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, 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

    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