Introduction

In this blog post, we will explore how to solve the LeetCode problem "451" 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 s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string. Return the sorted string. If there are multiple answers, return any of them. Example 1: Input: s = "tree" Output: "eert" Explanation: 'e' appears twice while 'r' and 't' both appear once. So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer. Example 2: Input: s = "cccaaa" Output: "aaaccc" Explanation: Both 'c' and 'a' appear three times, so both "cccaaa" and "aaaccc" are valid answers. Note that "cacaca" is incorrect, as the same characters must be together. Example 3: Input: s = "Aabb" Output: "bbAa" Explanation: "bbaA" is also a valid answer, but "Aabb" is incorrect. Note that 'A' and 'a' are treated as two different characters. Constraints: 1 <= s.length <= 5 * 105 s consists of uppercase and lowercase English letters and digits.

Explanation

Here's a breakdown of the solution and the corresponding Python code:

• High-Level Approach:

  • Count character frequencies using a dictionary (or Counter).
  • Sort characters based on their frequencies in descending order.
  • Build the sorted string by repeating each character by its frequency.

• Complexity:

  • Runtime: O(N log N), where N is the length of the string (due to sorting).
  • Storage: O(N), where N is the number of unique characters in the string (in the worst case).

Code

    from collections import Counter

def frequency_sort(s: str) -> str:
    """
    Sorts a string in decreasing order based on the frequency of characters.

    Args:
        s: The input string.

    Returns:
        The sorted string.
    """

    char_counts = Counter(s)
    sorted_chars = sorted(char_counts.items(), key=lambda item: item[1], reverse=True)
    result = ""
    for char, count in sorted_chars:
        result += char * count
    return result