Introduction

In this blog post, we will explore how to solve the LeetCode problem "1002" 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

Given a string array words, return an array of all characters that show up in all strings within the words (including duplicates). You may return the answer in any order. Example 1: Input: words = ["bella","label","roller"] Output: ["e","l","l"] Example 2: Input: words = ["cool","lock","cook"] Output: ["c","o"] Constraints: 1 <= words.length <= 100 1 <= words[i].length <= 100 words[i] consists of lowercase English letters.

Explanation

Here's the approach, followed by the Python code:

• Frequency Counting: For each word, count the frequency of each character.
• Intersection of Frequencies: Find the minimum frequency of each character across all words. This represents the common characters and their counts.

• Build Result: Construct the result array by including each character as many times as its minimum frequency.

• Runtime Complexity: O(N M), where N is the number of words and M is the maximum length of a word. *Storage Complexity: O(1), as we use a fixed-size array (26) to store character frequencies.

Code

    def commonChars(words):
    """
    Finds the common characters (including duplicates) in all strings within the input array.

    Args:
        words: A list of strings.

    Returns:
        A list of strings, representing the common characters.
    """

    if not words:
        return []

    min_freq = [float('inf')] * 26  # Initialize with infinity

    for word in words:
        freq = [0] * 26
        for char in word:
            freq[ord(char) - ord('a')] += 1

        for i in range(26):
            min_freq[i] = min(min_freq[i], freq[i])

    result = []
    for i in range(26):
        for _ in range(int(min_freq[i])):
            result.append(chr(ord('a') + i))

    return result