Introduction

In this blog post, we will explore how to solve the LeetCode problem "2103" 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

There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9. You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where: The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B'). The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9'). For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1. Return the number of rods that have all three colors of rings on them. Example 1: Input: rings = "B0B6G0R6R0R6G9" Output: 1 Explanation: - The rod labeled 0 holds 3 rings with all colors: red, green, and blue. - The rod labeled 6 holds 3 rings, but it only has red and blue. - The rod labeled 9 holds only a green ring. Thus, the number of rods with all three colors is 1. Example 2: Input: rings = "B0R0G0R9R0B0G0" Output: 1 Explanation: - The rod labeled 0 holds 6 rings with all colors: red, green, and blue. - The rod labeled 9 holds only a red ring. Thus, the number of rods with all three colors is 1. Example 3: Input: rings = "G4" Output: 0 Explanation: Only one ring is given. Thus, no rods have all three colors. Constraints: rings.length == 2 * n 1 <= n <= 100 rings[i] where i is even is either 'R', 'G', or 'B' (0-indexed). rings[i] where i is odd is a digit from '0' to '9' (0-indexed).

Explanation

• Data Structure: Use a list of sets, where each set represents a rod and stores the colors of the rings on that rod.
  • Iteration: Iterate through the input string rings in pairs, extracting the color and rod number for each ring.
  • Check for All Colors: After processing all rings, iterate through the rods and count the number of rods that contain all three colors (R, G, B).

• Runtime Complexity: O(n), where n is the length of the input string rings.
• Storage Complexity: O(1), because the maximum size of the rods list is fixed at 10, and each set within it can contain at most 3 elements.

Code

    def count_points(rings: str) -> int:
    """
    Counts the number of rods that have rings of all three colors (red, green, blue).

    Args:
        rings: A string describing the rings and their positions on the rods.

    Returns:
        The number of rods with rings of all three colors.
    """

    rods = [set() for _ in range(10)]  # Initialize a list of sets for each rod

    for i in range(0, len(rings), 2):
        color = rings[i]
        rod_number = int(rings[i + 1])
        rods[rod_number].add(color)

    count = 0
    for rod in rods:
        if 'R' in rod and 'G' in rod and 'B' in rod:
            count += 1

    return count