feat: update solutions to lc problems: No.2164,2732

.github/workflows/deploy.yml

@@ -69,7 +69,7 @@ jobs:
           mkdocs build -f mkdocs-en.yml
 
       - name: Deploy
-        uses: peaceiris/actions-gh-pages@v3
+        uses: peaceiris/actions-gh-pages@v4
         with:
           github_token: ${{ secrets.GITHUB_TOKEN }}
           publish_dir: ./site

solution/2100-2199/2163.Minimum Difference in Sums After Removal of Elements/README_EN.md

@@ -78,7 +78,15 @@ It can be shown that it is not possible to obtain a difference smaller than 1.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Priority Queue (Max and Min Heap) + Prefix and Suffix Sum + Enumeration of Split Points
+
+The problem is essentially equivalent to finding a split point in $nums$, dividing the array into two parts. In the first part, select the smallest $n$ elements, and in the second part, select the largest $n$ elements, so that the difference between the sums of the two parts is minimized.
+
+We can use a max heap to maintain the smallest $n$ elements in the prefix, and a min heap to maintain the largest $n$ elements in the suffix. We define $pre[i]$ as the sum of the smallest $n$ elements among the first $i$ elements of the array $nums$, and $suf[i]$ as the sum of the largest $n$ elements from the $i$-th element to the last element of the array. During the process of maintaining the max and min heaps, update the values of $pre[i]$ and $suf[i]$.
+
+Finally, we enumerate the split points in the range of $i \in [n, 2n]$, calculate the value of $pre[i] - suf[i + 1]$, and take the minimum value.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
 
 <!-- tabs:start -->
 

solution/2100-2199/2164.Sort Even and Odd Indices Independently/README.md

@@ -78,7 +78,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：排序
+
+我们可以将奇数下标和偶数下标分别取出来，然后对奇数下标的数组进行非递增排序，对偶数下标的数组进行非递减排序，最后再将两个数组合并即可。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\text{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -133,16 +137,21 @@ public:
         vector<int> a;
         vector<int> b;
         for (int i = 0; i < n; ++i) {
-            if (i % 2 == 0)
+            if (i % 2 == 0) {
                 a.push_back(nums[i]);
-            else
+            } else {
                 b.push_back(nums[i]);
+            }
         }
         sort(a.begin(), a.end());
-        sort(b.begin(), b.end(), greater<int>());
+        sort(b.rbegin(), b.rend());
         vector<int> ans(n);
-        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) ans[i] = a[j];
-        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) ans[i] = b[j];
+        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) {
+            ans[i] = a[j];
+        }
+        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) {
+            ans[i] = b[j];
+        }
         return ans;
     }
 };
@@ -177,6 +186,33 @@ func sortEvenOdd(nums []int) []int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function sortEvenOdd(nums: number[]): number[] {
+    const n = nums.length;
+    const a: number[] = [];
+    const b: number[] = [];
+    for (let i = 0; i < n; ++i) {
+        if (i % 2 === 0) {
+            a.push(nums[i]);
+        } else {
+            b.push(nums[i]);
+        }
+    }
+    a.sort((x, y) => x - y);
+    b.sort((x, y) => y - x);
+    const ans: number[] = [];
+    for (let i = 0, j = 0; j < a.length; i += 2, ++j) {
+        ans[i] = a[j];
+    }
+    for (let i = 1, j = 0; j < b.length; i += 2, ++j) {
+        ans[i] = b[j];
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2100-2199/2164.Sort Even and Odd Indices Independently/README_EN.md

@@ -76,7 +76,11 @@ The resultant array formed is [2,1], which is the same as the initial array.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting
+
+We can extract the elements at odd and even indices separately, then sort the array of odd indices in non-increasing order and the array of even indices in non-decreasing order. Finally, merge the two arrays back together.
+
+The time complexity is $O(n \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\text{nums}$.
 
 <!-- tabs:start -->
 
@@ -131,16 +135,21 @@ public:
         vector<int> a;
         vector<int> b;
         for (int i = 0; i < n; ++i) {
-            if (i % 2 == 0)
+            if (i % 2 == 0) {
                 a.push_back(nums[i]);
-            else
+            } else {
                 b.push_back(nums[i]);
+            }
         }
         sort(a.begin(), a.end());
-        sort(b.begin(), b.end(), greater<int>());
+        sort(b.rbegin(), b.rend());
         vector<int> ans(n);
-        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) ans[i] = a[j];
-        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) ans[i] = b[j];
+        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) {
+            ans[i] = a[j];
+        }
+        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) {
+            ans[i] = b[j];
+        }
         return ans;
     }
 };
@@ -175,6 +184,33 @@ func sortEvenOdd(nums []int) []int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function sortEvenOdd(nums: number[]): number[] {
+    const n = nums.length;
+    const a: number[] = [];
+    const b: number[] = [];
+    for (let i = 0; i < n; ++i) {
+        if (i % 2 === 0) {
+            a.push(nums[i]);
+        } else {
+            b.push(nums[i]);
+        }
+    }
+    a.sort((x, y) => x - y);
+    b.sort((x, y) => y - x);
+    const ans: number[] = [];
+    for (let i = 0, j = 0; j < a.length; i += 2, ++j) {
+        ans[i] = a[j];
+    }
+    for (let i = 1, j = 0; j < b.length; i += 2, ++j) {
+        ans[i] = b[j];
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2100-2199/2164.Sort Even and Odd Indices Independently/Solution.cpp

@@ -5,16 +5,21 @@ class Solution {
         vector<int> a;
         vector<int> b;
         for (int i = 0; i < n; ++i) {
-            if (i % 2 == 0)
+            if (i % 2 == 0) {
                 a.push_back(nums[i]);
-            else
+            } else {
                 b.push_back(nums[i]);
+            }
         }
         sort(a.begin(), a.end());
-        sort(b.begin(), b.end(), greater<int>());
+        sort(b.rbegin(), b.rend());
         vector<int> ans(n);
-        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) ans[i] = a[j];
-        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) ans[i] = b[j];
+        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) {
+            ans[i] = a[j];
+        }
+        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) {
+            ans[i] = b[j];
+        }
         return ans;
     }
 };

solution/2100-2199/2164.Sort Even and Odd Indices Independently/Solution.ts

@@ -0,0 +1,22 @@
+function sortEvenOdd(nums: number[]): number[] {
+    const n = nums.length;
+    const a: number[] = [];
+    const b: number[] = [];
+    for (let i = 0; i < n; ++i) {
+        if (i % 2 === 0) {
+            a.push(nums[i]);
+        } else {
+            b.push(nums[i]);
+        }
+    }
+    a.sort((x, y) => x - y);
+    b.sort((x, y) => y - x);
+    const ans: number[] = [];
+    for (let i = 0, j = 0; j < a.length; i += 2, ++j) {
+        ans[i] = a[j];
+    }
+    for (let i = 1, j = 0; j < b.length; i += 2, ++j) {
+        ans[i] = b[j];
+    }
+    return ans;
+}

solution/2700-2799/2732.Find a Good Subset of the Matrix/README.md

@@ -84,7 +84,19 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：分情况讨论
+
+我们可以从小到大考虑答案选择的行数 $k$。
+
+-   如果 $k = 1$，每一列的和最大为 $0$，那么必须满足有一行的所有元素都是 $0$，否则无法满足条件。
+-   如果 $k = 2$，每一列的和最大为 $1$，那么必须存在有两行，且这两行的元素按位或之后的结果是 $0$，否则无法满足条件。
+-   如果 $k = 3$，每一列的和最大也是 $1$。如果 $k = 2$ 不满足条件，那么 $k = 3$ 也一定不满足条件，所以我们不需要考虑所有 $k \gt 2$ 且 $k$ 为奇数的情况。
+-   如果 $k = 4$，每一列的和最大为 $2$，此时一定是 $k = 2$ 不满足条件，也就是说，任意选取两行，都存在至少一个列的和为 $2$。我们在 $4$ 行中任意选取 $2$ 行，一共有 $C_4^2 = 6$ 种选法，那么存在至少 $6$ 个 $2$ 的列。由于列数 $n \le 5$，所以一定存在至少一列的和大于 $2$，所以 $k = 4$ 也不满足条件。
+-   对于 $k \gt 4$ 且 $k$ 为偶数的情况，我们可以得出同样的结论，即 $k$ 一定不满足条件。
+
+综上所述，我们只需要考虑 $k = 1$ 和 $k = 2$ 的情况即可。即判断是否有一行全为 $0$，或者是否存在两行按位或之后的结果为 $0$。
+
+时间复杂度 $O(m \times n + 4^n)$，空间复杂度 $O(2^n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
 
 <!-- tabs:start -->
 

solution/2700-2799/2732.Find a Good Subset of the Matrix/README_EN.md

@@ -82,7 +82,19 @@ The length of the chosen subset is 1.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Case Analysis
+
+We can consider the number of rows $k$ chosen for the answer from smallest to largest.
+
+-   If $k = 1$, the maximum sum of each column is $0$. Therefore, there must be a row where all elements are $0$, otherwise, the condition cannot be met.
+-   If $k = 2$, the maximum sum of each column is $1$. There must exist two rows, and the bitwise OR result of these two rows' elements is $0$, otherwise, the condition cannot be met.
+-   If $k = 3$, the maximum sum of each column is also $1$. If the condition for $k = 2$ is not met, then the condition for $k = 3$ will definitely not be met either. Therefore, we do not need to consider any case where $k > 2$ and $k$ is odd.
+-   If $k = 4$, the maximum sum of each column is $2$. This situation definitely occurs when the condition for $k = 2$ is not met, meaning that for any two selected rows, there exists at least one column with a sum of $2$. When choosing any 2 rows out of 4, there are a total of $C_4^2 = 6$ ways to choose, so there are at least $6$ columns with a sum of $2$. Since the number of columns $n \le 5$, there must be at least one column with a sum greater than $2$, so the condition for $k = 4$ is also not met.
+-   For $k > 4$ and $k$ being even, we can draw the same conclusion, that $k$ definitely does not meet the condition.
+
+In summary, we only need to consider the cases of $k = 1$ and $k = 2$. That is, to check whether there is a row entirely composed of $0$s, or whether there exist two rows whose bitwise OR result is $0$.
+
+The time complexity is $O(m \times n + 4^n)$, and the space complexity is $O(2^n)$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively.
 
 <!-- tabs:start -->