xxyyy' → 1

Delete one y to get "xxyy", where x and y each appear twice.

Example 2

text='mmmmnn' → 1

Replace one m with the next letter, n: in "mmmnnn" both letters appear three times. One action is enough.

Example 3

text='kkmm' → 0

The letters k and m already appear twice each, so nothing needs to change.

Constraints

3 ≤ text.length ≤ 2 · 10⁴
text consists of lowercase English letters only

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Topics

Hash Table String Dynamic Programming Counting Enumeration

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Input

text='xxyyy'

Expected

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