Leetcode #2103: Rings and Rods
In this guide, we solve Leetcode #2103 Rings and Rods in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def countPoints(self, rings: str) -> int:
mask = [0] * 10
d = {"R": 1, "G": 2, "B": 4}
for i in range(0, len(rings), 2):
c = rings[i]
j = int(rings[i + 1])
mask[j] |= d[c]
return mask.count(7)
Complexity
The time complexity is , and the space complexity is , where represents the length of the string , and represents the size of the character set. The space complexity is , where represents the length of the string , and represents the size of the character set.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.