feat: add solutions to lc problems: No.2824~2831

solution/2800-2899/2823.Deep Object Filter/README_EN.md

@@ -55,7 +55,7 @@ fn = (x) => Array.isArray(x)
 <ul>
 	<li><code>fn</code> is a function that returns a boolean value</li>
 	<li><code>obj</code> is a valid JSON object</li>
-	<li><code>2 &lt;= JSON.stringify(obj).length &lt;= 10**5</code></li>
+	<li><code>2 &lt;= JSON.stringify(obj).length &lt;= 10<sup>5</sup></code></li>
 </ul>
 
 ## Solutions

solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/README.md

@@ -0,0 +1,175 @@
+# [2824. 统计和小于目标的下标对数目](https://leetcode.cn/problems/count-pairs-whose-sum-is-less-than-target)
+
+[English Version](/solution/2800-2899/2824.Count%20Pairs%20Whose%20Sum%20is%20Less%20than%20Target/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+给你一个下标从 <strong>0</strong>&nbsp;开始长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>target</code>&nbsp;，请你返回满足&nbsp;<code>0 &lt;= i &lt; j &lt; n</code> 且 <code>nums[i] + nums[j] &lt; target</code>&nbsp;的下标对&nbsp;<code>(i, j)</code>&nbsp;的数目。
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [-1,1,2,3,1], target = 2
+<b>输出：</b>3
+<b>解释：</b>总共有 3 个下标对满足题目描述：
+- (0, 1) ，0 &lt; 1 且 nums[0] + nums[1] = 0 &lt; target
+- (0, 2) ，0 &lt; 2 且 nums[0] + nums[2] = 1 &lt; target 
+- (0, 4) ，0 &lt; 4 且 nums[0] + nums[4] = 0 &lt; target
+注意 (0, 3) 不计入答案因为 nums[0] + nums[3] 不是严格小于 target 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [-6,2,5,-2,-7,-1,3], target = -2
+<b>输出：</b>10
+<b>解释：</b>总共有 10 个下标对满足题目描述：
+- (0, 1) ，0 &lt; 1 且 nums[0] + nums[1] = -4 &lt; target
+- (0, 3) ，0 &lt; 3 且 nums[0] + nums[3] = -8 &lt; target
+- (0, 4) ，0 &lt; 4 且 nums[0] + nums[4] = -13 &lt; target
+- (0, 5) ，0 &lt; 5 且 nums[0] + nums[5] = -7 &lt; target
+- (0, 6) ，0 &lt; 6 且 nums[0] + nums[6] = -3 &lt; target
+- (1, 4) ，1 &lt; 4 且 nums[1] + nums[4] = -5 &lt; target
+- (3, 4) ，3 &lt; 4 且 nums[3] + nums[4] = -9 &lt; target
+- (3, 5) ，3 &lt; 5 且 nums[3] + nums[5] = -3 &lt; target
+- (4, 5) ，4 &lt; 5 且 nums[4] + nums[5] = -8 &lt; target
+- (4, 6) ，4 &lt; 6 且 nums[4] + nums[6] = -4 &lt; target
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length == n &lt;= 50</code></li>
+	<li><code>-50 &lt;= nums[i], target &lt;= 50</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：排序 + 二分查找**
+
+我们先对数组 $nums$ 进行排序，然后枚举 $j$，在 $[0, j)$ 的范围内使用二分查找第一个大于等于 $target - nums[j]$ 的下标 $i$，那么 $[0, i)$ 的范围内的所有下标 $k$ 都满足条件，因此答案增加 $i$。
+
+遍历结束后，我们返回答案。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 $nums$ 的长度。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def countPairs(self, nums: List[int], target: int) -> int:
+        nums.sort()
+        ans = 0
+        for j, x in enumerate(nums):
+            i = bisect_left(nums, target - x, hi=j)
+            ans += i
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int countPairs(List<Integer> nums, int target) {
+        Collections.sort(nums);
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int x = nums.get(j);
+            int i = search(nums, target - x, j);
+            ans += i;
+        }
+        return ans;
+    }
+
+    private int search(List<Integer> nums, int x, int r) {
+        int l = 0;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (nums.get(mid) >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int countPairs(vector<int>& nums, int target) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int i = lower_bound(nums.begin(), nums.begin() + j, target - nums[j]) - nums.begin();
+            ans += i;
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func countPairs(nums []int, target int) (ans int) {
+	sort.Ints(nums)
+	for j, x := range nums {
+		i := sort.SearchInts(nums[:j], target-x)
+		ans += i
+	}
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function countPairs(nums: number[], target: number): number {
+    nums.sort((a, b) => a - b);
+    let ans = 0;
+    const search = (x: number, r: number): number => {
+        let l = 0;
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (nums[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    for (let j = 0; j < nums.length; ++j) {
+        const i = search(target - nums[j], j);
+        ans += i;
+    }
+    return ans;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/README_EN.md

@@ -0,0 +1,157 @@
+# [2824. Count Pairs Whose Sum is Less than Target](https://leetcode.com/problems/count-pairs-whose-sum-is-less-than-target)
+
+[中文文档](/solution/2800-2899/2824.Count%20Pairs%20Whose%20Sum%20is%20Less%20than%20Target/README.md)
+
+## Description
+
+Given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>target</code>, return <em>the number of pairs</em> <code>(i, j)</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; n</code> <em>and</em> <code>nums[i] + nums[j] &lt; target</code>.
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-1,1,2,3,1], target = 2
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> There are 3 pairs of indices that satisfy the conditions in the statement:
+- (0, 1) since 0 &lt; 1 and nums[0] + nums[1] = 0 &lt; target
+- (0, 2) since 0 &lt; 2 and nums[0] + nums[2] = 1 &lt; target 
+- (0, 4) since 0 &lt; 4 and nums[0] + nums[4] = 0 &lt; target
+Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-6,2,5,-2,-7,-1,3], target = -2
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> There are 10 pairs of indices that satisfy the conditions in the statement:
+- (0, 1) since 0 &lt; 1 and nums[0] + nums[1] = -4 &lt; target
+- (0, 3) since 0 &lt; 3 and nums[0] + nums[3] = -8 &lt; target
+- (0, 4) since 0 &lt; 4 and nums[0] + nums[4] = -13 &lt; target
+- (0, 5) since 0 &lt; 5 and nums[0] + nums[5] = -7 &lt; target
+- (0, 6) since 0 &lt; 6 and nums[0] + nums[6] = -3 &lt; target
+- (1, 4) since 1 &lt; 4 and nums[1] + nums[4] = -5 &lt; target
+- (3, 4) since 3 &lt; 4 and nums[3] + nums[4] = -9 &lt; target
+- (3, 5) since 3 &lt; 5 and nums[3] + nums[5] = -3 &lt; target
+- (4, 5) since 4 &lt; 5 and nums[4] + nums[5] = -8 &lt; target
+- (4, 6) since 4 &lt; 6 and nums[4] + nums[6] = -4 &lt; target
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length == n &lt;= 50</code></li>
+	<li><code>-50 &lt;= nums[i], target &lt;= 50</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def countPairs(self, nums: List[int], target: int) -> int:
+        nums.sort()
+        ans = 0
+        for j, x in enumerate(nums):
+            i = bisect_left(nums, target - x, hi=j)
+            ans += i
+        return ans
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int countPairs(List<Integer> nums, int target) {
+        Collections.sort(nums);
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int x = nums.get(j);
+            int i = search(nums, target - x, j);
+            ans += i;
+        }
+        return ans;
+    }
+
+    private int search(List<Integer> nums, int x, int r) {
+        int l = 0;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (nums.get(mid) >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int countPairs(vector<int>& nums, int target) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int i = lower_bound(nums.begin(), nums.begin() + j, target - nums[j]) - nums.begin();
+            ans += i;
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func countPairs(nums []int, target int) (ans int) {
+	sort.Ints(nums)
+	for j, x := range nums {
+		i := sort.SearchInts(nums[:j], target-x)
+		ans += i
+	}
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function countPairs(nums: number[], target: number): number {
+    nums.sort((a, b) => a - b);
+    let ans = 0;
+    const search = (x: number, r: number): number => {
+        let l = 0;
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (nums[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    for (let j = 0; j < nums.length; ++j) {
+        const i = search(target - nums[j], j);
+        ans += i;
+    }
+    return ans;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/Solution.cpp

@@ -0,0 +1,12 @@
+class Solution {
+public:
+    int countPairs(vector<int>& nums, int target) {
+        sort(nums.begin(), nums.end());
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int i = lower_bound(nums.begin(), nums.begin() + j, target - nums[j]) - nums.begin();
+            ans += i;
+        }
+        return ans;
+    }
+};

solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/Solution.go

@@ -0,0 +1,8 @@
+func countPairs(nums []int, target int) (ans int) {
+	sort.Ints(nums)
+	for j, x := range nums {
+		i := sort.SearchInts(nums[:j], target-x)
+		ans += i
+	}
+	return
+}

solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    public int countPairs(List<Integer> nums, int target) {
+        Collections.sort(nums);
+        int ans = 0;
+        for (int j = 0; j < nums.size(); ++j) {
+            int x = nums.get(j);
+            int i = search(nums, target - x, j);
+            ans += i;
+        }
+        return ans;
+    }
+
+    private int search(List<Integer> nums, int x, int r) {
+        int l = 0;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (nums.get(mid) >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}