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>