2103. Rings and Rods

문제)

2103. Rings and Rods

 

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).

솔루션1)

- 해시 테이블(dict())과 집합(set)을 이용해서 쉽게 풀이 가능

class Solution:
    def countPoints(self, rings: str) -> int:
        # rod num, rgb
        rod = defaultdict(set)
        
        for i in range(0, len(rings), 2):
            color = rings[i]
            rod_num = rings[i+1]
            rod[rod_num].add(color)
        
        
        return sum([1 for colors in rod.values() if len(colors) == 3])