Solving the Minimum Domino Rotations Problem: A Step-by-Step Guide

In the world of algorithmic challenges, the “Minimum Domino Rotations for Equal Row” problem presents a unique puzzle that blends mathematical insight with strategic problem-solving. Found on competitive programming platforms like LeetCode, this problem requires you to think critically about how to manipulate rows of dominos to achieve a uniform result with the least number of rotations. In this blog, we’ll break down the problem, explore various approaches to solve it, and provide detailed examples and code implementations to help you master this challenge.

Problem Statement

You are given two arrays, tops and bottoms, each representing the top and bottom halves of a row of dominos. Your goal is to make all the values in one of these rows equal by performing the minimum number of rotations.

A rotation involves swapping the top and bottom values of a domino at a given index. Your task is to determine the minimum number of rotations required to make all elements in either the tops or bottoms array the same. If it’s not possible, return -1.

Example:

Consider the following input:

Input:
tops = [2, 1, 2, 4, 2, 2]
bottoms = [5, 2, 6, 2, 3, 2]

Output:
2

Explanation:
- Rotate dominos at index 1 (swap 1 and 2).
- Rotate dominos at index 3 (swap 4 and 2).
After these rotations, all elements in `tops` will be 2.

Analyzing the Problem

To solve this problem, the key insight is to focus on specific values that could potentially make all elements in either the tops or bottoms row the same. Specifically, we should consider the value at tops[0] and bottoms[0] as potential candidates, as these are the only numbers that could possibly fill the entire row with a minimum number of rotations.

Approach to Solve the Problem

The problem can be approached in the following steps:

1. Identify Candidate Values: Consider the values tops[0] and bottoms[0] as potential candidates that could fill the entire row. These are the only values that can be swapped across the entire array to achieve uniformity.
2. Check Feasibility: For each candidate value, check if it is possible to make all the elements in either the tops or bottoms row equal to that candidate value. If not, the candidate is not viable.
3. Count Rotations: For each viable candidate, count the minimum number of rotations required to achieve the uniform row. Track the minimum rotations needed among all viable candidates.
4. Return Result: If a viable candidate is found, return the minimum number of rotations required. If no candidate is viable, return -1.

Detailed Example Walkthrough

Let’s walk through a detailed example to illustrate the solution:

Example 2:

Input:
tops = [3, 5, 1, 2, 3]
bottoms = [3, 6, 3, 3, 4]

Identify Candidate Values:

• Candidate values are tops[0] = 3 and bottoms[0] = 3.

Check Feasibility for Candidate 3:

• For tops: The value 3 can be placed at all positions except for index 3, where the value is 2 and a rotation will be needed.
• For bottoms: The value 3 can be placed at all positions except for index 4, where the value is 4 and a rotation will be needed.

Count Rotations:

• Rotations required to make all values in tops equal to 3: 1 (index 3).
• Rotations required to make all values in bottoms equal to 3: 1 (index 4).

Return Result:

• The minimum number of rotations required is 1, as only one rotation is needed to make either row uniform.

Python Code Implementation

def min_domino_rotations(tops, bottoms):
def check(x):
rotations_top = rotations_bottom = 0
for i in range(len(tops)):
if tops[i] != x and bottoms[i] != x:
return float('inf')elif tops[i] != x:
rotations_top += 1elif bottoms[i] != x:
rotations_bottom += 1return min(rotations_top, rotations_bottom)
rotations = min(check(tops[0]), check(bottoms[0]))
return -1 if rotations == float('inf') else rotations

Example usage:

tops = [2, 1, 2, 4, 2, 2]
bottoms = [5, 2, 6, 2, 3, 2]
print(min_domino_rotations(tops, bottoms)) # Output: 2

Explanation of the Code:

• Helper Function check(x): This function calculates the minimum rotations required to make all elements equal to x in either tops or bottoms. It returns float('inf') if it’s not possible to achieve uniformity with x.
• Rotation Calculation: We calculate the rotations needed for both tops[0] and bottoms[0], and return the minimum of these values.

Complexity Analysis

Time Complexity: The time complexity is O(n) because we iterate through the arrays a fixed number of times (once for each candidate value).

Space Complexity: The space complexity is O(1) as we only use a few extra variables for counting rotations.

Edge Cases

When solving the Minimum Domino Rotations problem, consider the following edge cases:

1. Single Element Arrays: If tops and bottoms have only one element, no rotations are needed, and the result should be 0.
2. No Possible Solution: If neither tops[0] nor bottoms[0] can fill the entire row, the function should return -1.
3. Identical Rows: If tops and bottoms are already identical, the result should be 0.

Practical Applications

Understanding the Minimum Domino Rotations problem can be beneficial beyond competitive programming. The principles of this problem, such as minimizing operations to achieve uniformity or consistency, can be applied in various fields, including data management, operations research, and system optimization.

Conclusion

The Minimum Domino Rotations problem is a great way to practice strategic thinking and algorithm design. By identifying key candidates and checking their viability, you can efficiently solve the problem and optimize your solution. Whether you’re preparing for coding interviews or enhancing your problem-solving skills, mastering this challenge will provide valuable insights into how to approach similar problems in the future.

Further Exploration

If you’re looking to deepen your understanding, try solving variations of this problem, such as considering more complex constraints or different types of operations. This will help you develop a broader perspective on algorithmic challenges and strengthen your coding abilities.

FAQs

What is the Minimum Domino Rotations problem?

The Minimum Domino Rotations problem involves two rows of dominos represented by two arrays, tops and bottoms. The goal is to determine the minimum number of rotations needed to make all the values in one of the rows the same. A rotation involves swapping the top and bottom values of a domino at a given index.

How do you solve the Minimum Domino Rotations problem?

To solve the problem, identify candidate values (usually tops[0] and bottoms[0]) that could potentially fill the entire row. Check if it’s possible to make all elements in either the tops or bottoms row equal to that candidate. Count the minimum number of rotations required for each candidate, and return the smallest value.

What is the significance of the candidate values in this problem?

The candidate values (usually the first elements of tops and bottoms) are crucial because they are the only values that could potentially fill the entire row with the least number of rotations. If neither candidate can fill the row, it’s impossible to solve the problem, and the answer is -1.

What should I do if no solution is possible?

If neither tops[0] nor bottoms[0] can be used to make all elements in one of the rows equal, the problem has no solution. In such cases, you should return -1.

What are the edge cases to consider for this problem?

Important edge cases include:

Arrays with only one element, where no rotations are needed.

Situations where tops and bottoms are already identical, so no rotations are required.

Scenarios where no candidate value can fill the entire row, resulting in a -1 return value.

What is the time complexity of the solution?

The time complexity of the solution is O(n), where n is the length of the tops and bottoms arrays. This is because the solution involves iterating through the arrays a fixed number of times.

Can the problem be solved with more than two candidate values?

Typically, only two candidate values are considered (tops[0] and bottoms[0]). If additional candidate values are viable (e.g., appearing frequently in both arrays), they could be considered, but this increases the complexity and is generally unnecessary for solving the problem efficiently.

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