LeetCode #1007 — MEDIUM

Minimum Domino Rotations For Equal Row

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the i^th domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)

We may rotate the i^th domino, so that tops[i] and bottoms[i] swap values.

Return the minimum number of rotations so that all the values in tops are the same, or all the values in bottoms are the same.

If it cannot be done, return -1.

Example 1:

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.

Example 2:

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

Step 01

Brute Force Baseline

Problem summary: 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. Return the minimum number of rotations so that all the values in tops are the same, or all the values in bottoms are the same. If it cannot be done, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[2,1,2,4,2,2]
[5,2,6,2,3,2]

Example 2

[3,5,1,2,3]
[3,6,3,3,4]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1007: Minimum Domino Rotations For Equal Row
class Solution {
    private int n;
    private int[] tops;
    private int[] bottoms;

    public int minDominoRotations(int[] tops, int[] bottoms) {
        n = tops.length;
        this.tops = tops;
        this.bottoms = bottoms;
        int ans = Math.min(f(tops[0]), f(bottoms[0]));
        return ans > n ? -1 : ans;
    }

    private int f(int x) {
        int cnt1 = 0, cnt2 = 0;
        for (int i = 0; i < n; ++i) {
            if (tops[i] != x && bottoms[i] != x) {
                return n + 1;
            }
            cnt1 += tops[i] == x ? 1 : 0;
            cnt2 += bottoms[i] == x ? 1 : 0;
        }
        return n - Math.max(cnt1, cnt2);
    }
}

// Accepted solution for LeetCode #1007: Minimum Domino Rotations For Equal Row
func minDominoRotations(tops []int, bottoms []int) int {
	n := len(tops)
	f := func(x int) int {
		cnt1, cnt2 := 0, 0
		for i, a := range tops {
			b := bottoms[i]
			if a != x && b != x {
				return n + 1
			}
			if a == x {
				cnt1++
			}
			if b == x {
				cnt2++
			}
		}
		return n - max(cnt1, cnt2)
	}
	ans := min(f(tops[0]), f(bottoms[0]))
	if ans > n {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #1007: Minimum Domino Rotations For Equal Row
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

// Accepted solution for LeetCode #1007: Minimum Domino Rotations For Equal Row
impl Solution {
    pub fn min_domino_rotations(tops: Vec<i32>, bottoms: Vec<i32>) -> i32 {
        let n = tops.len() as i32;
        let f = |x: i32| -> i32 {
            let mut cnt1 = 0;
            let mut cnt2 = 0;
            for i in 0..n as usize {
                if tops[i] != x && bottoms[i] != x {
                    return n + 1;
                }
                if tops[i] == x {
                    cnt1 += 1;
                }
                if bottoms[i] == x {
                    cnt2 += 1;
                }
            }
            n - cnt1.max(cnt2)
        };

        let ans = f(tops[0]).min(f(bottoms[0]));
        if ans > n {
            -1
        } else {
            ans
        }
    }
}

// Accepted solution for LeetCode #1007: Minimum Domino Rotations For Equal Row
function minDominoRotations(tops: number[], bottoms: number[]): number {
    const n = tops.length;
    const f = (x: number): number => {
        let [cnt1, cnt2] = [0, 0];
        for (let i = 0; i < n; ++i) {
            if (tops[i] !== x && bottoms[i] !== x) {
                return n + 1;
            }
            cnt1 += tops[i] === x ? 1 : 0;
            cnt2 += bottoms[i] === x ? 1 : 0;
        }
        return n - Math.max(cnt1, cnt2);
    };
    const ans = Math.min(f(tops[0]), f(bottoms[0]));
    return ans > n ? -1 : ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.