How do you solve Minimum Domino Rotations For Equal Row on LeetCode?

Minimum Domino Rotations For Equal Row (LeetCode #1007) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The first domino's values are crucial as they determine potential candidates for uniformity.

What is the time complexity of Minimum Domino Rotations For Equal Row?

The optimal solution for Minimum Domino Rotations For Equal Row runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only iterate through the arrays a constant number of times (twice at most), making it efficient.

What is the brute force solution for Minimum Domino Rotations For Equal Row?

The brute force approach for Minimum Domino Rotations For Equal Row is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible value from the tops and bottoms arrays to see if we can make all values in either row equal by rotating the dominoes. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Domino Rotations For Equal Row?

Minimum Domino Rotations For Equal Row uses the following concepts: Array, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Domino Rotations For Equal Row?

Always consider edge cases, such as when all values are already the same. Think about the implications of each domino's values on the overall solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Domino Rotations For Equal Row?

Failing to check both rows for potential candidates, leading to missed solutions. Not handling cases where no rotations can achieve uniformity properly.

#1007

Minimum Domino Rotations For Equal Row

Medium

ArrayGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we only need to check the values on the first domino. If we can make all dominoes match that value, we can calculate the minimum rotations efficiently.

⚙️

Algorithm

3 steps

1Step 1: Identify the first domino's top and bottom values as potential candidates.
2Step 2: For each candidate, count how many rotations are needed to make all tops or bottoms equal to that candidate.
3Step 3: Return the minimum rotations found for either candidate, or -1 if neither is possible.

solution.py14 lines

1def minDominoRotations(tops, bottoms):
2    def countRotations(value):
3        top_rotations = bottom_rotations = 0
4        for i in range(len(tops)):
5            if tops[i] != value and bottoms[i] != value:
6                return float('inf')
7            elif tops[i] != value:
8                top_rotations += 1
9            elif bottoms[i] != value:
10                bottom_rotations += 1
11        return min(top_rotations, bottom_rotations)
12
13    min_rotations = min(countRotations(tops[0]), countRotations(bottoms[0]))
14    return min_rotations if min_rotations != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the arrays a constant number of times (twice at most), making it efficient.

1The first domino's values are crucial as they determine potential candidates for uniformity.
2If a value cannot be made uniform across either row, it is impossible to achieve the goal.

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