# 1007. Minimum Domino Rotations For Equal Row

## 題目敘述

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 * 10**4`
• `bottoms.length == tops.length`
• `1 <= tops[i], bottoms[i] <= 6`

## 解法解析

1. 先判斷在不在 `tops``bottoms`
2. 分別計算在 `tops``bottoms` 需要替換的次數
3. 回傳 step2 中的最小值(`tops` 還是 `bottoms` 的替換次數最少的)
4. 如果是 `-1` 那就換成用 `bottoms[0]` 當代表判斷

### 解法範例

#### Go

``````func minDominoRotations(tops []int, bottoms []int) int {
rotations := check(tops, bottoms, tops[0])
if rotations != -1 || tops[0] == bottoms[0] {
return rotations
}
return check(tops, bottoms, bottoms[0])
}

func check(tops, bottoms []int, x int) int {
var rotationsA, rotationsB int

for i := 0; i < len(tops); i++ {
if tops[i] != x && bottoms[i] != x {
return -1
} else if tops[i] != x {
rotationsA++
} else if bottoms[i] != x {
rotationsB++
}
}

if rotationsA > rotationsB {
return rotationsB
}
return rotationsA
}
``````

#### JavaScript

``````/**
* @param {number[]} tops
* @param {number[]} bottoms
* @return {number}
*/
var minDominoRotations = function (tops, bottoms) {
const check = (x) => {
let rotationsA = 0,
rotationsB = 0;
for (let i = 0; i < tops.length; i++) {
if (tops[i] != x && bottoms[i] != x) return -1;
else if (tops[i] != x) rotationsA++;
else if (bottoms[i] != x) rotationsB++;
}
return Math.min(rotationsA, rotationsB);
};

const rotations = check(tops[0]);
return rotations !== -1 || bottoms[0] === tops[0]
? rotations
: check(bottoms[0]);
};
``````

#### Kotlin

``````class Solution {
fun minDominoRotations(tops: IntArray, bottoms: IntArray): Int {
var rotations = check(tops, bottoms, tops[0])
if (rotations != -1 || tops[0] == bottoms[0]) {
return rotations

}
return check(tops, bottoms, bottoms[0])
}

private fun check(tops: IntArray, bottoms: IntArray, x: Int): Int {
var rotationsA = 0
var rotationsB = 0
for (i in tops.indices) {
if (tops[i] != x && bottoms[i] != x) return -1
if (tops[i] != x) rotationsA++
if (bottoms[i] != x) rotationsB++
}
return Math.min(rotationsA, rotationsB)
}
}
``````

#### PHP

``````class Solution
{

/**
* @param Integer[] \$tops
* @param Integer[] \$bottoms
* @return Integer
*/
function minDominoRotations(\$tops, \$bottoms)
{
\$rotations = \$this->check(\$tops, \$bottoms, \$tops[0]);
if (\$rotations !== -1 || \$tops[0] == \$bottoms[0]) {
return \$rotations;
}
return \$this->check(\$tops, \$bottoms, \$bottoms[0]);
}

function check(\$tops, \$bottoms, \$x)
{
\$rotations_a = \$rotations_b = 0;
for (\$i = 0; \$i < count(\$tops); \$i++) {
if (\$tops[\$i] != \$x && \$bottoms[\$i] != \$x) {
return -1;
} else if (\$tops[\$i] != \$x) {
\$rotations_a++;
} else if (\$bottoms[\$i] != \$x) {
\$rotations_b++;
}
}
return min(\$rotations_a, \$rotations_b);
}
}
``````

#### Python

``````class Solution:
def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:
def check(x):
"""
Return min number of swaps
if one could make all elements in tops or bottoms equal to x.
Else return -1.
"""
# how many rotations should be done
# to have all elements in tops equal to x
# and to have all elements in bottoms equal to x
rotations_a = rotations_b = 0
for i in range(n):
# rotations couldn't be done
if tops[i] != x and bottoms[i] != x:
return -1
# tops[i] != x and bottoms[i] == x
elif tops[i] != x:
rotations_a += 1
# tops[i] == x and bottoms[i] != x
elif bottoms[i] != x:
rotations_b += 1
# min number of rotations to have all
# elements equal to x in tops or bottoms
return min(rotations_a, rotations_b)

n = len(tops)
rotations = check(tops[0])
# If one could make all elements in tops or bottoms equal to tops[0]
if rotations != -1 or tops[0] == bottoms[0]:
return rotations
# If one could make all elements in tops or bottoms equal to bottoms[0]
return check(bottoms[0])
``````

#### Rust

``````
``````

#### Swift

``````class Solution {
func minDominoRotations(_ tops: [Int], _ bottoms: [Int]) -> Int {
let rotations = check(tops[0], tops, bottoms)
if rotations != -1 || tops[0] == bottoms[0] {
return rotations
}
return check(bottoms[0], tops, bottoms)
}

private func check(_ x: Int, _ tops: [Int], _ bottoms: [Int]) -> Int {
var rotationsA = 0, rotationsB = 0

for i in 0..<tops.count {
if tops[i] != x, bottoms[i] != x {
return -1
} else if  tops[i] != x {
rotationsA += 1
} else if bottoms[i] != x {
rotationsB += 1
}
}
return min(rotationsA, rotationsB)
}
}
``````