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Commit 408589c

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feat: add solutions to lc problem: No.0517 (#1589)
No.0517.Super Washing Machines
1 parent 0ad4f70 commit 408589c

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7 files changed

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‎solution/0500-0599/0517.Super Washing Machines/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:贪心**
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如果洗衣机内的衣服总数不能被洗衣机的数量整除,那么不可能使得每台洗衣机内的衣服数量相等,直接返回 $-1$。
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否则,假设洗衣机内的衣服总数为 $s,ドル那么最终每台洗衣机内的衣服数量都会变为 $k = s / n$。
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我们定义 $a_i$ 为第 $i$ 台洗衣机内的衣服数量与 $k$ 的差值,即 $a_i = \text{machines}[i] - k$。若 $a_i > 0,ドル则表示第 $i$ 台洗衣机内有多余的衣服,需要向相邻的洗衣机传递;若 $a_i < 0,ドル则表示第 $i$ 台洗衣机内缺少衣服,需要从相邻的洗衣机获得。
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我们将前 $i$ 台洗衣机的衣服数量差值之和定义为 $s_i = \sum_{j=0}^{i-1} a_j,ドル如果把前 $i$ 台洗衣机视为一组,其余的洗衣机视为另一组。那么若 $s_i$ 为正数,表示第一组洗衣机内有多余的衣服,需要向第二组洗衣机传递;若 $s_i$ 为负数,表示第一组洗衣机内缺少衣服,需要从第二组洗衣机获得。
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那么有以下两种情况:
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1. 两组之间的衣服,最多需要移动的次数为 $\max_{i=0}^{n-1} \lvert s_i \rvert$;
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1. 组内某一台洗衣机的衣服数量过多,需要向左右两侧移出衣服,最多需要移动的次数为 $\max_{i=0}^{n-1} a_i$。
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我们取两者的最大值即可。
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时间复杂度 $O(n),ドル其中 $n$ 为洗衣机的数量。空间复杂度 $O(1)$。
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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 findMinMoves(self, machines: List[int]) -> int:
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n = len(machines)
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k, mod = divmod(sum(machines), n)
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if mod:
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return -1
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ans = s = 0
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for x in machines:
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x -= k
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s += x
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ans = max(ans, abs(s), x)
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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 findMinMoves(int[] machines) {
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int n = machines.length;
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int s = 0;
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for (int x : machines) {
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s += x;
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}
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if (s % n != 0) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.max(Math.abs(s), x));
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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 findMinMoves(vector<int>& machines) {
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int n = machines.size();
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int s = accumulate(machines.begin(), machines.end(), 0);
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if (s % n) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = max({ans, abs(s), x});
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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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### **Go**
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```go
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func findMinMoves(machines []int) (ans int) {
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n := len(machines)
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s := 0
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for _, x := range machines {
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s += x
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}
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if s%n != 0 {
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return -1
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}
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k := s / n
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s = 0
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for _, x := range machines {
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x -= k
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s += x
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ans = max(ans, max(abs(s), x))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function findMinMoves(machines: number[]): number {
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const n = machines.length;
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let s = machines.reduce((a, b) => a + b);
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if (s % n !== 0) {
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return -1;
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}
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const k = Math.floor(s / n);
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s = 0;
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let ans = 0;
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for (let x of machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.abs(s), x);
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}
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return ans;
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}
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```
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### **...**

‎solution/0500-0599/0517.Super Washing Machines/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -57,13 +57,126 @@ It&#39;s impossible to make all three washing machines have the same number of d
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### **Python3**
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```python
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class Solution:
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def findMinMoves(self, machines: List[int]) -> int:
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n = len(machines)
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k, mod = divmod(sum(machines), n)
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if mod:
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return -1
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ans = s = 0
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for x in machines:
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x -= k
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s += x
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ans = max(ans, abs(s), x)
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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 findMinMoves(int[] machines) {
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int n = machines.length;
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int s = 0;
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for (int x : machines) {
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s += x;
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}
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if (s % n != 0) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.max(Math.abs(s), x));
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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 findMinMoves(vector<int>& machines) {
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int n = machines.size();
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int s = accumulate(machines.begin(), machines.end(), 0);
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if (s % n) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = max({ans, abs(s), x});
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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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### **Go**
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```go
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func findMinMoves(machines []int) (ans int) {
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n := len(machines)
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s := 0
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for _, x := range machines {
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s += x
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}
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if s%n != 0 {
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return -1
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}
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k := s / n
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s = 0
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for _, x := range machines {
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x -= k
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s += x
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ans = max(ans, max(abs(s), x))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function findMinMoves(machines: number[]): number {
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const n = machines.length;
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let s = machines.reduce((a, b) => a + b);
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if (s % n !== 0) {
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return -1;
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}
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const k = Math.floor(s / n);
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s = 0;
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let ans = 0;
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for (let x of machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.abs(s), x);
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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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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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int findMinMoves(vector<int>& machines) {
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int n = machines.size();
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int s = accumulate(machines.begin(), machines.end(), 0);
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if (s % n) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = max({ans, abs(s), x});
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,32 @@
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func findMinMoves(machines []int) (ans int) {
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n := len(machines)
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s := 0
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for _, x := range machines {
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s += x
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}
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if s%n != 0 {
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return -1
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}
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k := s / n
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s = 0
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for _, x := range machines {
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x -= k
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s += x
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ans = max(ans, max(abs(s), x))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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class Solution {
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public int findMinMoves(int[] machines) {
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int n = machines.length;
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int s = 0;
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for (int x : machines) {
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s += x;
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}
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if (s % n != 0) {
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return -1;
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}
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int k = s / n;
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s = 0;
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int ans = 0;
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for (int x : machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.max(Math.abs(s), x));
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}
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return ans;
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
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class Solution:
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def findMinMoves(self, machines: List[int]) -> int:
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n = len(machines)
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k, mod = divmod(sum(machines), n)
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if mod:
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return -1
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ans = s = 0
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for x in machines:
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x -= k
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s += x
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ans = max(ans, abs(s), x)
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return ans
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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function findMinMoves(machines: number[]): number {
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const n = machines.length;
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let s = machines.reduce((a, b) => a + b);
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if (s % n !== 0) {
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return -1;
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}
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const k = Math.floor(s / n);
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s = 0;
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let ans = 0;
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for (let x of machines) {
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x -= k;
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s += x;
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ans = Math.max(ans, Math.abs(s), x);
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}
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return ans;
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}

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