feat: add solutions to lc problems: No.2221,2233 (doocs#3971)

Author: yanglbme

solution/0500-0599/0554.Brick Wall/images/a.png

@@ -69,7 +69,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们可以直接模拟题目描述的操作，对数组 $\textit{nums}$ 进行 $n - 1$ 轮操作，每轮操作都按照题目描述的规则更新数组 $\textit{nums}$。最后返回数组 $\textit{nums}$ 中剩下的唯一元素即可。
+
+时间复杂度 $O(n^2)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -78,10 +82,9 @@ tags:
 ```python
 class Solution:
     def triangularSum(self, nums: List[int]) -> int:
-        n = len(nums)
-        for i in range(n, 0, -1):
-            for j in range(i - 1):
-                nums[j] = (nums[j] + nums[j + 1]) % 10
+        for k in range(len(nums) - 1, 0, -1):
+            for i in range(k):
+                nums[i] = (nums[i] + nums[i + 1]) % 10
         return nums[0]
 ```
 
@@ -90,10 +93,9 @@ class Solution:
 ```java
 class Solution {
     public int triangularSum(int[] nums) {
-        int n = nums.length;
-        for (int i = n; i >= 0; --i) {
-            for (int j = 0; j < i - 1; ++j) {
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.length - 1; k > 0; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
             }
         }
         return nums[0];
@@ -107,10 +109,11 @@ class Solution {
 class Solution {
 public:
     int triangularSum(vector<int>& nums) {
-        int n = nums.size();
-        for (int i = n; i >= 0; --i)
-            for (int j = 0; j < i - 1; ++j)
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.size() - 1; k; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
+            }
+        }
         return nums[0];
     }
 };
@@ -120,16 +123,28 @@ public:
 
 ```go
 func triangularSum(nums []int) int {
-	n := len(nums)
-	for i := n; i >= 0; i-- {
-		for j := 0; j < i-1; j++ {
-			nums[j] = (nums[j] + nums[j+1]) % 10
+	for k := len(nums) - 1; k > 0; k-- {
+		for i := 0; i < k; i++ {
+			nums[i] = (nums[i] + nums[i+1]) % 10
 		}
 	}
 	return nums[0]
 }
 ```
 
+#### TypeScript
+
+```ts
+function triangularSum(nums: number[]): number {
+    for (let k = nums.length - 1; k; --k) {
+        for (let i = 0; i < k; ++i) {
+            nums[i] = (nums[i] + nums[i + 1]) % 10;
+        }
+    }
+    return nums[0];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2221.Find Triangular Sum of an Array/README.md

@@ -65,7 +65,11 @@ Since there is only one element in nums, the triangular sum is the value of that
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We can directly simulate the operations described in the problem. Perform $n - 1$ rounds of operations on the array $\textit{nums}$, updating the array $\textit{nums}$ according to the rules described in the problem for each round. Finally, return the only remaining element in the array $\textit{nums}$.
+
+The time complexity is $O(n^2)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -74,10 +78,9 @@ Since there is only one element in nums, the triangular sum is the value of that
 ```python
 class Solution:
     def triangularSum(self, nums: List[int]) -> int:
-        n = len(nums)
-        for i in range(n, 0, -1):
-            for j in range(i - 1):
-                nums[j] = (nums[j] + nums[j + 1]) % 10
+        for k in range(len(nums) - 1, 0, -1):
+            for i in range(k):
+                nums[i] = (nums[i] + nums[i + 1]) % 10
         return nums[0]
 ```
 
@@ -86,10 +89,9 @@ class Solution:
 ```java
 class Solution {
     public int triangularSum(int[] nums) {
-        int n = nums.length;
-        for (int i = n; i >= 0; --i) {
-            for (int j = 0; j < i - 1; ++j) {
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.length - 1; k > 0; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
             }
         }
         return nums[0];
@@ -103,10 +105,11 @@ class Solution {
 class Solution {
 public:
     int triangularSum(vector<int>& nums) {
-        int n = nums.size();
-        for (int i = n; i >= 0; --i)
-            for (int j = 0; j < i - 1; ++j)
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.size() - 1; k; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
+            }
+        }
         return nums[0];
     }
 };
@@ -116,16 +119,28 @@ public:
 
 ```go
 func triangularSum(nums []int) int {
-	n := len(nums)
-	for i := n; i >= 0; i-- {
-		for j := 0; j < i-1; j++ {
-			nums[j] = (nums[j] + nums[j+1]) % 10
+	for k := len(nums) - 1; k > 0; k-- {
+		for i := 0; i < k; i++ {
+			nums[i] = (nums[i] + nums[i+1]) % 10
 		}
 	}
 	return nums[0]
 }
 ```
 
+#### TypeScript
+
+```ts
+function triangularSum(nums: number[]): number {
+    for (let k = nums.length - 1; k; --k) {
+        for (let i = 0; i < k; ++i) {
+            nums[i] = (nums[i] + nums[i + 1]) % 10;
+        }
+    }
+    return nums[0];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2221.Find Triangular Sum of an Array/README_EN.md

@@ -1,10 +1,11 @@
 class Solution {
 public:
     int triangularSum(vector<int>& nums) {
-        int n = nums.size();
-        for (int i = n; i >= 0; --i)
-            for (int j = 0; j < i - 1; ++j)
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.size() - 1; k; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
+            }
+        }
         return nums[0];
     }
-};
+};

solution/2200-2299/2221.Find Triangular Sum of an Array/Solution.cpp

@@ -1,9 +1,8 @@
 func triangularSum(nums []int) int {
-	n := len(nums)
-	for i := n; i >= 0; i-- {
-		for j := 0; j < i-1; j++ {
-			nums[j] = (nums[j] + nums[j+1]) % 10
+	for k := len(nums) - 1; k > 0; k-- {
+		for i := 0; i < k; i++ {
+			nums[i] = (nums[i] + nums[i+1]) % 10
 		}
 	}
 	return nums[0]
-}
+}

solution/2200-2299/2221.Find Triangular Sum of an Array/Solution.go

@@ -1,11 +1,10 @@
 class Solution {
     public int triangularSum(int[] nums) {
-        int n = nums.length;
-        for (int i = n; i >= 0; --i) {
-            for (int j = 0; j < i - 1; ++j) {
-                nums[j] = (nums[j] + nums[j + 1]) % 10;
+        for (int k = nums.length - 1; k > 0; --k) {
+            for (int i = 0; i < k; ++i) {
+                nums[i] = (nums[i] + nums[i + 1]) % 10;
             }
         }
         return nums[0];
     }
-}
+}

solution/2200-2299/2221.Find Triangular Sum of an Array/Solution.java

@@ -1,7 +1,6 @@
 class Solution:
     def triangularSum(self, nums: List[int]) -> int:
-        n = len(nums)
-        for i in range(n, 0, -1):
-            for j in range(i - 1):
-                nums[j] = (nums[j] + nums[j + 1]) % 10
+        for k in range(len(nums) - 1, 0, -1):
+            for i in range(k):
+                nums[i] = (nums[i] + nums[i + 1]) % 10
         return nums[0]

solution/2200-2299/2221.Find Triangular Sum of an Array/Solution.py

@@ -0,0 +1,8 @@
+function triangularSum(nums: number[]): number {
+    for (let k = nums.length - 1; k; --k) {
+        for (let i = 0; i < k; ++i) {
+            nums[i] = (nums[i] + nums[i + 1]) % 10;
+        }
+    }
+    return nums[0];
+}

solution/2200-2299/2221.Find Triangular Sum of an Array/Solution.ts

@@ -63,9 +63,9 @@ tags:
 
 ### 方法一：贪心 + 优先队列（小根堆）
 
-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。
+根据题目描述，要使得乘积最大，我们需要尽量增大较小的数，因此我们可以使用小根堆来维护数组 $\textit{nums}$。每次从小根堆中取出最小的数，将其增加 $1$，然后重新放回小根堆中。重复这个过程 $k$ 次后，我们将当前小根堆中的所有数相乘，即可得到答案。
 
-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。
+时间复杂度 $O(k \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -76,31 +76,27 @@ class Solution:
     def maximumProduct(self, nums: List[int], k: int) -> int:
         heapify(nums)
         for _ in range(k):
-            heappush(nums, heappop(nums) + 1)
-        ans = 1
+            heapreplace(nums, nums[0] + 1)
         mod = 10**9 + 7
-        for v in nums:
-            ans = (ans * v) % mod
-        return ans
+        return reduce(lambda x, y: x * y % mod, nums)
 ```
 
 #### Java
 
 ```java
 class Solution {
-    private static final int MOD = (int) 1e9 + 7;
-
     public int maximumProduct(int[] nums, int k) {
-        PriorityQueue<Integer> q = new PriorityQueue<>();
-        for (int v : nums) {
-            q.offer(v);
+        PriorityQueue<Integer> pq = new PriorityQueue<>();
+        for (int x : nums) {
+            pq.offer(x);
         }
         while (k-- > 0) {
-            q.offer(q.poll() + 1);
+            pq.offer(pq.poll() + 1);
         }
+        final int mod = (int) 1e9 + 7;
         long ans = 1;
-        while (!q.isEmpty()) {
-            ans = (ans * q.poll()) % MOD;
+        for (int x : pq) {
+            ans = (ans * x) % mod;
         }
         return (int) ans;
     }
@@ -113,16 +109,22 @@ class Solution {
 class Solution {
 public:
     int maximumProduct(vector<int>& nums, int k) {
-        int mod = 1e9 + 7;
-        make_heap(nums.begin(), nums.end(), greater<int>());
-        while (k--) {
-            pop_heap(nums.begin(), nums.end(), greater<int>());
-            ++nums.back();
-            push_heap(nums.begin(), nums.end(), greater<int>());
+        priority_queue<int, vector<int>, greater<int>> pq;
+        for (int x : nums) {
+            pq.push(x);
+        }
+        while (k-- > 0) {
+            int smallest = pq.top();
+            pq.pop();
+            pq.push(smallest + 1);
         }
+        const int mod = 1e9 + 7;
         long long ans = 1;
-        for (int v : nums) ans = (ans * v) % mod;
-        return ans;
+        while (!pq.empty()) {
+            ans = (ans * pq.top()) % mod;
+            pq.pop();
+        }
+        return static_cast<int>(ans);
     }
 };
 ```
@@ -137,8 +139,8 @@ func maximumProduct(nums []int, k int) int {
 		heap.Fix(&h, 0)
 	}
 	ans := 1
-	for _, v := range nums {
-		ans = (ans * v) % (1e9 + 7)
+	for _, x := range nums {
+		ans = (ans * x) % (1e9 + 7)
 	}
 	return ans
 }
@@ -149,6 +151,25 @@ func (hp) Push(any)     {}
 func (hp) Pop() (_ any) { return }
 ```
 
+#### TypeScript
+
+```ts
+function maximumProduct(nums: number[], k: number): number {
+    const pq = new MinPriorityQueue();
+    nums.forEach(x => pq.enqueue(x));
+    while (k--) {
+        const x = pq.dequeue().element;
+        pq.enqueue(x + 1);
+    }
+    let ans = 1;
+    const mod = 10 ** 9 + 7;
+    while (!pq.isEmpty()) {
+        ans = (ans * pq.dequeue().element) % mod;
+    }
+    return ans;
+}
+```
+
 #### JavaScript
 
 ```js
@@ -158,18 +179,16 @@ func (hp) Pop() (_ any) { return }
  * @return {number}
  */
 var maximumProduct = function (nums, k) {
-    const n = nums.length;
-    let pq = new MinPriorityQueue();
-    for (let i = 0; i < n; i++) {
-        pq.enqueue(nums[i]);
-    }
-    for (let i = 0; i < k; i++) {
-        pq.enqueue(pq.dequeue().element + 1);
+    const pq = new MinPriorityQueue();
+    nums.forEach(x => pq.enqueue(x));
+    while (k--) {
+        const x = pq.dequeue().element;
+        pq.enqueue(x + 1);
     }
     let ans = 1;
-    const limit = 10 ** 9 + 7;
-    for (let i = 0; i < n; i++) {
-        ans = (ans * pq.dequeue().element) % limit;
+    const mod = 10 ** 9 + 7;
+    while (!pq.isEmpty()) {
+        ans = (ans * pq.dequeue().element) % mod;
     }
     return ans;
 };