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feat: add solutions to lc problem: No.1675
No.1675.Minimize Deviation in Array
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‎solution/1600-1699/1675.Minimize Deviation in Array/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 + 优先队列**
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直观上，为了得到数组的最小偏移量，我们需要将减小数组的最大值，增大数组的最小值。
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由于每次可以执行乘、除两种操作：将奇数乘以 $2$；将偶数除以 $2$，情况较为复杂，我们可以将奇数统一乘以 $2$，转成偶数，这样就等价于只有一种除法操作。除法操作只能减少某个数，而只有减少最大值，结果才可能更优。
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因此，我们用优先队列（大根堆）维护数组的最大值，每次取出堆顶元素做除法操作，将新值放入堆中，并且更新最小值以及堆顶元素与最小值的差值的最小值。
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当堆顶元素为奇数时，操作停止。
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时间复杂度 $O(n\log n \times \log m)$。其中 $n$, $m$ 分别是数组 `nums` 的长度以及数组的最大元素。由于数组中的最大元素除以 $2$ 的操作最多有 $O(\log m)$ 次，因此全部元素除以 $2$ 的操作最多有 $O(n\log m)$ 次。每次弹出、放入堆的操作，时间复杂度为 $O(\log n)$。因此，总的时间复杂度为 $O(n\log n \times \log m)$。
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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 minimumDeviation(self, nums: List[int]) -> int:
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h = []
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mi = inf
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for v in nums:
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if v & 1:
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v <<= 1
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h.append(-v)
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mi = min(mi, v)
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heapify(h)
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ans = -h[0] - mi
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while h[0] % 2 == 0:
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x = heappop(h) // 2
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heappush(h, x)
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mi = min(mi, -x)
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ans = min(ans, -h[0] - mi)
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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 minimumDeviation(int[] nums) {
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PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
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int mi = Integer.MAX_VALUE;
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for (int v : nums) {
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if (v % 2 == 1) {
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v <<= 1;
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}
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q.offer(v);
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mi = Math.min(mi, v);
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}
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int ans = q.peek() - mi;
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while (q.peek() % 2 == 0) {
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int x = q.poll() / 2;
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q.offer(x);
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mi = Math.min(mi, x);
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ans = Math.min(ans, q.peek() - mi);
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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 minimumDeviation(vector<int>& nums) {
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int mi = INT_MAX;
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priority_queue<int> pq;
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for (int v : nums) {
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if (v & 1) v <<= 1;
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pq.push(v);
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mi = min(mi, v);
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}
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int ans = pq.top() - mi;
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while (pq.top() % 2 == 0) {
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int x = pq.top() >> 1;
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pq.pop();
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pq.push(x);
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mi = min(mi, x);
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ans = min(ans, pq.top() - mi);
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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 minimumDeviation(nums []int) int {
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q := hp{}
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mi := math.MaxInt32
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for _, v := range nums {
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if v%2 == 1 {
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v <<= 1
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}
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heap.Push(&q, v)
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mi = min(mi, v)
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}
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ans := q.IntSlice[0] - mi
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for q.IntSlice[0]%2 == 0 {
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x := heap.Pop(&q).(int) >> 1
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heap.Push(&q, x)
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mi = min(mi, x)
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ans = min(ans, q.IntSlice[0]-mi)
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}
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return ans
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}
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func min(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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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
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```
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### **...**

‎solution/1600-1699/1675.Minimize Deviation in Array/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minimumDeviation(self, nums: List[int]) -> int:
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h = []
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mi = inf
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for v in nums:
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if v & 1:
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v <<= 1
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h.append(-v)
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mi = min(mi, v)
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heapify(h)
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ans = -h[0] - mi
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while h[0] % 2 == 0:
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x = heappop(h) // 2
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heappush(h, x)
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mi = min(mi, -x)
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ans = min(ans, -h[0] - mi)
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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 minimumDeviation(int[] nums) {
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PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
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int mi = Integer.MAX_VALUE;
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for (int v : nums) {
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if (v % 2 == 1) {
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v <<= 1;
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}
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q.offer(v);
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mi = Math.min(mi, v);
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}
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int ans = q.peek() - mi;
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while (q.peek() % 2 == 0) {
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int x = q.poll() / 2;
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q.offer(x);
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mi = Math.min(mi, x);
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ans = Math.min(ans, q.peek() - mi);
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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 minimumDeviation(vector<int>& nums) {
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int mi = INT_MAX;
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priority_queue<int> pq;
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for (int v : nums) {
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if (v & 1) v <<= 1;
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pq.push(v);
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mi = min(mi, v);
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}
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int ans = pq.top() - mi;
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while (pq.top() % 2 == 0) {
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int x = pq.top() >> 1;
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pq.pop();
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pq.push(x);
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mi = min(mi, x);
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ans = min(ans, pq.top() - mi);
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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 minimumDeviation(nums []int) int {
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q := hp{}
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mi := math.MaxInt32
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for _, v := range nums {
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if v%2 == 1 {
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v <<= 1
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}
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heap.Push(&q, v)
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mi = min(mi, v)
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}
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ans := q.IntSlice[0] - mi
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for q.IntSlice[0]%2 == 0 {
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x := heap.Pop(&q).(int) >> 1
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heap.Push(&q, x)
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mi = min(mi, x)
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ans = min(ans, q.IntSlice[0]-mi)
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}
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return ans
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}
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func min(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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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
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```
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### **...**

‎solution/1600-1699/1675.Minimize Deviation in Array/Solution.cpp

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class Solution {
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public:
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int minimumDeviation(vector<int>& nums) {
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int mi = INT_MAX;
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priority_queue<int> pq;
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for (int v : nums) {
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if (v & 1) v <<= 1;
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pq.push(v);
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mi = min(mi, v);
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}
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int ans = pq.top() - mi;
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while (pq.top() % 2 == 0) {
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int x = pq.top() >> 1;
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pq.pop();
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pq.push(x);
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mi = min(mi, x);
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ans = min(ans, pq.top() - mi);
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}
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return ans;
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}
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};

‎solution/1600-1699/1675.Minimize Deviation in Array/Solution.go

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func minimumDeviation(nums []int) int {
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q := hp{}
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mi := math.MaxInt32
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for _, v := range nums {
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if v%2 == 1 {
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v <<= 1
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}
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heap.Push(&q, v)
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mi = min(mi, v)
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}
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ans := q.IntSlice[0] - mi
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for q.IntSlice[0]%2 == 0 {
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x := heap.Pop(&q).(int) >> 1
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heap.Push(&q, x)
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mi = min(mi, x)
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ans = min(ans, q.IntSlice[0]-mi)
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}
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return ans
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}
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func min(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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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }

‎solution/1600-1699/1675.Minimize Deviation in Array/Solution.java

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class Solution {
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public int minimumDeviation(int[] nums) {
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PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
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int mi = Integer.MAX_VALUE;
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for (int v : nums) {
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if (v % 2 == 1) {
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v <<= 1;
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}
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q.offer(v);
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mi = Math.min(mi, v);
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}
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int ans = q.peek() - mi;
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while (q.peek() % 2 == 0) {
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int x = q.poll() / 2;
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q.offer(x);
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mi = Math.min(mi, x);
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ans = Math.min(ans, q.peek() - mi);
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}
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return ans;
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}
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}

‎solution/1600-1699/1675.Minimize Deviation in Array/Solution.py

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class Solution:
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def minimumDeviation(self, nums: List[int]) -> int:
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h = []
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mi = inf
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for v in nums:
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if v & 1:
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v <<= 1
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h.append(-v)
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mi = min(mi, v)
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heapify(h)
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ans = -h[0] - mi
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while h[0] % 2 == 0:
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x = heappop(h) // 2
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heappush(h, x)
15+
mi = min(mi, -x)
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ans = min(ans, -h[0] - mi)
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return ans

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