Introduction

In this blog post, we will explore how to solve the LeetCode problem "2516" 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

You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s. Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character. Example 1: Input: s = "aabaaaacaabc", k = 2 Output: 8 Explanation: Take three characters from the left of s. You now have two 'a' characters, and one 'b' character. Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters. A total of 3 + 5 = 8 minutes is needed. It can be proven that 8 is the minimum number of minutes needed. Example 2: Input: s = "a", k = 1 Output: -1 Explanation: It is not possible to take one 'b' or 'c' so return -1. Constraints: 1 <= s.length <= 105 s consists of only the letters 'a', 'b', and 'c'. 0 <= k <= s.length

Explanation

Here's the breakdown of the solution:

• Prefix and Suffix Counts: Calculate prefix and suffix counts of 'a', 'b', and 'c' in the string. This will allow us to quickly determine the counts of each character within any prefix or suffix.
• Iterate and Check: Iterate through all possible lengths of prefixes. For each prefix length, calculate the corresponding suffix length needed to cover the entire string. Check if the combined counts of 'a', 'b', and 'c' in the prefix and suffix meet the requirement of at least k for each character.

• Minimize Length: Keep track of the minimum length (prefix length + suffix length) that satisfies the condition. If no combination satisfies the condition, return -1.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def min_minutes(s: str, k: int) -> int:
    """
    Given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k.
    Each minute, you may take either the leftmost character of s, or the rightmost character of s.
    Return the minimum number of minutes needed for you to take at least k of each character,
    or return -1 if it is not possible to take k of each character.

    Example 1:
    Input: s = "aabaaaacaabc", k = 2
    Output: 8
    Explanation:
    Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
    Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
    A total of 3 + 5 = 8 minutes is needed. It can be proven that 8 is the minimum number of minutes needed.

    Example 2:
    Input: s = "a", k = 1
    Output: -1
    Explanation: It is not possible to take one 'b' or 'c' so return -1.

    Constraints:
    1 <= s.length <= 105
    s consists of only the letters 'a', 'b', and 'c'.
    0 <= k <= s.length
    """

    n = len(s)
    prefix_counts = [[0] * 3 for _ in range(n + 1)]  # prefix_counts[i][0] = num of a's in s[:i]
    suffix_counts = [[0] * 3 for _ in range(n + 1)]  # suffix_counts[i][0] = num of a's in s[n-i:]

    for i in range(1, n + 1):
        prefix_counts[i][0] = prefix_counts[i - 1][0]
        prefix_counts[i][1] = prefix_counts[i - 1][1]
        prefix_counts[i][2] = prefix_counts[i - 1][2]
        if s[i - 1] == 'a':
            prefix_counts[i][0] += 1
        elif s[i - 1] == 'b':
            prefix_counts[i][1] += 1
        else:
            prefix_counts[i][2] += 1

    for i in range(1, n + 1):
        suffix_counts[i][0] = suffix_counts[i - 1][0]
        suffix_counts[i][1] = suffix_counts[i - 1][1]
        suffix_counts[i][2] = suffix_counts[i - 1][2]
        if s[n - i] == 'a':
            suffix_counts[i][0] += 1
        elif s[n - i] == 'b':
            suffix_counts[i][1] += 1
        else:
            suffix_counts[i][2] += 1

    min_len = float('inf')

    for prefix_len in range(n + 1):
        suffix_len = n - prefix_len
        a_count = prefix_counts[prefix_len][0] + suffix_counts[suffix_len][0]
        b_count = prefix_counts[prefix_len][1] + suffix_counts[suffix_len][1]
        c_count = prefix_counts[prefix_len][2] + suffix_counts[suffix_len][2]

        if a_count >= k and b_count >= k and c_count >= k:
            min_len = min(min_len, prefix_len + suffix_len)

    if min_len == float('inf'):
        return -1
    else:
        return min_len