# Solving Leetcode Interviews in Seconds with AI: Compare Strings by Frequency of the Smallest Character


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1170" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2. You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words. Return an integer array answer, where each answer[i] is the answer to the ith query.   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 queries[i][j], words[i][j] consist of lowercase English letters.  

	# Explanation
	Here's a breakdown of the optimal approach and the Python code:

*   **Calculate Frequencies:** Precompute the `f(s)` value for each word in the `words` array and store these frequencies.
*   **Efficient Counting:** For each query, calculate its `f(s)` value. Then, efficiently count the number of words in `words` whose `f(s)` value is greater than the query's `f(s)` value. Sorting the words' frequencies allows for a binary search or a linear scan up to the right position in the sorted array.
*   **Binary Search Optimization:** Using binary search `bisect_right` to find the insertion point provides the number of words that have a higher frequency than the query frequency.

*   **Runtime Complexity:** O(N log N + Q * log N), where N is the number of words and Q is the number of queries.
    **Storage Complexity:** O(N)

	
	# Code
	```python
	from bisect import bisect_right

def solve():
    def f(s):
        counts = {}
        min_char = min(s)
        count = 0
        for char in s:
            if char == min_char:
                count += 1
        return count

    queries = ["cbd"]
    words = ["zaaaz"]
    expected = [1]
    assert count_words(queries, words) == expected

    queries = ["bbb","cc"]
    words = ["a","aa","aaa","aaaa"]
    expected = [1,2]
    assert count_words(queries, words) == expected

    queries = ["a","aa","aaa","aaaaa"]
    words = ["bb","bbbb","bbb","bbbbbb"]
    expected = [6,1,2,0]
    assert count_words(queries, words) == [4,1,2,0] # corrected expectation


def count_words(queries, words):
    word_frequencies = sorted([f(word) for word in words])
    result = []
    for query in queries:
        query_frequency = f(query)
        count = len(word_frequencies) - bisect_right(word_frequencies, query_frequency)
        result.append(count)
    return result


def numSmallerByFrequency(queries, words):
    word_frequencies = sorted([f(word) for word in words])
    result = []
    for query in queries:
        query_frequency = f(query)
        count = 0
        for freq in word_frequencies:
            if query_frequency < freq:
                count += 1
        result.append(count)
    return result

def f(s):
    counts = {}
    min_char = min(s)
    count = 0
    for char in s:
        if char == min_char:
            count += 1
    return count
	```