Add solution and test-cases for problem 1007

Author: 0xff-dev

leetcode/1001-1100/1007.Minimum-Domino-Rotations-For-Equal-Row/README.md

@@ -1,28 +1,32 @@
 # [1007.Minimum Domino Rotations For Equal Row][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+In a row of dominoes, `tops[i]` and `bottoms[i]` represent the top and bottom halves of the __i<sup>th</sup>__ domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)
 
-**Example 1:**
+We may rotate the __i<sup>th</sup>__ domino, so that `tops[i]` and `bottoms[i]` swap values.
 
-```
-Input: a = "11", b = "1"
-Output: "100"
-```
+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.
 
-## 题意
-> ...
+If it cannot be done, return `-1`.
 
-## 题解
+**Example 1:**  
+![domino](./domino.png)
 
-### 思路1
-> ...
-Minimum Domino Rotations For Equal Row
-```go
+```
+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.
+```
 
 ## 结语
 

leetcode/1001-1100/1007.Minimum-Domino-Rotations-For-Equal-Row/Solution.go

@@ -1,5 +1,48 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func topCanImpl(elem int, tops, bottoms []int) (bool, int) {
+	count := 0
+	for idx := 1; idx < len(tops); idx++ {
+		if tops[idx] == elem {
+			continue
+		}
+		if bottoms[idx] != elem {
+			return false, count
+		}
+		count++
+	}
+
+	return true, count
+}
+
+func bottomCanImpl(elem int, tops, bottoms []int) (bool, int) {
+	count := 0
+	for idx := 1; idx < len(bottoms); idx++ {
+		if bottoms[idx] == elem {
+			continue
+		}
+		if tops[idx] != elem {
+			return false, count
+		}
+		count++
+	}
+
+	return true, count
+}
+
+func Solution(tops []int, bottoms []int) int {
+	minCount := -1
+	if ttOk, ttCount := topCanImpl(tops[0], tops, bottoms); ttOk && (minCount == -1 || minCount > ttCount) {
+		minCount = ttCount
+	}
+	if tbOk, tbCount := topCanImpl(bottoms[0], tops, bottoms); tbOk && (minCount == -1 || minCount > tbCount+1) {
+		minCount = tbCount + 1
+	}
+	if bbOk, bbCount := bottomCanImpl(bottoms[0], tops, bottoms); bbOk && (minCount == -1 || minCount > bbCount) {
+		minCount = bbCount
+	}
+	if btOk, btCount := bottomCanImpl(tops[0], tops, bottoms); btOk && (minCount == -1 || minCount > btCount+1) {
+		minCount = btCount + 1
+	}
+	return minCount
 }