Leetcode #1007: Minimum Domino Rotations For Equal Row
In this guide, we solve Leetcode #1007 Minimum Domino Rotations For Equal Row 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
In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the ith domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.) We may rotate the ith domino, so that tops[i] and bottoms[i] swap values.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2]
Output: 2
Explanation:
The first figure represents the dominoes as given by tops and bottoms: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.
Python Solution
class Solution:
def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:
def f(x: int) -> int:
cnt1 = cnt2 = 0
for a, b in zip(tops, bottoms):
if x not in (a, b):
return inf
cnt1 += a == x
cnt2 += b == x
return len(tops) - max(cnt1, cnt2)
ans = min(f(tops[0]), f(bottoms[0]))
return -1 if ans == inf else ans
Complexity
The time complexity is , where is the length of the array. The space complexity is .
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.