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feat: add solutions to lc problem: No.1962 #2144

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Dec 23, 2023
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feat: add solutions to lc problem: No.1962 #2144

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94 changes: 61 additions & 33 deletions solution/1900-1999/1962.Remove Stones to Minimize the Total/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,7 +57,17 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：优先队列（大根堆）**
**方法一：贪心 + 优先队列（大根堆）**

根据题目描述，为了使得剩下的石子总数最小，我们需要尽可能多地移除石子堆中的石子。因此，每次应该选择数量最多的石子堆进行移除。

我们创建一个优先队列（大根堆） $pq$，用于存储石子堆的数量。初始时，将所有石子堆的数量加入优先队列。

接下来，我们进行 $k$ 次操作。在每一次操作中，我们取出优先队列的堆顶元素 $x$，将 $x$ 减半后重新加入优先队列。

在进行了 $k$ 次操作后，优先队列中所有元素的和即为答案。

时间复杂度 $O(n + k \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 `piles` 的长度。

<!-- tabs:start -->

Expand All @@ -68,13 +78,11 @@
```python
class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
h = []
for p in piles:
heappush(h, -p)
pq = [-x for x in piles]
heapify(pq)
for _ in range(k):
p = -heappop(h)
heappush(h, -((p + 1) >> 1))
return -sum(h)
heapreplace(pq, pq[0] // 2)
return -sum(pq)
```

### **Java**
Expand All @@ -84,17 +92,17 @@ class Solution:
```java
class Solution {
public int minStoneSum(int[] piles, int k) {
PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> (b - a));
for (int p : piles) {
q.offer(p);
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int x : piles) {
pq.offer(x);
}
while (k-- > 0) {
int p = q.poll();
q.offer((p + 1) >> 1);
int x = pq.poll();
pq.offer(x - x / 2);
}
int ans = 0;
while (!q.isEmpty()) {
ans += q.poll();
while (!pq.isEmpty()) {
ans += pq.poll();
}
return ans;
}
Expand All @@ -107,17 +115,19 @@ class Solution {
class Solution {
public:
int minStoneSum(vector<int>& piles, int k) {
priority_queue<int> q;
for (int& p : piles) q.push(p);
priority_queue<int> pq;
for (int x : piles) {
pq.push(x);
}
while (k--) {
int p = q.top();
q.pop();
q.push((p + 1) >> 1);
int x = pq.top();
pq.pop();
pq.push(x - x / 2);
}
int ans = 0;
while (!q.empty()) {
ans += q.top();
q.pop();
while (!pq.empty()) {
ans += pq.top();
pq.pop();
}
return ans;
}
Expand All @@ -127,19 +137,17 @@ public:
### **Go**

```go
func minStoneSum(piles []int, k int) int {
q := &hp{piles}
heap.Init(q)
for k > 0 {
p := q.pop()
q.push((p + 1) >> 1)
k--
func minStoneSum(piles []int, k int) (ans int) {
pq := &hp{piles}
heap.Init(pq)
for ; k > 0; k-- {
x := pq.pop()
pq.push(x - x/2)
}
ans := 0
for q.Len() > 0 {
ans += q.pop()
for pq.Len() > 0 {
ans += pq.pop()
}
return ans
return
}

type hp struct{ sort.IntSlice }
Expand All @@ -156,6 +164,26 @@ func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int { return heap.Pop(h).(int) }
```

### **TypeScript**

```ts
function minStoneSum(piles: number[], k: number): number {
const pq = new MaxPriorityQueue();
for (const x of piles) {
pq.enqueue(x);
}
while (k--) {
const x = pq.dequeue().element;
pq.enqueue(x - ((x / 2) | 0));
}
let ans = 0;
while (pq.size()) {
ans += pq.dequeue().element;
}
return ans;
}
```

### **...**

```
Expand Down
94 changes: 62 additions & 32 deletions solution/1900-1999/1962.Remove Stones to Minimize the Total/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -51,38 +51,48 @@ The total number of stones in [2,3,3,4] is 12.

## Solutions

**Solution 1: Greedy + Priority Queue (Max Heap)**

According to the problem description, in order to minimize the total number of remaining stones, we need to remove as many stones as possible from the stone piles. Therefore, we should always choose the pile with the most stones for removal.

We create a priority queue (max heap) $pq$ to store the number of stones in each pile. Initially, we add the number of stones in all piles to the priority queue.

Next, we perform $k$ operations. In each operation, we take out the top element $x$ of the priority queue, halve $x$, and then add it back to the priority queue.

After performing $k$ operations, the sum of all elements in the priority queue is the answer.

The time complexity is $O(n + k \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array `piles`.

<!-- tabs:start -->

### **Python3**

```python
class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
h = []
for p in piles:
heappush(h, -p)
pq = [-x for x in piles]
heapify(pq)
for _ in range(k):
p = -heappop(h)
heappush(h, -((p + 1) >> 1))
return -sum(h)
heapreplace(pq, pq[0] // 2)
return -sum(pq)
```

### **Java**

```java
class Solution {
public int minStoneSum(int[] piles, int k) {
PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> (b - a));
for (int p : piles) {
q.offer(p);
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int x : piles) {
pq.offer(x);
}
while (k-- > 0) {
int p = q.poll();
q.offer((p + 1) >> 1);
int x = pq.poll();
pq.offer(x - x / 2);
}
int ans = 0;
while (!q.isEmpty()) {
ans += q.poll();
while (!pq.isEmpty()) {
ans += pq.poll();
}
return ans;
}
Expand All @@ -95,17 +105,19 @@ class Solution {
class Solution {
public:
int minStoneSum(vector<int>& piles, int k) {
priority_queue<int> q;
for (int& p : piles) q.push(p);
priority_queue<int> pq;
for (int x : piles) {
pq.push(x);
}
while (k--) {
int p = q.top();
q.pop();
q.push((p + 1) >> 1);
int x = pq.top();
pq.pop();
pq.push(x - x / 2);
}
int ans = 0;
while (!q.empty()) {
ans += q.top();
q.pop();
while (!pq.empty()) {
ans += pq.top();
pq.pop();
}
return ans;
}
Expand All @@ -115,19 +127,17 @@ public:
### **Go**

```go
func minStoneSum(piles []int, k int) int {
q := &hp{piles}
heap.Init(q)
for k > 0 {
p := q.pop()
q.push((p + 1) >> 1)
k--
func minStoneSum(piles []int, k int) (ans int) {
pq := &hp{piles}
heap.Init(pq)
for ; k > 0; k-- {
x := pq.pop()
pq.push(x - x/2)
}
ans := 0
for q.Len() > 0 {
ans += q.pop()
for pq.Len() > 0 {
ans += pq.pop()
}
return ans
return
}

type hp struct{ sort.IntSlice }
Expand All @@ -144,6 +154,26 @@ func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int { return heap.Pop(h).(int) }
```

### **TypeScript**

```ts
function minStoneSum(piles: number[], k: number): number {
const pq = new MaxPriorityQueue();
for (const x of piles) {
pq.enqueue(x);
}
while (k--) {
const x = pq.dequeue().element;
pq.enqueue(x - ((x / 2) | 0));
}
let ans = 0;
while (pq.size()) {
ans += pq.dequeue().element;
}
return ans;
}
```

### **...**

```
Expand Down
36 changes: 19 additions & 17 deletions solution/1900-1999/1962.Remove Stones to Minimize the Total/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,20 @@
class Solution {
public:
int minStoneSum(vector<int>& piles, int k) {
priority_queue<int> q;
for (int& p : piles) q.push(p);
while (k--) {
int p = q.top();
q.pop();
q.push((p + 1) >> 1);
}
int ans = 0;
while (!q.empty()) {
ans += q.top();
q.pop();
}
return ans;
}
class Solution {
public:
int minStoneSum(vector<int>& piles, int k) {
priority_queue<int> pq;
for (int x : piles) {
pq.push(x);
}
while (k--) {
int x = pq.top();
pq.pop();
pq.push(x - x / 2);
}
int ans = 0;
while (!pq.empty()) {
ans += pq.top();
pq.pop();
}
return ans;
}
};
20 changes: 9 additions & 11 deletions solution/1900-1999/1962.Remove Stones to Minimize the Total/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,16 +1,14 @@
func minStoneSum(piles []int, k int) int {
q := &hp{piles}
heap.Init(q)
for k > 0 {
p := q.pop()
q.push((p + 1) >> 1)
k--
func minStoneSum(piles []int, k int) (ans int) {
pq := &hp{piles}
heap.Init(pq)
for ; k > 0; k-- {
x := pq.pop()
pq.push(x - x/2)
}
ans := 0
for q.Len() > 0 {
ans += q.pop()
for pq.Len() > 0 {
ans += pq.pop()
}
return ans
return
}

type hp struct{ sort.IntSlice }
Expand Down
32 changes: 16 additions & 16 deletions solution/1900-1999/1962.Remove Stones to Minimize the Total/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,17 +1,17 @@
class Solution {
public int minStoneSum(int[] piles, int k) {
PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> (b - a));
for (int p : piles) {
q.offer(p);
}
while (k-- > 0) {
int p = q.poll();
q.offer((p + 1) >> 1);
}
int ans = 0;
while (!q.isEmpty()) {
ans += q.poll();
}
return ans;
}
class Solution {
public int minStoneSum(int[] piles, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int x : piles) {
pq.offer(x);
}
while (k-- > 0) {
int x = pq.poll();
pq.offer(x - x / 2);
}
int ans = 0;
while (!pq.isEmpty()) {
ans += pq.poll();
}
return ans;
}
}
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