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feat: add solutions to lc problem: No.2233
No.2233.Maximum Product After K Increments
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‎solution/2200-2299/2233.Maximum Product After K Increments/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（小根堆）**
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每次操作，贪心地选择最小的元素进行加 $1$，共进行 $k$ 次操作。最后累乘所有元素得到结果。注意取模操作。
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时间复杂度 $O(n+klogn)$。其中，$n$ 表示 $nums$ 的长度。建堆的时间复杂度为 $O(n)$，每次弹出最小元素进行加 $1$，再放回堆中，时间复杂度为 $O(logn)$，共进行 $k$ 次操作。
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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 maximumProduct(self, nums: List[int], k: int) -> int:
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heapify(nums)
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for _ in range(k):
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heappush(nums, heappop(nums) + 1)
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ans = 1
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mod = 10**9 + 7
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for v in nums:
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ans = (ans * v) % mod
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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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private static final int MOD = (int) 1e9 + 7;
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public int maximumProduct(int[] nums, int k) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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for (int v : nums) {
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q.offer(v);
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}
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while (k-- > 0) {
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q.offer(q.poll() + 1);
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}
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long ans = 1;
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while (!q.isEmpty()) {
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ans = (ans * q.poll()) % MOD;
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}
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return (int) 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 maximumProduct(vector<int>& nums, int k) {
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int mod = 1e9 + 7;
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make_heap(nums.begin(), nums.end(), greater<int>());
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while (k--)
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{
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pop_heap(nums.begin(), nums.end(), greater<int>());
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++nums.back();
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push_heap(nums.begin(), nums.end(), greater<int>());
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}
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long long ans = 1;
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for (int v : nums) ans = (ans * v) % mod;
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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 maximumProduct(nums []int, k int) int {
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h := hp{nums}
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for heap.Init(&h); k > 0; k-- {
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h.IntSlice[0]++
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heap.Fix(&h, 0)
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}
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ans := 1
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for _, v := range nums {
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ans = (ans * v) % (1e9 + 7)
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}
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return ans
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}
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type hp struct{ sort.IntSlice }
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func (hp) Push(interface{}) {}
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func (hp) Pop() (_ interface{}) { return }
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```
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### **JavaScript**

‎solution/2200-2299/2233.Maximum Product After K Increments/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maximumProduct(self, nums: List[int], k: int) -> int:
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heapify(nums)
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for _ in range(k):
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heappush(nums, heappop(nums) + 1)
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ans = 1
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mod = 10**9 + 7
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for v in nums:
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ans = (ans * v) % mod
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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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private static final int MOD = (int) 1e9 + 7;
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public int maximumProduct(int[] nums, int k) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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for (int v : nums) {
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q.offer(v);
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}
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while (k-- > 0) {
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q.offer(q.poll() + 1);
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}
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long ans = 1;
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while (!q.isEmpty()) {
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ans = (ans * q.poll()) % MOD;
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}
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return (int) 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 maximumProduct(vector<int>& nums, int k) {
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int mod = 1e9 + 7;
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make_heap(nums.begin(), nums.end(), greater<int>());
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while (k--)
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{
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pop_heap(nums.begin(), nums.end(), greater<int>());
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++nums.back();
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push_heap(nums.begin(), nums.end(), greater<int>());
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}
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long long ans = 1;
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for (int v : nums) ans = (ans * v) % mod;
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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 maximumProduct(nums []int, k int) int {
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h := hp{nums}
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for heap.Init(&h); k > 0; k-- {
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h.IntSlice[0]++
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heap.Fix(&h, 0)
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}
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ans := 1
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for _, v := range nums {
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ans = (ans * v) % (1e9 + 7)
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}
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return ans
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}
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type hp struct{ sort.IntSlice }
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func (hp) Push(interface{}) {}
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func (hp) Pop() (_ interface{}) { return }
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```
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### **JavaScript**

‎solution/2200-2299/2233.Maximum Product After K Increments/Solution.cpp

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class Solution {
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public:
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int maximumProduct(vector<int>& nums, int k) {
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int mod = 1e9 + 7;
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make_heap(nums.begin(), nums.end(), greater<int>());
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while (k--)
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{
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pop_heap(nums.begin(), nums.end(), greater<int>());
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++nums.back();
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push_heap(nums.begin(), nums.end(), greater<int>());
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}
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long long ans = 1;
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for (int v : nums) ans = (ans * v) % mod;
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return ans;
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}
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};

‎solution/2200-2299/2233.Maximum Product After K Increments/Solution.go

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func maximumProduct(nums []int, k int) int {
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h := hp{nums}
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for heap.Init(&h); k > 0; k-- {
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h.IntSlice[0]++
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heap.Fix(&h, 0)
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}
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ans := 1
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for _, v := range nums {
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ans = (ans * v) % (1e9 + 7)
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}
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return ans
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}
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type hp struct{ sort.IntSlice }
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func (hp) Push(interface{}) {}
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func (hp) Pop() (_ interface{}) { return }

‎solution/2200-2299/2233.Maximum Product After K Increments/Solution.java

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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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public int maximumProduct(int[] nums, int k) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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for (int v : nums) {
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q.offer(v);
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}
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while (k-- > 0) {
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q.offer(q.poll() + 1);
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}
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long ans = 1;
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while (!q.isEmpty()) {
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ans = (ans * q.poll()) % MOD;
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}
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return (int) ans;
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}
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}

‎solution/2200-2299/2233.Maximum Product After K Increments/Solution.py

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@@ -0,0 +1,10 @@
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class Solution:
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def maximumProduct(self, nums: List[int], k: int) -> int:
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heapify(nums)
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for _ in range(k):
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heappush(nums, heappop(nums) + 1)
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ans = 1
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mod = 10**9 + 7
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for v in nums:
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ans = (ans * v) % mod
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return ans

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