How do you solve Rings and Rods on LeetCode?

Rings and Rods (LeetCode #2103) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set to track colors simplifies the checking process.

What is the brute force solution for Rings and Rods?

The brute force approach for Rings and Rods is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each rod individually to see if it contains all three colors of rings. This approach is straightforward but inefficient as it requires checking each rod multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rings and Rods?

Rings and Rods uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Rings and Rods?

Clarify the problem requirements before jumping into coding. Think about edge cases, such as empty strings or strings with only one color. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Rings and Rods?

Not considering the case where rods may have fewer than three colors. Overlooking the need to check all rods individually.

#2103

Rings and Rods

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking each rod multiple times, we can use an array to track the presence of colors for each rod in a single pass through the rings string. This reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Create an array of size 10, where each element is a set to track colors for each rod.
2Step 2: Iterate through the rings string, updating the corresponding rod's set with the color of the ring.
3Step 3: Count how many rods have all three colors by checking the size of each set.

solution.py12 lines

1# Full working Python code
2rings = 'B0B6G0R6R0R6G9'
3
4def countRodsWithAllColors(rings):
5    color_map = [set() for _ in range(10)]
6    for i in range(0, len(rings), 2):
7        color = rings[i]
8        position = int(rings[i + 1])
9        color_map[position].add(color)
10    return sum(1 for colors in color_map if len(colors) == 3)
11
12print(countRodsWithAllColors(rings))

ℹ

Complexity note: The time complexity is O(n) because we traverse the rings string once, and the space complexity is O(n) due to the storage of color sets for each rod.

1Using a set to track colors simplifies the checking process.
2Iterating through the rings only once is more efficient.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.