Solving Compare Strings by Frequency of the Smallest Character in go. Please try yourself first to solve the problem and submit your implementation to LeetCode before looking into solution.

## Problem Description

Example 1:

``````Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").``````

Example 2:

``````Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").``````

Constraints:

• `1 <= queries.length <= 2000`

• `1 <= words.length <= 2000`

• `1 <= queries[i].length, words[i].length <= 10`

• 1 <= queries.length <= 2000

• 1 <= words.length <= 2000

• 1 <= queries[i].length, words[i].length <= 10

See the full details of the problem Compare Strings by Frequency of the Smallest Character at LeetCode

Originally posted at: @github.com/halfrost/LeetCode-Go