feat: add solutions to lc problem: No.3231

solution/0500-0599/0545.Boundary of Binary Tree/README.md

@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>二叉树的 <strong>边界</strong> 是由 <strong>根节点 </strong>, <strong>左边界</strong> , 按从左到右顺序的<strong> 叶节点</strong> 和 <strong>逆序的右边界</strong> ，按顺序依次连接组成。</p>
+<p>二叉树的 <strong>边界</strong> 是由 <strong>根节点 </strong>、<strong>左边界</strong> 、按从左到右顺序的<strong> 叶节点</strong> 和 <strong>逆序的右边界</strong> ，按顺序依次连接组成。</p>
 
 <p><strong>左边界 </strong>是满足下述定义的节点集合：</p>
 

solution/3200-3299/3231.Minimum Number of Increasing Subsequence to Be Removed/README.md

@@ -0,0 +1,238 @@
+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3231.Minimum%20Number%20of%20Increasing%20Subsequence%20to%20Be%20Removed/README.md
+---
+
+<!-- problem:start -->
+
+# [3231. Minimum Number of Increasing Subsequence to Be Removed 🔒](https://leetcode.cn/problems/minimum-number-of-increasing-subsequence-to-be-removed)
+
+[English Version](/solution/3200-3299/3231.Minimum%20Number%20of%20Increasing%20Subsequence%20to%20Be%20Removed/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>Given an array of integers <code>nums</code>, you are allowed to perform the following operation any number of times:</p>
+
+<ul>
+	<li>Remove a <strong>strictly increasing</strong> <span data-keyword="subsequence-array">subsequence</span> from the array.</li>
+</ul>
+
+<p>Your task is to find the <strong>minimum</strong> number of operations required to make the array <strong>empty</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,3,1,4,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We remove subsequences <code>[1, 2]</code>, <code>[3, 4]</code>, <code>[5]</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,4,3,2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：贪心 + 二分查找
+
+我们从左到右遍历数组 $\textit{nums}$，对于每个元素 $x$，我们需要贪心地将其追加到前面序列中最后一个元素小于 $x$ 的最大值后面。如果找不到这样的元素，则说明当前元素 $x$ 比前面序列中的所有元素都小，我们需要新开辟一个序列，将 $x$ 放入其中。
+
+这样分析下来，我们可以发现，前面序列中的最后一个元素呈单调递减的状态。因此，我们可以使用二分查找来找到前面序列中最后一个元素小于 $x$ 的第一个元素位置，然后将 $x$ 放入该位置。
+
+最终，我们返回序列的个数即可。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minOperations(self, nums: List[int]) -> int:
+        g = []
+        for x in nums:
+            l, r = 0, len(g)
+            while l < r:
+                mid = (l + r) >> 1
+                if g[mid] < x:
+                    r = mid
+                else:
+                    l = mid + 1
+            if l == len(g):
+                g.append(x)
+            else:
+                g[l] = x
+        return len(g)
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minOperations(int[] nums) {
+        List<Integer> g = new ArrayList<>();
+        for (int x : nums) {
+            int l = 0, r = g.size();
+            while (l < r) {
+                int mid = (l + r) >> 1;
+                if (g.get(mid) < x) {
+                    r = mid;
+                } else {
+                    l = mid + 1;
+                }
+            }
+            if (l == g.size()) {
+                g.add(x);
+            } else {
+                g.set(l, x);
+            }
+        }
+        return g.size();
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minOperations(vector<int>& nums) {
+        vector<int> g;
+        for (int x : nums) {
+            int l = 0, r = g.size();
+            while (l < r) {
+                int mid = (l + r) >> 1;
+                if (g[mid] < x) {
+                    r = mid;
+                } else {
+                    l = mid + 1;
+                }
+            }
+            if (l == g.size()) {
+                g.push_back(x);
+            } else {
+                g[l] = x;
+            }
+        }
+        return g.size();
+    }
+};
+```
+
+#### Go
+
+```go
+func minOperations(nums []int) int {
+	g := []int{}
+	for _, x := range nums {
+		l, r := 0, len(g)
+		for l < r {
+			mid := (l + r) >> 1
+			if g[mid] < x {
+				r = mid
+			} else {
+				l = mid + 1
+			}
+		}
+		if l == len(g) {
+			g = append(g, x)
+		} else {
+			g[l] = x
+		}
+	}
+	return len(g)
+}
+```
+
+#### TypeScript
+
+```ts
+function minOperations(nums: number[]): number {
+    const g: number[] = [];
+    for (const x of nums) {
+        let [l, r] = [0, g.length];
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (g[mid] < x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        if (l === g.length) {
+            g.push(x);
+        } else {
+            g[l] = x;
+        }
+    }
+    return g.length;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+    pub fn min_operations(nums: Vec<i32>) -> i32 {
+        let mut g = Vec::new();
+        for &x in nums.iter() {
+            let mut l = 0;
+            let mut r = g.len();
+            while l < r {
+                let mid = (l + r) / 2;
+                if g[mid] < x {
+                    r = mid;
+                } else {
+                    l = mid + 1;
+                }
+            }
+            if l == g.len() {
+                g.push(x);
+            } else {
+                g[l] = x;
+            }
+        }
+        g.len() as i32
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->