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1737. Change Minimum Characters to Satisfy One of Three Conditions

You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter.

Your goal is to satisfy one of the following three conditions:

• Every letter in a is strictly less than every letter in b in the alphabet.
• Every letter in b is strictly less than every letter in a in the alphabet.
• Botha and b consist of only one distinct letter.

Return the minimum number of operations needed to achieve your goal.

Example 1:

Input: a = "aba", b = "caa" Output: 2 Explanation: Consider the best way to make each condition true: 1) Change b to "ccc" in 2 operations, then every letter in a is less than every letter in b. 2) Change a to "bbb" and b to "aaa" in 3 operations, then every letter in b is less than every letter in a. 3) Change a to "aaa" and b to "aaa" in 2 operations, then a and b consist of one distinct letter. The best way was done in 2 operations (either condition 1 or condition 3). 

Example 2:

Input: a = "dabadd", b = "cda" Output: 3 Explanation: The best way is to make condition 1 true by changing b to "eee". 

Constraints:

• 1 <= a.length, b.length <= 105
• a and b consist only of lowercase letters.

Solutions (Rust)

1. Solution

implSolution{pubfnmin_characters(a:String,b:String) -> i32{letmut ret = a .chars().chain(b.chars()).filter(|&chab| chab != 'a').count();for ch in'b'..='z'{ ret = ret.min( a.chars().filter(|&cha| cha >= ch).count() + b.chars().filter(|&chb| chb < ch).count(),); ret = ret.min( a.chars().filter(|&cha| cha < ch).count() + b.chars().filter(|&chb| chb >= ch).count(),); ret = ret.min( a.chars().chain(b.chars()).filter(|&chab| chab != ch).count(),);} ret asi32}}