feat: add solutions to lc problems: No.3319,3322 (doocs#3639)

Author: yanglbme

solution/0100-0199/0112.Path Sum/README_EN.md

@@ -37,7 +37,7 @@ tags:
 <pre>
 <strong>Input:</strong> root = [1,2,3], targetSum = 5
 <strong>Output:</strong> false
-<strong>Explanation:</strong> There two root-to-leaf paths in the tree:
+<strong>Explanation:</strong> There are two root-to-leaf paths in the tree:
 (1 --&gt; 2): The sum is 3.
 (1 --&gt; 3): The sum is 4.
 There is no root-to-leaf path with sum = 5.

solution/0400-0499/0442.Find All Duplicates in an Array/README_EN.md

@@ -17,7 +17,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Given an integer array <code>nums</code> of length <code>n</code> where all the integers of <code>nums</code> are in the range <code>[1, n]</code> and each integer appears <strong>once</strong> or <strong>twice</strong>, return <em>an array of all the integers that appears <strong>twice</strong></em>.</p>
+<p>Given an integer array <code>nums</code> of length <code>n</code> where all the integers of <code>nums</code> are in the range <code>[1, n]</code> and each integer appears <strong>at most</strong> <strong>twice</strong>, return <em>an array of all the integers that appears <strong>twice</strong></em>.</p>
 
 <p>You must write an algorithm that runs in <code>O(n)</code> time and uses only <em>constant</em> auxiliary space, excluding the space needed to store the output</p>
 

solution/0800-0899/0887.Super Egg Drop/README.md

@@ -158,7 +158,7 @@ public:
     int superEggDrop(int k, int n) {
         int f[n + 1][k + 1];
         memset(f, 0, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i < 1) {
                 return 0;
             }
@@ -171,17 +171,17 @@ public:
             int l = 1, r = i;
             while (l < r) {
                 int mid = (l + r + 1) >> 1;
-                int a = dfs(mid - 1, j - 1);
-                int b = dfs(i - mid, j);
+                int a = dfs(dfs, mid - 1, j - 1);
+                int b = dfs(dfs, i - mid, j);
                 if (a <= b) {
                     l = mid;
                 } else {
                     r = mid - 1;
                 }
             }
-            return f[i][j] = max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1;
+            return f[i][j] = max(dfs(dfs, l - 1, j - 1), dfs(dfs, i - l, j)) + 1;
         };
-        return dfs(n, k);
+        return dfs(dfs, n, k);
     }
 };
 ```

solution/0800-0899/0887.Super Egg Drop/README_EN.md

@@ -141,7 +141,7 @@ public:
     int superEggDrop(int k, int n) {
         int f[n + 1][k + 1];
         memset(f, 0, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i < 1) {
                 return 0;
             }
@@ -154,17 +154,17 @@ public:
             int l = 1, r = i;
             while (l < r) {
                 int mid = (l + r + 1) >> 1;
-                int a = dfs(mid - 1, j - 1);
-                int b = dfs(i - mid, j);
+                int a = dfs(dfs, mid - 1, j - 1);
+                int b = dfs(dfs, i - mid, j);
                 if (a <= b) {
                     l = mid;
                 } else {
                     r = mid - 1;
                 }
             }
-            return f[i][j] = max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1;
+            return f[i][j] = max(dfs(dfs, l - 1, j - 1), dfs(dfs, i - l, j)) + 1;
         };
-        return dfs(n, k);
+        return dfs(dfs, n, k);
     }
 };
 ```

solution/0800-0899/0887.Super Egg Drop/Solution.cpp

@@ -3,7 +3,7 @@ class Solution {
     int superEggDrop(int k, int n) {
         int f[n + 1][k + 1];
         memset(f, 0, sizeof(f));
-        function<int(int, int)> dfs = [&](int i, int j) -> int {
+        auto dfs = [&](auto&& dfs, int i, int j) -> int {
             if (i < 1) {
                 return 0;
             }
@@ -16,16 +16,16 @@ class Solution {
             int l = 1, r = i;
             while (l < r) {
                 int mid = (l + r + 1) >> 1;
-                int a = dfs(mid - 1, j - 1);
-                int b = dfs(i - mid, j);
+                int a = dfs(dfs, mid - 1, j - 1);
+                int b = dfs(dfs, i - mid, j);
                 if (a <= b) {
                     l = mid;
                 } else {
                     r = mid - 1;
                 }
             }
-            return f[i][j] = max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1;
+            return f[i][j] = max(dfs(dfs, l - 1, j - 1), dfs(dfs, i - l, j)) + 1;
         };
-        return dfs(n, k);
+        return dfs(dfs, n, k);
     }
-};
+};

solution/3000-3099/3016.Minimum Number of Pushes to Type Word II/README_EN.md

@@ -49,7 +49,7 @@ It can be shown that no other mapping can provide a lower cost.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/images/keypadv2e2.png" style="width: 329px; height: 313px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/images/edited.png" style="width: 329px; height: 313px;" />
 <pre>
 <strong>Input:</strong> word = &quot;xyzxyzxyzxyz&quot;
 <strong>Output:</strong> 12

solution/3000-3099/3095.Shortest Subarray With OR at Least K I/README_EN.md

@@ -37,6 +37,8 @@ tags:
 <p><strong>Explanation:</strong></p>
 
 <p>The subarray <code>[3]</code> has <code>OR</code> value of <code>3</code>. Hence, we return <code>1</code>.</p>
+
+<p>Note that <code>[2]</code> is also a special subarray.</p>
 </div>
 
 <p><strong class="example">Example 2:</strong></p>

solution/3300-3399/3314.Construct the Minimum Bitwise Array I/README_EN.md

@@ -22,8 +22,6 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3314.Co
 
 <p>If it is <em>not possible</em> to find such a value for <code>ans[i]</code> that satisfies the <strong>condition</strong>, then set <code>ans[i] = -1</code>.</p>
 
-<p>A <strong>prime number</strong> is a natural number greater than 1 with only two factors, 1 and itself.</p>
-
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/3300-3399/3315.Construct the Minimum Bitwise Array II/README_EN.md

@@ -22,8 +22,6 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3315.Co
 
 <p>If it is <em>not possible</em> to find such a value for <code>ans[i]</code> that satisfies the <strong>condition</strong>, then set <code>ans[i] = -1</code>.</p>
 
-<p>A <strong>prime number</strong> is a natural number greater than 1 with only two factors, 1 and itself.</p>
-
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/3300-3399/3316.Find Maximum Removals From Source String/README_EN.md

@@ -27,8 +27,6 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3316.Fi
 
 <p>Return the <strong>maximum</strong> number of <em>operations</em> that can be performed.</p>
 
-<p>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</p>
-
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/3300-3399/3319.K-th Largest Perfect Subtree Size in Binary Tree/README.md

@@ -83,32 +83,213 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3319.K-
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：DFS + 排序
+
+我们定义一个函数 $\textit{dfs}$，用于计算以当前节点为根节点的完美二叉子树的大小，用一个数组 $\textit{nums}$ 记录所有完美二叉子树的大小。如果以当前节点为根节点的子树不是完美二叉子树，则返回 $-1$。
+
+函数 $\textit{dfs}$ 的执行过程如下：
+
+1. 如果当前节点为空，则返回 $0$；
+2. 递归计算左子树和右子树的完美二叉子树的大小，分别记为 $l$ 和 $r$；
+3. 如果左子树和右子树的大小不相等，或者左子树和右子树的大小小于 $0$，则返回 $-1$；
+4. 计算当前节点的完美二叉子树的大小 $\textit{cnt} = l + r + 1$，并将 $\textit{cnt}$ 添加到数组 $\textit{nums}$ 中；
+5. 返回 $\textit{cnt}$。
+
+我们调用 $\textit{dfs}$ 函数计算出所有完美二叉子树的大小，如果数组 $\textit{nums}$ 的长度小于 $k$，则返回 $-1$，否则对数组 $\textit{nums}$ 进行降序排序，返回第 $k$ 大的完美二叉子树的大小。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def kthLargestPerfectSubtree(self, root: Optional[TreeNode], k: int) -> int:
+        def dfs(root: Optional[TreeNode]) -> int:
+            if root is None:
+                return 0
+            l, r = dfs(root.left), dfs(root.right)
+            if l < 0 or l != r:
+                return -1
+            cnt = l + r + 1
+            nums.append(cnt)
+            return cnt
+
+        nums = []
+        dfs(root)
+        if len(nums) < k:
+            return -1
+        nums.sort(reverse=True)
+        return nums[k - 1]
 ```
 
 #### Java
 
 ```java
-
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    private List<Integer> nums = new ArrayList<>();
+
+    public int kthLargestPerfectSubtree(TreeNode root, int k) {
+        dfs(root);
+        if (nums.size() < k) {
+            return -1;
+        }
+        nums.sort(Comparator.reverseOrder());
+        return nums.get(k - 1);
+    }
+
+    private int dfs(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        int l = dfs(root.left);
+        int r = dfs(root.right);
+        if (l < 0 || l != r) {
+            return -1;
+        }
+        int cnt = l + r + 1;
+        nums.add(cnt);
+        return cnt;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int kthLargestPerfectSubtree(TreeNode* root, int k) {
+        vector<int> nums;
+        auto dfs = [&](auto&& dfs, TreeNode* root) -> int {
+            if (!root) {
+                return 0;
+            }
+            int l = dfs(dfs, root->left);
+            int r = dfs(dfs, root->right);
+            if (l < 0 || l != r) {
+                return -1;
+            }
+            int cnt = l + r + 1;
+            nums.push_back(cnt);
+            return cnt;
+        };
+        dfs(dfs, root);
+        if (nums.size() < k) {
+            return -1;
+        }
+        ranges::sort(nums, greater<int>());
+        return nums[k - 1];
+    }
+};
 ```
 
 #### Go
 
 ```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func kthLargestPerfectSubtree(root *TreeNode, k int) int {
+	nums := []int{}
+	var dfs func(*TreeNode) int
+	dfs = func(root *TreeNode) int {
+		if root == nil {
+			return 0
+		}
+		l, r := dfs(root.Left), dfs(root.Right)
+		if l < 0 || l != r {
+			return -1
+		}
+		cnt := l + r + 1
+		nums = append(nums, cnt)
+		return cnt
+	}
+	dfs(root)
+	if len(nums) < k {
+		return -1
+	}
+	sort.Sort(sort.Reverse(sort.IntSlice(nums)))
+	return nums[k-1]
+}
+```
 
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function kthLargestPerfectSubtree(root: TreeNode | null, k: number): number {
+    const nums: number[] = [];
+    const dfs = (root: TreeNode | null): number => {
+        if (!root) {
+            return 0;
+        }
+        const l = dfs(root.left);
+        const r = dfs(root.right);
+        if (l < 0 || l !== r) {
+            return -1;
+        }
+        const cnt = l + r + 1;
+        nums.push(cnt);
+        return cnt;
+    };
+    dfs(root);
+    if (nums.length < k) {
+        return -1;
+    }
+    return nums.sort((a, b) => b - a)[k - 1];
+}
 ```
 
 <!-- tabs:end -->