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feat: add solutions to lc problem: No.2163 #1504

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Expand Up @@ -62,28 +62,231 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一:优先队列(大小根堆)+ 前后缀和 + 枚举分割点**

题目实际上等价于在 $nums$ 中找到一个分割点,将数组分成左右两部分,在前一部分中选取最小的 $n$ 个元素,在后一部分中选取最大的 $n$ 个元素,使得两部分和的差值最小。

我们可以用一个大根堆维护前缀中最小的 $n$ 个元素,用一个小根堆维护后缀中最大的 $n$ 个元素。我们定义 $pre[i]$ 表示在数组 $nums$ 的前 $i$ 个元素中选择最小的 $n$ 个元素的和,定义 $suf[i]$ 表示从数组第 $i$ 个元素到最后一个元素中选择最大的 $n$ 个元素的和。在维护大小根堆的过程中,更新 $pre[i]$ 和 $suf[i]$ 的值。

最后,我们在 $i \in [n, 2n]$ 的范围内枚举分割点,计算 $pre[i] - suf[i + 1]$ 的值,取最小值即可。

时间复杂度 $O(n \times \log n)$,空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def minimumDifference(self, nums: List[int]) -> int:
m = len(nums)
n = m // 3

s = 0
pre = [0] * (m + 1)
q1 = []
for i, x in enumerate(nums[: n * 2], 1):
s += x
heappush(q1, -x)
if len(q1) > n:
s -= -heappop(q1)
pre[i] = s

s = 0
suf = [0] * (m + 1)
q2 = []
for i in range(m, n, -1):
x = nums[i - 1]
s += x
heappush(q2, x)
if len(q2) > n:
s -= heappop(q2)
suf[i] = s

return min(pre[i] - suf[i + 1] for i in range(n, n * 2 + 1))
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public long minimumDifference(int[] nums) {
int m = nums.length;
int n = m / 3;
long s = 0;
long[] pre = new long[m + 1];
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int i = 1; i <= n * 2; ++i) {
int x = nums[i - 1];
s += x;
pq.offer(x);
if (pq.size() > n) {
s -= pq.poll();
}
pre[i] = s;
}
s = 0;
long[] suf = new long[m + 1];
pq = new PriorityQueue<>();
for (int i = m; i > n; --i) {
int x = nums[i - 1];
s += x;
pq.offer(x);
if (pq.size() > n) {
s -= pq.poll();
}
suf[i] = s;
}
long ans = 1L << 60;
for (int i = n; i <= n * 2; ++i) {
ans = Math.min(ans, pre[i] - suf[i + 1]);
}
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
long long minimumDifference(vector<int>& nums) {
int m = nums.size();
int n = m / 3;

using ll = long long;
ll s = 0;
ll pre[m + 1];
priority_queue<int> q1;
for (int i = 1; i <= n * 2; ++i) {
int x = nums[i - 1];
s += x;
q1.push(x);
if (q1.size() > n) {
s -= q1.top();
q1.pop();
}
pre[i] = s;
}
s = 0;
ll suf[m + 1];
priority_queue<int, vector<int>, greater<int>> q2;
for (int i = m; i > n; --i) {
int x = nums[i - 1];
s += x;
q2.push(x);
if (q2.size() > n) {
s -= q2.top();
q2.pop();
}
suf[i] = s;
}
ll ans = 1e18;
for (int i = n; i <= n * 2; ++i) {
ans = min(ans, pre[i] - suf[i + 1]);
}
return ans;
}
};
```

### **Go**

```go
func minimumDifference(nums []int) int64 {
m := len(nums)
n := m / 3
s := 0
pre := make([]int, m+1)
q1 := hp{}
for i := 1; i <= n*2; i++ {
x := nums[i-1]
s += x
heap.Push(&q1, -x)
if q1.Len() > n {
s -= -heap.Pop(&q1).(int)
}
pre[i] = s
}
s = 0
suf := make([]int, m+1)
q2 := hp{}
for i := m; i > n; i-- {
x := nums[i-1]
s += x
heap.Push(&q2, x)
if q2.Len() > n {
s -= heap.Pop(&q2).(int)
}
suf[i] = s
}
ans := int64(1e18)
for i := n; i <= n*2; i++ {
ans = min(ans, int64(pre[i]-suf[i+1]))
}
return ans
}

func min(a, b int64) int64 {
if a < b {
return a
}
return b
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() interface{} {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
```

### **TypeScript**

```ts

function minimumDifference(nums: number[]): number {
const m = nums.length;
const n = Math.floor(m / 3);
let s = 0;
const pre: number[] = Array(m + 1);
const q1 = new MaxPriorityQueue();
for (let i = 1; i <= n * 2; ++i) {
const x = nums[i - 1];
s += x;
q1.enqueue(x, x);
if (q1.size() > n) {
s -= q1.dequeue().element;
}
pre[i] = s;
}
s = 0;
const suf: number[] = Array(m + 1);
const q2 = new MinPriorityQueue();
for (let i = m; i > n; --i) {
const x = nums[i - 1];
s += x;
q2.enqueue(x, x);
if (q2.size() > n) {
s -= q2.dequeue().element;
}
suf[i] = s;
}
let ans = Number.MAX_SAFE_INTEGER;
for (let i = n; i <= n * 2; ++i) {
ans = Math.min(ans, pre[i] - suf[i + 1]);
}
return ans;
}
```

### **...**
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