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Commit f02a68f

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feat: add solutions to lc problem: No.1551.Minimum Operations to Make Array Equal
1 parent 7fd5bf8 commit f02a68f

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6 files changed

+103
-22
lines changed

6 files changed

+103
-22
lines changed

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

Lines changed: 41 additions & 2 deletions
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@@ -37,27 +37,66 @@
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<li><code>1 &lt;= n &lt;= 10^4</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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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)`。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minOperations(self, n: int) -> int:
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ans = 0
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for i in range(n >> 1):
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ans += (n - (2 * i + 1))
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < (n >> 1); i++) {
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ans += (n - (2 * i + 1));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minOperations(n int) int {
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ans := 0
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for i := 0; i < (n >> 1); i++ {
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ans += (n - (2*i + 1))
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}
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return ans
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}
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```
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### **...**

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

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@@ -6,22 +6,14 @@
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<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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 3
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 6
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 10^4</code></li>
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</ul>
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### **Python3**
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```python
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class Solution:
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def minOperations(self, n: int) -> int:
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ans = 0
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for i in range(n >> 1):
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ans += (n - (2 * i + 1))
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < (n >> 1); i++) {
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ans += (n - (2 * i + 1));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minOperations(n int) int {
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ans := 0
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for i := 0; i < (n >> 1); i++ {
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ans += (n - (2*i + 1))
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}
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return ans
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}
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```
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### **...**
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class Solution {
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public:
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int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < (n >> 1); ++i) ans += (n - (2 * i + 1));
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return ans;
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}
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};
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func minOperations(n int) int {
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ans := 0
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for i := 0; i < (n >> 1); i++ {
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ans += (n - (2*i + 1))
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}
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return ans
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}

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

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Original file line numberDiff line numberDiff line change
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class Solution {
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public int minOperations(int n) {
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int ans = 0;
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for (int i = 0; i < n / 2; i++) {
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int curr = 2 * i + 1;
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ans += Math.abs(n - curr);
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for (int i = 0; i < (n >> 1); i++) {
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ans += (n - (2 * i + 1));
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}
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return ans;
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}
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class Solution:
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def minOperations(self, n: int) -> int:
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ans = 0
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for i in range(n >> 1):
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ans += (n - (2 * i + 1))
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return ans

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