feat: add solutions to lc problem: No.1217

No.1217.Minimum Cost to Move Chips to The Same Position

Author: yanglbme

solution/0500-0599/0591.Tag Validator/README.md

@@ -108,9 +108,7 @@ class Solution:
     def isValid(self, code: str) -> bool:
         def check(tag):
             n = len(tag)
-            if n < 1 or n > 9:
-                return False
-            return all(c.isupper() for c in tag)
+            return 1 <= n <= 9 and all(c.isupper() for c in tag)
 
         stk = []
         i, n = 0, len(code)
@@ -128,9 +126,8 @@ class Solution:
                 if i < 0:
                     return False
                 t = code[j: i]
-                if not check(t) or not stk or stk[-1] != t:
+                if not check(t) or not stk or stk.pop() != t:
                     return False
-                stk.pop()
             elif code[i] == '<':
                 j = i + 1
                 i = code.find('>', j)

solution/0500-0599/0591.Tag Validator/README_EN.md

@@ -76,9 +76,7 @@ class Solution:
     def isValid(self, code: str) -> bool:
         def check(tag):
             n = len(tag)
-            if n < 1 or n > 9:
-                return False
-            return all(c.isupper() for c in tag)
+            return 1 <= n <= 9 and all(c.isupper() for c in tag)
 
         stk = []
         i, n = 0, len(code)
@@ -96,9 +94,8 @@ class Solution:
                 if i < 0:
                     return False
                 t = code[j: i]
-                if not check(t) or not stk or stk[-1] != t:
+                if not check(t) or not stk or stk.pop() != t:
                     return False
-                stk.pop()
             elif code[i] == '<':
                 j = i + 1
                 i = code.find('>', j)

solution/0500-0599/0591.Tag Validator/Solution.py

@@ -2,9 +2,7 @@ class Solution:
     def isValid(self, code: str) -> bool:
         def check(tag):
             n = len(tag)
-            if n < 1 or n > 9:
-                return False
-            return all(c.isupper() for c in tag)
+            return 1 <= n <= 9 and all(c.isupper() for c in tag)
 
         stk = []
         i, n = 0, len(code)
@@ -22,9 +20,8 @@ def check(tag):
                 if i < 0:
                     return False
                 t = code[j: i]
-                if not check(t) or not stk or stk[-1] != t:
+                if not check(t) or not stk or stk.pop() != t:
                     return False
-                stk.pop()
             elif code[i] == '<':
                 j = i + 1
                 i = code.find('>', j)

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/README.md

@@ -63,22 +63,69 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：脑筋急转弯**
+
+将所有偶数下标的芯片移动到 0 号位置，所有奇数下标的芯片移动到 1 号位置，所有的代价为 0，接下来只需要在 0/1 号位置中选择其中一个较小数量的芯片，移动到另一个位置。所需的最小代价就是那个较小的数量。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minCostToMoveChips(self, position: List[int]) -> int:
+        a = sum(p % 2 for p in position)
+        b = len(position) - a
+        return min(a, b)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int minCostToMoveChips(int[] position) {
+        int a = 0;
+        for (int p : position) {
+            a += p % 2;
+        }
+        int b = position.length - a;
+        return Math.min(a, b);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minCostToMoveChips(vector<int>& position) {
+        int a = 0;
+        for (auto& p : position) a += p & 1;
+        int b = position.size() - a;
+        return min(a, b);
+    }
+};
+```
 
+### **Go**
+
+```go
+func minCostToMoveChips(position []int) int {
+	a := 0
+	for _, p := range position {
+		a += p & 1
+	}
+	b := len(position) - a
+	if a < b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/README_EN.md

@@ -56,13 +56,56 @@ Total cost is 1.
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minCostToMoveChips(self, position: List[int]) -> int:
+        a = sum(p % 2 for p in position)
+        b = len(position) - a
+        return min(a, b)
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int minCostToMoveChips(int[] position) {
+        int a = 0;
+        for (int p : position) {
+            a += p % 2;
+        }
+        int b = position.length - a;
+        return Math.min(a, b);
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minCostToMoveChips(vector<int>& position) {
+        int a = 0;
+        for (auto& p : position) a += p & 1;
+        int b = position.size() - a;
+        return min(a, b);
+    }
+};
+```
 
+### **Go**
+
+```go
+func minCostToMoveChips(position []int) int {
+	a := 0
+	for _, p := range position {
+		a += p & 1
+	}
+	b := len(position) - a
+	if a < b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/Solution.cpp

@@ -0,0 +1,9 @@
+class Solution {
+public:
+    int minCostToMoveChips(vector<int>& position) {
+        int a = 0;
+        for (auto& p : position) a += p & 1;
+        int b = position.size() - a;
+        return min(a, b);
+    }
+};

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/Solution.go

@@ -0,0 +1,11 @@
+func minCostToMoveChips(position []int) int {
+	a := 0
+	for _, p := range position {
+		a += p & 1
+	}
+	b := len(position) - a
+	if a < b {
+		return a
+	}
+	return b
+}

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/Solution.java

@@ -0,0 +1,10 @@
+class Solution {
+    public int minCostToMoveChips(int[] position) {
+        int a = 0;
+        for (int p : position) {
+            a += p % 2;
+        }
+        int b = position.length - a;
+        return Math.min(a, b);
+    }
+}

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/Solution.py

@@ -0,0 +1,5 @@
+class Solution:
+    def minCostToMoveChips(self, position: List[int]) -> int:
+        a = sum(p % 2 for p in position)
+        b = len(position) - a
+        return min(a, b)