Rings and Rods

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Strings

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 i^th pair denotes the i^th ring's color ('R', 'G', 'B').
The second character of the i^th pair denotes the rod that the i^th 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).

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible characters for the colors, and what is the range for the rod numbers?
Can the input string be empty or null? If so, what should I return?
Is the order of color-rod pairs in the input string significant? For example, does 'R1G1B1' differ from 'B1G1R1'?
If a rod has multiple occurrences of the same color, should it still be considered as having that color? (e.g., 'RR1' would count as having the color 'R'?)
Is the rod number guaranteed to be a single digit?

Brute Force Solution

Approach

The brute force strategy examines every single possible arrangement of rings on the rods to find the rods with all three colors. We essentially check each rod individually to see if it meets the criteria.

Here's how the algorithm would work step-by-step:

Look at each rod, one at a time.
For each rod, check if it has a red ring.
If it doesn't have a red ring, move to the next rod.
If it does, check if that same rod also has a green ring.
If it doesn't have a green ring, move to the next rod.
If it has both a red and green ring, check if that rod also has a blue ring.
If the rod has all three colors, mark it as a 'good' rod.
After checking all the rods, count how many 'good' rods there are.

Code Implementation

def count_rods_with_all_colors(rings):
    rod_colors = {}
    for i in range(0, len(rings), 2):
        color = rings[i]
        rod = int(rings[i+1])

        if rod not in rod_colors:
            rod_colors[rod] = set()
        rod_colors[rod].add(color)

    count = 0
    for rod in rod_colors:
        if 'R' in rod_colors[rod] and 'G' in rod_colors[rod] and 'B' in rod_colors[rod]:
            count += 1
    return count

def count_rods_with_all_colors_brute_force(rings):
    good_rod_count = 0
    
    # Check each rod from 0 to 9
    for rod_number in range(10):
        has_red = False
        has_green = False
        has_blue = False
        
        # Iterate through the rings string
        for i in range(0, len(rings), 2):
            color = rings[i]
            rod = int(rings[i + 1])
            
            # Check if the current rod matches the rod number
            if rod == rod_number:
                if color == 'R':
                    has_red = True
                    
                if color == 'G':
                    has_green = True
                    
                if color == 'B':
                    has_blue = True

        #Check if rod has all three colors.
        if has_red and has_green and has_blue:
            #Increment when rod has all three.
            good_rod_count += 1
            
    return good_rod_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n characters in the input string 'rings' once. Inside the loop, it performs a constant amount of work for each rod by checking for the presence of 'R', 'G', and 'B'. This check takes constant time. Since the dominant operation is a single loop through the input string, the time complexity is O(n).

Space Complexity

O(1) – The provided plain English explanation describes iterating through rods and checking for the presence of red, green, and blue rings on each. It does not mention the creation of any auxiliary data structures like arrays, lists, or hash maps. The algorithm only needs to keep track of a few boolean flags (or similar) to indicate the presence of each color on the current rod being examined. Thus, the space used is constant, irrespective of the number of rods or rings (input size).

Optimal Solution

Approach

The problem involves figuring out which rods have all three colors of rings on them. The key is to efficiently track which colors are present on each rod and then check which rods satisfy the condition of having all colors.

Here's how the algorithm would work step-by-step:

Imagine you have a way to remember which colors are on each rod. Think of each rod as having a checklist for red, green, and blue.
Go through the provided information about the rings one by one.
For each ring, determine which rod it's on and what color it is.
Mark that color as present on that particular rod's checklist.
After processing all rings, go through the rods and see which ones have all three colors marked on their checklists.
The number of rods with all three colors is your answer.

Code Implementation

def count_rods_with_all_colors(rings):
    # rods_colors will track which colors are on each rod.
    rods_colors = {}

    for i in range(0, len(rings), 2):
        ring_color = rings[i]
        rod_number = int(rings[i+1])

        # Initialize the set for the rod if it's not already present.
        if rod_number not in rods_colors:
            rods_colors[rod_number] = set()

        # Add the current ring color to the set of colors for that rod.
        rods_colors[rod_number].add(ring_color)

    count_of_valid_rods = 0
    # Count rods with all three colors
    for rod_colors in rods_colors.values():
        #This is the logic that decides if all three colors are present
        if len(rod_colors) == 3:
            count_of_valid_rods += 1

    return count_of_valid_rods

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of length n (number of rings) once to record the presence of colors on each rod. Then, it iterates through a fixed number of rods (10) to check if each rod has all three colors. Therefore, the dominant operation is the initial iteration through the input string, making the time complexity O(n).

Space Complexity

O(1) – The plain English solution describes using a checklist for each rod to track the presence of red, green, and blue colors. Since there is a fixed number of rods (10), this checklist takes constant space. Specifically, we need to store the presence/absence of 3 colors for each of the 10 rods, which can be represented using a fixed-size data structure. Therefore, the auxiliary space required does not depend on the input size N (length of the rings string) and remains constant.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 immediately, as there are no rods or rings.
Input string with odd length	Return 0 immediately because the input string is invalid based on the r_i p_i pattern.
Input string with only one 'ring-rod' pair (length 2)	Process this single pair and update the rod's state; ensure the rod is properly initialized.
Input string with all identical ring colors	The solution should correctly update rod states without issues, potentially leading to all rods having all colors.
Input string with all rings on the same rod	The solution should correctly update the one rod's state, potentially leading to one rod having all colors.
Rings are not in order	The order should not matter for this problem because we are processing each pair one by one.
Repeated ring-rod pairs	Duplicate ring-rod pairs will result in redundant processing; the solution needs to handle this by correctly updating rod status appropriately.
Invalid rod index (not between 0 and 9 inclusive)	The solution should either ignore or throw an error when encountering an invalid rod index.

Null or empty input string

How to Handle:

Return 0 immediately, as there are no rods or rings.

Input string with odd length

How to Handle:

Return 0 immediately because the input string is invalid based on the r_i p_i pattern.

Input string with only one 'ring-rod' pair (length 2)

How to Handle:

Process this single pair and update the rod's state; ensure the rod is properly initialized.

Input string with all identical ring colors

How to Handle:

The solution should correctly update rod states without issues, potentially leading to all rods having all colors.

Input string with all rings on the same rod

How to Handle:

The solution should correctly update the one rod's state, potentially leading to one rod having all colors.

Rings are not in order

How to Handle:

The order should not matter for this problem because we are processing each pair one by one.

Repeated ring-rod pairs

How to Handle:

Duplicate ring-rod pairs will result in redundant processing; the solution needs to handle this by correctly updating rod status appropriately.

Invalid rod index (not between 0 and 9 inclusive)

How to Handle:

The solution should either ignore or throw an error when encountering an invalid rod index.