feat: add solutions to lc problem: No.1551.Minimum Operations to Make Array Equal

Author: yanglbme

solution/1500-1599/1551.Minimum Operations to Make Array Equal/README.md

@@ -37,27 +37,66 @@
 	<li><code>1 &lt;= n &lt;= 10^4</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+数组 arr 的前 n 项和为 `(1 + (2 * n - 1)) * n / 2 = n * n`，若数组所有元素相等，那么每一项元素应该都是 n，因此只需累计数组前半部分的元素操作次数 `n - (2 * i + 1)` 即可，即 n ∈ `[0, n / 2)`。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minOperations(self, n: int) -> int:
+        ans = 0
+        for i in range(n >> 1):
+            ans += (n - (2 * i + 1))
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int minOperations(int n) {
+        int ans = 0;
+        for (int i = 0; i < (n >> 1); i++) {
+            ans += (n - (2 * i + 1));
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minOperations(int n) {
+        int ans = 0;
+        for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
+        return ans;
+    }
+};
+```
+
+### **Go**
 
+```go
+func minOperations(n int) int {
+	ans := 0
+	for i := 0; i < (n >> 1); i++ {
+		ans += (n - (2*i + 1))
+	}
+	return ans
+}
 ```
 
 ### **...**

solution/1500-1599/1551.Minimum Operations to Make Array Equal/README_EN.md

@@ -6,22 +6,14 @@
 
 <p>You have an array <code>arr</code> of length <code>n</code> where <code>arr[i] = (2 * i) + 1</code> for all valid values of <code>i</code> (i.e. <code>0 &lt;= i &lt; n</code>).</p>
 
-
-
 <p>In one operation, you can select two indices <code>x</code>&nbsp;and <code>y</code> where <code>0 &lt;= x, y &lt; n</code> and subtract <code>1</code> from <code>arr[x]</code> and add <code>1</code> to <code>arr[y]</code>&nbsp;(i.e. perform <code>arr[x] -=1&nbsp;</code>and <code>arr[y] += 1</code>).&nbsp;The goal is to make all the elements of the array <strong>equal</strong>. It is <strong>guaranteed</strong> that all the elements of the array can be made equal using some operations.</p>
 
-
-
 <p>Given an integer <code>n</code>, the length of the array. Return <em>the minimum number of operations</em> needed to make&nbsp;all the elements of arr equal.</p>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> n = 3
@@ -36,12 +28,8 @@ In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
 
 </pre>
 
-
-
 <p><strong>Example 2:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> n = 6
@@ -50,14 +38,10 @@ In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li><code>1 &lt;= n &lt;= 10^4</code></li>
 </ul>
@@ -69,13 +53,51 @@ In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minOperations(self, n: int) -> int:
+        ans = 0
+        for i in range(n >> 1):
+            ans += (n - (2 * i + 1))
+        return ans
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int minOperations(int n) {
+        int ans = 0;
+        for (int i = 0; i < (n >> 1); i++) {
+            ans += (n - (2 * i + 1));
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minOperations(int n) {
+        int ans = 0;
+        for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
+        return ans;
+    }
+};
+```
+
+### **Go**
 
+```go
+func minOperations(n int) int {
+	ans := 0
+	for i := 0; i < (n >> 1); i++ {
+		ans += (n - (2*i + 1))
+	}
+	return ans
+}
 ```
 
 ### **...**

solution/1500-1599/1551.Minimum Operations to Make Array Equal/Solution.cpp

@@ -0,0 +1,8 @@
+class Solution {
+public:
+    int minOperations(int n) {
+        int ans = 0;
+        for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
+        return ans;
+    }
+};

solution/1500-1599/1551.Minimum Operations to Make Array Equal/Solution.go

@@ -0,0 +1,7 @@
+func minOperations(n int) int {
+	ans := 0
+	for i := 0; i < (n >> 1); i++ {
+		ans += (n - (2*i + 1))
+	}
+	return ans
+}

solution/1500-1599/1551.Minimum Operations to Make Array Equal/Solution.java

@@ -1,9 +1,8 @@
 class Solution {
     public int minOperations(int n) {
         int ans = 0;
-        for (int i = 0; i < n / 2; i++) {
-            int curr = 2 * i + 1;
-            ans += Math.abs(n - curr);
+        for (int i = 0; i < (n >> 1); i++) {
+            ans += (n - (2 * i + 1));
         }
         return ans;
     }

solution/1500-1599/1551.Minimum Operations to Make Array Equal/Solution.py

@@ -0,0 +1,6 @@
+class Solution:
+    def minOperations(self, n: int) -> int:
+        ans = 0
+        for i in range(n >> 1):
+            ans += (n - (2 * i + 1))
+        return ans