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Commit 9480261

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feat: add solutions to lc problems: No.2824~2867 (doocs#2007)
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‎solution/2800-2899/2824.Count Pairs Whose Sum is Less than Target/README_EN.md‎

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## Solutions
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**Solution 1: Sorting + Binary Search**
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First, we sort the array $nums$. Then, for each $j,ドル we use binary search in the range $[0, j)$ to find the first index $i$ that is greater than or equal to $target - nums[j]$. All indices $k$ in the range $[0, i)$ meet the condition, so the answer increases by $i$.
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After the traversal, we return the answer.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.
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‎solution/2800-2899/2825.Make String a Subsequence Using Cyclic Increments/README_EN.md‎

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## Solutions
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**Solution 1: Two Pointers**
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This problem actually requires us to determine whether a string $s$ is a subsequence of another string $t$. However, the characters do not have to match exactly. If two characters are the same, or one character is the next character of the other, they can match.
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The time complexity is $O(m + n),ドル where $m$ and $n$ are the lengths of the strings $str1$ and $str2$ respectively. The space complexity is $O(1)$.
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‎solution/2800-2899/2826.Sorting Three Groups/README.md‎

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**方法一:动态规划**
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我们定义 $f[i][j]$ 表示将前 $i$ 个数变成美丽数组,并且第 $i$ 个数变成 $j+1$ 的最少操作次数。那么答案就是 $min(f[n][0], f[n][1], f[n][2])$。
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我们定义 $f[i][j]$ 表示将前 $i$ 个数变成美丽数组,并且第 $i$ 个数变成 $j+1$ 的最少操作次数。那么答案就是 $\min(f[n][0], f[n][1], f[n][2])$。
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我们可以枚举第 $i$ 个数变成 $j+1$ 的所有情况,然后取最小值。这里我们可以用滚动数组优化空间复杂度。
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‎solution/2800-2899/2826.Sorting Three Groups/README_EN.md‎

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## Solutions
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**Solution 1: Dynamic Programming**
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We define $f[i][j]$ as the minimum number of operations to turn the first $i$ numbers into a beautiful array, and the $i$th number is changed to $j+1$. The answer is $\min(f[n][0], f[n][1], f[n][2])$.
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We can enumerate all cases where the $i$th number is changed to $j+1,ドル and then take the minimum value. Here, we can use a rolling array to optimize the space complexity.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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‎solution/2800-2899/2828.Check if a String Is an Acronym of Words/README_EN.md‎

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## Solutions
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**Solution 1: Simulation**
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We can traverse each string in the array $words,ドル concatenate their first letters to form a new string $t,ドル and then check if $t$ is equal to $s$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array $words$.
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‎solution/2800-2899/2829.Determine the Minimum Sum of a k-avoiding Array/README_EN.md‎

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## Solutions
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**Solution 1: Greedy + Simulation**
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We start from the positive integer $i=1,ドル and judge whether $i$ can be added to the array in turn. If it can be added, we add $i$ to the array, accumulate it to the answer, and then mark $k-i$ as visited, indicating that $k-i$ cannot be added to the array. The loop continues until the length of the array is $n$.
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The time complexity is $O(n^2),ドル and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.
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‎solution/2800-2899/2830.Maximize the Profit as the Salesman/README_EN.md‎

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## Solutions
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**Solution 1: Sorting + Binary Search + Dynamic Programming**
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We sort all the purchase offers by $end$ in ascending order, and then use dynamic programming to solve the problem.
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Define $f[i]$ to represent the maximum amount of gold we can get from the first $i$ purchase offers. The answer is $f[n]$.
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For $f[i],ドル we can choose not to sell the $i$th purchase offer, in which case $f[i] = f[i - 1]$; or we can choose to sell the $i$th purchase offer, in which case $f[i] = f[j] + gold_i,ドル where $j$ is the largest index that satisfies $end_j \leq start_i$.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(n)$. Here, $n$ is the number of purchase offers.
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‎solution/2800-2899/2831.Find the Longest Equal Subarray/README_EN.md‎

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## Solutions
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**Solution 1: Hash Table + Two Pointers**
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We use two pointers to maintain a monotonically variable length window, and a hash table to maintain the number of occurrences of each element in the window.
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The number of all elements in the window minus the number of the most frequently occurring element in the window is the number of elements that need to be deleted from the window.
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Each time, we add the element pointed to by the right pointer to the window, then update the hash table, and also update the number of the most frequently occurring element in the window. When the number of elements that need to be deleted from the window exceeds $k,ドル we move the left pointer once, and then update the hash table.
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After the traversal, return the number of the most frequently occurring element.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array.
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‎solution/2800-2899/2832.Maximal Range That Each Element Is Maximum in It/README_EN.md‎

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## Solutions
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**Solution 1: Monotonic Stack**
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This problem is a template for monotonic stack. We only need to use the monotonic stack to find the position of the first element larger than $nums[i]$ on the left and right, denoted as $left[i]$ and $right[i]$. Then, the interval length with $nums[i]$ as the maximum value is $right[i] - left[i] - 1$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array.
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‎solution/2800-2899/2833.Furthest Point From Origin/README_EN.md‎

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## Solutions
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**Solution 1: Greedy**
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When encountering the character '_', we can choose to move left or right. The problem requires us to find the farthest point from the origin. Therefore, we can first traverse once, greedily move all '_' to the left, and find the farthest point from the origin at this time. Then traverse again, greedily move all '\_' to the right, and find the farthest point from the origin at this time. Finally, take the maximum of the two traversals.
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Further, we only need to calculate the difference between the number of 'L' and 'R' in the string, and then add the number of '\_'.
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The time complexity is $O(n),ドル where $n$ is the length of the string. The space complexity is $O(1)$.
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