#1007

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Domino Rotations For Equal Row

Minimize domino rotations to make either row identical in the problem of Minimum Domino Rotations For Equal Row.

Problem Statement

In the Minimum Domino Rotations For Equal Row problem, you are given two arrays representing the top and bottom halves of dominoes. You can rotate any domino, swapping the values between the top and bottom. The goal is to find the minimum number of rotations required to make all values in the tops array identical or all values in the bottoms array identical.

Return the minimum number of rotations required. If it is not possible to achieve either, return -1. You need to account for edge cases where no solution can exist, such as when the numbers in both arrays conflict and cannot align with a single value.

Examples

Example 1

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

Output: 2

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.

Example 2

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

Output: -1

In this case, it is not possible to rotate the dominoes to make one row of values equal.

Constraints

2 <= tops.length <= 2 * 104
bottoms.length == tops.length
1 <= tops[i], bottoms[i] <= 6

Solution Approach

Greedy Choice with Invariant Validation

To solve this problem, we can leverage a greedy approach, trying to align all values either in the tops or bottoms array. Start by testing the most frequent values in the arrays, and check if they can form an invariant across all dominoes. For each value, count how many rotations would be needed and pick the one with the least rotations.

Checking for Impossible Cases

Before attempting rotations, check if a solution is even possible. If there is no common value that exists in both rows (tops and bottoms), it is impossible to make the rows identical. Return -1 early in such cases to optimize performance.

Efficient Iteration and Count Management

Iterate through the dominoes, checking for each potential target value in both the tops and bottoms. Track how many changes are required for each domino to match the target value, updating the count efficiently as you go along.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n) because we iterate through the arrays once to count occurrences and check possible rotations. Space complexity is O(1) as we only need a constant amount of extra space for counting rotations, regardless of the array size.

What Interviewers Usually Probe

Look for the candidate's ability to identify potential targets and manage rotations efficiently.
Evaluate how well the candidate handles edge cases like impossible configurations.
Check for the candidate’s understanding of how greedy strategies and invariant validation apply to array problems.

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases where no solution exists and returning -1 when necessary.
Overlooking the need for efficient iteration, which can lead to unnecessary time complexity increases.
Assuming the problem can always be solved when the rows are of equal length, without considering that values may not align.

Follow-up variants

Instead of focusing on rotations, the problem could be modified to look for the minimum swaps needed.
Consider a variant where the allowed operation is limited to one row (either tops or bottoms), reducing the problem's complexity.
Introduce additional constraints, such as larger numbers in the arrays, requiring more careful space or time complexity optimizations.

FAQ

What is the pattern behind the Minimum Domino Rotations For Equal Row problem?

The problem follows a greedy choice strategy, aiming to minimize rotations to make one row of dominoes identical. The key is identifying common values and validating potential solutions with invariant checks.

How do I know when it's impossible to solve this problem?

It's impossible to solve when no common value exists in both the tops and bottoms arrays, meaning no amount of rotations can make one row identical.

How can I optimize my solution for the Minimum Domino Rotations For Equal Row problem?

Optimize by using early exits when checking for common values and by iterating efficiently through the arrays, minimizing unnecessary checks or rotations.

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

The time complexity is O(n) because we iterate through the arrays once, checking each potential target value and counting rotations.

Can this problem be solved without using greedy algorithms?

While it is possible to approach this problem with brute force or dynamic programming, using a greedy strategy with invariant validation is the most efficient solution.

terminal

Solution

Solution 1: Greedy

According to the problem description, we know that in order to make all values in $tops$ or all values in $bottoms$ the same, the value must be one of $tops[0]$ or $bottoms[0]$.

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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

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