DEV Community

Abhishek Chaudhary
Abhishek Chaudhary

Posted on

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).

SOLUTION:

class Solution:
    def countPoints(self, rings: str) -> int:
        n = len(rings)
        rods = [set(), set(), set()]
        for i in range(0, n, 2):
            if rings[i] == 'R':
                rods[0].add(int(rings[i + 1]))
            if rings[i] == 'G':
                rods[1].add(int(rings[i + 1]))
            if rings[i] == 'B':
                rods[2].add(int(rings[i + 1]))
        return len(set.intersection(*rods))
Enter fullscreen mode Exit fullscreen mode

Top comments (0)