feat: add solutions to lc problem: No.2536

solution/2500-2599/2527.Find Xor-Beauty of Array/README.md

@@ -51,7 +51,7 @@ tags:
 - (1,0,0) 有效值为 ((4 | 1) & 1) = 1
 - (1,0,1) 有效值为 ((4 | 1) & 4) = 4
 - (1,1,0) 有效值为 ((4 | 4) & 1) = 0
-- (1,1,1) 有效值为 ((4 | 4) & 4) = 4 
+- (1,1,1) 有效值为 ((4 | 4) & 4) = 4
 数组的异或美丽值为所有有效值的按位异或 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4 ^ 0 ^ 4 = 5 。</pre>
 
 <p><strong>示例 2：</strong></p>
@@ -79,9 +79,9 @@ tags:
 
 ### 方法一：位运算
 
-我们首先考虑 $i$ 与 $j$ 不相等的情况，此时 $(nums[i] | nums[j]) \& nums[k]$ 与 $(nums[j] | nums[i]) \& nums[k]$ 的结果是相同的，两者的异或结果为 $0$。
+我们首先考虑 $i$ 与 $j$ 不相等的情况，此时 `((nums[i] | nums[j]) & nums[k])` 与 `((nums[j] | nums[i]) & nums[k])` 的结果是相同的，两者的异或结果为 $0$。
 
-因此，我们只需要考虑 $i$ 与 $j$ 相等的情况。此时 $(nums[i] | nums[j]) \& nums[k] = nums[i] \& nums[k]$，如果 $i \neq k$，那么与 $nums[k] \& nums[i]$ 的结果是相同的，这些值的异或结果为 $0$。
+因此，我们只需要考虑 $i$ 与 $j$ 相等的情况。此时 `((nums[i] | nums[j]) & nums[k]) = (nums[i] & nums[k])`，如果 $i \neq k$，那么与 `nums[k] & nums[i]` 的结果是相同的，这些值的异或结果为 $0$。
 
 因此，我们最终只需要考虑 $i = j = k$ 的情况，那么答案就是所有 $nums[i]$ 的异或结果。
 

solution/2500-2599/2527.Find Xor-Beauty of Array/README_EN.md

@@ -41,7 +41,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [1,4]
 <strong>Output:</strong> 5
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 The triplets and their corresponding effective values are listed below:
 - (0,0,0) with effective value ((1 | 1) &amp; 1) = 1
 - (0,0,1) with effective value ((1 | 1) &amp; 4) = 0
@@ -50,7 +50,7 @@ The triplets and their corresponding effective values are listed below:
 - (1,0,0) with effective value ((4 | 1) &amp; 1) = 1
 - (1,0,1) with effective value ((4 | 1) &amp; 4) = 4
 - (1,1,0) with effective value ((4 | 4) &amp; 1) = 0
-- (1,1,1) with effective value ((4 | 4) &amp; 4) = 4 
+- (1,1,1) with effective value ((4 | 4) &amp; 4) = 4
 Xor-beauty of array will be bitwise XOR of all beauties = 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4 ^ 0 ^ 4 = 5.</pre>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -75,7 +75,15 @@ Xor-beauty of array will be bitwise XOR of all beauties = 1 ^ 0 ^ 1 ^ 4 ^ 1 ^ 4
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Bit Manipulation
+
+We first consider the case where $i$ and $j$ are not equal. In this case, `((nums[i] | nums[j]) & nums[k])` and `((nums[j] | nums[i]) & nums[k])` produce the same result, and their XOR result is $0$.
+
+Therefore, we only need to consider the case where $i$ and $j$ are equal. In this case, `((nums[i] | nums[j]) & nums[k]) = (nums[i] & nums[k])`. If $i \neq k$, then this is the same as the result of `nums[k] & nums[i]`, and the XOR result of these values is $0$.
+
+Therefore, we ultimately only need to consider the case where $i = j = k$, and the answer is the XOR result of all $nums[i]$.
+
+The time complexity is $O(n)$ and the space complexity is $O(1)$, where $n$ is the length of the array.
 
 <!-- tabs:start -->
 

solution/2500-2599/2530.Maximal Score After Applying K Operations/README.md

@@ -139,28 +139,21 @@ public:
 
 ```go
 func maxKelements(nums []int, k int) (ans int64) {
-	h := &hp{nums}
-	heap.Init(h)
+	h := hp{nums}
+	heap.Init(&h)
 	for ; k > 0; k-- {
-		v := h.pop()
-		ans += int64(v)
-		h.push((v + 2) / 3)
+		ans += int64(h.IntSlice[0])
+		h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
+		heap.Fix(&h, 0)
 	}
 	return
 }
 
 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 any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
-func (h *hp) Pop() any {
-	a := h.IntSlice
-	v := a[len(a)-1]
-	h.IntSlice = a[:len(a)-1]
-	return v
-}
-func (h *hp) push(v int) { heap.Push(h, v) }
-func (h *hp) pop() int   { return heap.Pop(h).(int) }
+func (hp) Push(any)             {}
+func (hp) Pop() (_ any)         { return }
 ```
 
 #### TypeScript
@@ -206,69 +199,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def maxKelements(self, nums: List[int], k: int) -> int:
-        for i, v in enumerate(nums):
-            nums[i] = -v
-        heapify(nums)
-        ans = 0
-        for _ in range(k):
-            ans -= heapreplace(nums, -ceil(-nums[0] / 3))
-        return ans
-```
-
-#### C++
-
-```cpp
-class Solution {
-public:
-    long long maxKelements(vector<int>& nums, int k) {
-        make_heap(nums.begin(), nums.end());
-        long long ans = 0;
-        while (k--) {
-            int v = nums[0];
-            ans += v;
-            pop_heap(nums.begin(), nums.end());
-            nums.back() = (v + 2) / 3;
-            push_heap(nums.begin(), nums.end());
-        }
-        return ans;
-    }
-};
-```
-
-#### Go
-
-```go
-func maxKelements(nums []int, k int) (ans int64) {
-	h := hp{nums}
-	heap.Init(&h)
-	for ; k > 0; k-- {
-		ans += int64(h.IntSlice[0])
-		h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
-		heap.Fix(&h, 0)
-	}
-	return
-}
-
-type hp struct{ sort.IntSlice }
-
-func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
-func (hp) Push(any)             {}
-func (hp) Pop() (_ any)         { return }
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2500-2599/2530.Maximal Score After Applying K Operations/README_EN.md

@@ -137,28 +137,21 @@ public:
 
 ```go
 func maxKelements(nums []int, k int) (ans int64) {
-	h := &hp{nums}
-	heap.Init(h)
+	h := hp{nums}
+	heap.Init(&h)
 	for ; k > 0; k-- {
-		v := h.pop()
-		ans += int64(v)
-		h.push((v + 2) / 3)
+		ans += int64(h.IntSlice[0])
+		h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
+		heap.Fix(&h, 0)
 	}
 	return
 }
 
 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 any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
-func (h *hp) Pop() any {
-	a := h.IntSlice
-	v := a[len(a)-1]
-	h.IntSlice = a[:len(a)-1]
-	return v
-}
-func (h *hp) push(v int) { heap.Push(h, v) }
-func (h *hp) pop() int   { return heap.Pop(h).(int) }
+func (hp) Push(any)             {}
+func (hp) Pop() (_ any)         { return }
 ```
 
 #### TypeScript
@@ -204,69 +197,4 @@ impl Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def maxKelements(self, nums: List[int], k: int) -> int:
-        for i, v in enumerate(nums):
-            nums[i] = -v
-        heapify(nums)
-        ans = 0
-        for _ in range(k):
-            ans -= heapreplace(nums, -ceil(-nums[0] / 3))
-        return ans
-```
-
-#### C++
-
-```cpp
-class Solution {
-public:
-    long long maxKelements(vector<int>& nums, int k) {
-        make_heap(nums.begin(), nums.end());
-        long long ans = 0;
-        while (k--) {
-            int v = nums[0];
-            ans += v;
-            pop_heap(nums.begin(), nums.end());
-            nums.back() = (v + 2) / 3;
-            push_heap(nums.begin(), nums.end());
-        }
-        return ans;
-    }
-};
-```
-
-#### Go
-
-```go
-func maxKelements(nums []int, k int) (ans int64) {
-	h := hp{nums}
-	heap.Init(&h)
-	for ; k > 0; k-- {
-		ans += int64(h.IntSlice[0])
-		h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
-		heap.Fix(&h, 0)
-	}
-	return
-}
-
-type hp struct{ sort.IntSlice }
-
-func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
-func (hp) Push(any)             {}
-func (hp) Pop() (_ any)         { return }
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/2500-2599/2530.Maximal Score After Applying K Operations/Solution.go

@@ -1,23 +1,16 @@
 func maxKelements(nums []int, k int) (ans int64) {
-	h := &hp{nums}
-	heap.Init(h)
+	h := hp{nums}
+	heap.Init(&h)
 	for ; k > 0; k-- {
-		v := h.pop()
-		ans += int64(v)
-		h.push((v + 2) / 3)
+		ans += int64(h.IntSlice[0])
+		h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
+		heap.Fix(&h, 0)
 	}
 	return
 }
 
 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 any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
-func (h *hp) Pop() any {
-	a := h.IntSlice
-	v := a[len(a)-1]
-	h.IntSlice = a[:len(a)-1]
-	return v
-}
-func (h *hp) push(v int) { heap.Push(h, v) }
-func (h *hp) pop() int   { return heap.Pop(h).(int) }
+func (hp) Push(any)             {}
+func (hp) Pop() (_ any)         { return }