feat: add solutions to lc problems: No.1572,1574 (#3818)

* No.1572.Matrix Diagonal Sum
* No.1574.Shortest Subarray to be Removed to Make Array Sorted

Author: yanglbme

solution/1500-1599/1572.Matrix Diagonal Sum/README.md

@@ -73,11 +73,10 @@ tags:
 
 ### 方法一：逐行遍历
 
-我们可以遍历矩阵的每一行 $row[i]$，对于每一行，我们可以计算出两个对角线上的元素，即 $row[i][i]$ 和 $row[i][n - i - 1]$，其中 $n$ 是矩阵的行数。如果 $i = n - i - 1$，则说明当前行的对角线上只有一个元素，否则有两个元素。我们将其加到答案中即可。
-
+我们可以遍历矩阵的每一行 $\textit{row}[i]$，对于每一行，我们可以计算出两个对角线上的元素，即 $\textit{row}[i][i]$ 和 $\textit{row}[i][n - i - 1]$，其中 $n$ 是矩阵的行数。如果 $i = n - i - 1$，则说明当前行的对角线上只有一个元素，否则有两个元素。我们将其加到答案中即可。
 遍历完所有行后，即可得到答案。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是矩阵的行数。
+时间复杂度 $O(n)$，其中 $n$ 是矩阵的行数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -163,12 +162,15 @@ impl Solution {
     pub fn diagonal_sum(mat: Vec<Vec<i32>>) -> i32 {
         let n = mat.len();
         let mut ans = 0;
+
         for i in 0..n {
-            ans += mat[i][i] + mat[n - 1 - i][i];
-        }
-        if (n & 1) == 1 {
-            ans -= mat[n >> 1][n >> 1];
+            ans += mat[i][i];
+            let j = n - i - 1;
+            if j != i {
+                ans += mat[i][j];
+            }
         }
+
         ans
     }
 }
@@ -179,37 +181,12 @@ impl Solution {
 ```c
 int diagonalSum(int** mat, int matSize, int* matColSize) {
     int ans = 0;
-    for (int i = 0; i < matSize; i++) {
-        ans += mat[i][i] + mat[i][matSize - 1 - i];
-    }
-    if (matSize & 1) {
-        ans -= mat[matSize >> 1][matSize >> 1];
-    }
-    return ans;
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function diagonalSum(mat: number[][]): number {
-    const n = mat.length;
-    let ans = 0;
-    for (let i = 0; i < n; i++) {
-        ans += mat[i][i] + mat[i][n - 1 - i];
-    }
-    if (n & 1) {
-        ans -= mat[n >> 1][n >> 1];
+    for (int i = 0; i < matSize; ++i) {
+        ans += mat[i][i];
+        int j = matSize - i - 1;
+        if (j != i) {
+            ans += mat[i][j];
+        }
     }
     return ans;
 }

solution/1500-1599/1572.Matrix Diagonal Sum/README_EN.md

@@ -67,7 +67,13 @@ Notice that element mat[1][1] = 5 is counted only once.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Row-by-Row Traversal
+
+We can traverse each row $\textit{row}[i]$ of the matrix. For each row, we calculate the elements on the two diagonals, i.e., $\textit{row}[i][i]$ and $\textit{row}[i][n - i - 1]$, where $n$ is the number of rows in the matrix. If $i = n - i - 1$, it means there is only one element on the diagonals of the current row; otherwise, there are two elements. We add these elements to the answer.
+
+After traversing all rows, we get the answer.
+
+The time complexity is $O(n)$, where $n$ is the number of rows in the matrix. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -153,12 +159,15 @@ impl Solution {
     pub fn diagonal_sum(mat: Vec<Vec<i32>>) -> i32 {
         let n = mat.len();
         let mut ans = 0;
+
         for i in 0..n {
-            ans += mat[i][i] + mat[n - 1 - i][i];
-        }
-        if (n & 1) == 1 {
-            ans -= mat[n >> 1][n >> 1];
+            ans += mat[i][i];
+            let j = n - i - 1;
+            if j != i {
+                ans += mat[i][j];
+            }
         }
+
         ans
     }
 }
@@ -169,37 +178,12 @@ impl Solution {
 ```c
 int diagonalSum(int** mat, int matSize, int* matColSize) {
     int ans = 0;
-    for (int i = 0; i < matSize; i++) {
-        ans += mat[i][i] + mat[i][matSize - 1 - i];
-    }
-    if (matSize & 1) {
-        ans -= mat[matSize >> 1][matSize >> 1];
-    }
-    return ans;
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function diagonalSum(mat: number[][]): number {
-    const n = mat.length;
-    let ans = 0;
-    for (let i = 0; i < n; i++) {
-        ans += mat[i][i] + mat[i][n - 1 - i];
-    }
-    if (n & 1) {
-        ans -= mat[n >> 1][n >> 1];
+    for (int i = 0; i < matSize; ++i) {
+        ans += mat[i][i];
+        int j = matSize - i - 1;
+        if (j != i) {
+            ans += mat[i][j];
+        }
     }
     return ans;
 }

solution/1500-1599/1572.Matrix Diagonal Sum/Solution.c

@@ -1,10 +1,11 @@
 int diagonalSum(int** mat, int matSize, int* matColSize) {
     int ans = 0;
-    for (int i = 0; i < matSize; i++) {
-        ans += mat[i][i] + mat[i][matSize - 1 - i];
-    }
-    if (matSize & 1) {
-        ans -= mat[matSize >> 1][matSize >> 1];
+    for (int i = 0; i < matSize; ++i) {
+        ans += mat[i][i];
+        int j = matSize - i - 1;
+        if (j != i) {
+            ans += mat[i][j];
+        }
     }
     return ans;
-}
+}

solution/1500-1599/1572.Matrix Diagonal Sum/Solution.rs

@@ -2,12 +2,15 @@ impl Solution {
     pub fn diagonal_sum(mat: Vec<Vec<i32>>) -> i32 {
         let n = mat.len();
         let mut ans = 0;
+
         for i in 0..n {
-            ans += mat[i][i] + mat[n - 1 - i][i];
-        }
-        if (n & 1) == 1 {
-            ans -= mat[n >> 1][n >> 1];
+            ans += mat[i][i];
+            let j = n - i - 1;
+            if j != i {
+                ans += mat[i][j];
+            }
         }
+
         ans
     }
 }

solution/1500-1599/1572.Matrix Diagonal Sum/Solution2.ts

@@ -68,7 +68,21 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Counting
+
+First, we traverse the string $s$ and count the number of characters $1$, denoted as $cnt$. If $cnt$ cannot be divided by $3$, then it is impossible to split the string, so we directly return $0$. If $cnt$ is $0$, it means there are no characters $1$ in the string. We can choose any two positions out of $n-1$ positions to split the string into three substrings, so the number of ways is $C_{n-1}^2$.
+
+If $cnt \gt 0$, we update $cnt$ to $\frac{cnt}{3}$, which is the number of characters $1$ in each substring.
+
+Next, we find the minimum index of the right boundary of the first substring, denoted as $i_1$, and the maximum index of the right boundary of the first substring (exclusive), denoted as $i_2$. Similarly, we find the minimum index of the right boundary of the second substring, denoted as $j_1$, and the maximum index of the right boundary of the second substring (exclusive), denoted as $j_2$. Then the number of ways is $(i_2 - i_1) \times (j_2 - j_1)$.
+
+Note that the answer may be very large, so we need to take the modulo $10^9+7$.
+
+The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the string $s$.
+
+Similar problems:
+
+-   [927. Three Equal Parts](https://github.com/doocs/leetcode/blob/main/solution/0900-0999/0927.Three%20Equal%20Parts/README_EN.md)
 
 <!-- tabs:start -->
 

solution/1500-1599/1573.Number of Ways to Split a String/README_EN.md

@@ -78,15 +78,15 @@ tags:
 
 ### 方法一：双指针 + 二分查找
 
-我们先找出数组的最长非递减前缀和最长非递减后缀，分别记为 $nums[0..i]$ 和 $nums[j..n-1]$。
+我们先找出数组的最长非递减前缀和最长非递减后缀，分别记为 $\textit{nums}[0..i]$ 和 $\textit{nums}[j..n-1]$。
 
 如果 $i \geq j$，说明数组本身就是非递减的，返回 $0$。
 
-否则，我们可以选择删除右侧后缀，也可以选择删除左侧前缀，因此初始时答案为 $min(n - i - 1, j)$。
+否则，我们可以选择删除右侧后缀，也可以选择删除左侧前缀，因此初始时答案为 $\min(n - i - 1, j)$。
 
-接下来，我们枚举左侧前缀的最右端点 $l$，对于每个 $l$，我们可以通过二分查找，在 $nums[j..n-1]$ 中找到第一个大于等于 $nums[l]$ 的位置，记为 $r$，此时我们可以删除 $nums[l+1..r-1]$，并且更新答案 $ans = min(ans, r - l - 1)$。继续枚举 $l$，最终得到答案。
+接下来，我们枚举左侧前缀的最右端点 $l$，对于每个 $l$，我们可以通过二分查找，在 $\textit{nums}[j..n-1]$ 中找到第一个大于等于 $\textit{nums}[l]$ 的位置，记为 $r$，此时我们可以删除 $\textit{nums}[l+1..r-1]$，并且更新答案 $\textit{ans} = \min(\textit{ans}, r - l - 1)$。继续枚举 $l$，最终得到答案。
 
-时间复杂度 $O(n \times \log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
+时间复杂度 $O(n \times \log n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -208,15 +208,15 @@ func findLengthOfShortestSubarray(arr []int) int {
 
 ### 方法二：双指针
 
-与方法一类似，我们先找出数组的最长非递减前缀和最长非递减后缀，分别记为 $nums[0..i]$ 和 $nums[j..n-1]$。
+与方法一类似，我们先找出数组的最长非递减前缀和最长非递减后缀，分别记为 $\textit{nums}[0..i]$ 和 $\textit{nums}[j..n-1]$。
 
 如果 $i \geq j$，说明数组本身就是非递减的，返回 $0$。
 
-否则，我们可以选择删除右侧后缀，也可以选择删除左侧前缀，因此初始时答案为 $min(n - i - 1, j)$。
+否则，我们可以选择删除右侧后缀，也可以选择删除左侧前缀，因此初始时答案为 $\min(n - i - 1, j)$。
 
-接下来，我们枚举左侧前缀的最右端点 $l$，对于每个 $l$，我们直接利用双指针找到第一个大于等于 $nums[l]$ 的位置，记为 $r$，此时我们可以删除 $nums[l+1..r-1]$，并且更新答案 $ans = min(ans, r - l - 1)$。继续枚举 $l$，最终得到答案。
+接下来，我们枚举左侧前缀的最右端点 $l$，对于每个 $l$，我们直接利用双指针找到第一个大于等于 $\textit{nums}[l]$ 的位置，记为 $r$，此时我们可以删除 $\textit{nums}[l+1..r-1]$，并且更新答案 $\textit{ans} = \min(\textit{ans}, r - l - 1)$。继续枚举 $l$，最终得到答案。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1500-1599/1574.Shortest Subarray to be Removed to Make Array Sorted/README.md

@@ -68,7 +68,17 @@ Another correct solution is to remove the subarray [3,10,4].
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers + Binary Search
+
+First, we find the longest non-decreasing prefix and the longest non-decreasing suffix of the array, denoted as $\textit{nums}[0..i]$ and $\textit{nums}[j..n-1]$, respectively.
+
+If $i \geq j$, it means the array is already non-decreasing, so we return $0$.
+
+Otherwise, we can choose to delete the right suffix or the left prefix. Therefore, initially, the answer is $\min(n - i - 1, j)$.
+
+Next, we enumerate the right endpoint $l$ of the left prefix. For each $l$, we can use binary search to find the first position greater than or equal to $\textit{nums}[l]$ in $\textit{nums}[j..n-1]$, denoted as $r$. At this point, we can delete $\textit{nums}[l+1..r-1]$ and update the answer $\textit{ans} = \min(\textit{ans}, r - l - 1)$. Continue enumerating $l$ to get the final answer.
+
+The time complexity is $O(n \times \log n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -188,7 +198,17 @@ func findLengthOfShortestSubarray(arr []int) int {
 
 <!-- solution:start -->
 
-### Solution 2
+### Solution 2: Two Pointers
+
+Similar to Solution 1, we first find the longest non-decreasing prefix and the longest non-decreasing suffix of the array, denoted as $\textit{nums}[0..i]$ and $\textit{nums}[j..n-1]$, respectively.
+
+If $i \geq j$, it means the array is already non-decreasing, so we return $0$.
+
+Otherwise, we can choose to delete the right suffix or the left prefix. Therefore, initially, the answer is $\min(n - i - 1, j)$.
+
+Next, we enumerate the right endpoint $l$ of the left prefix. For each $l$, we directly use two pointers to find the first position greater than or equal to $\textit{nums}[l]$ in $\textit{nums}[j..n-1]$, denoted as $r$. At this point, we can delete $\textit{nums}[l+1..r-1]$ and update the answer $\textit{ans} = \min(\textit{ans}, r - l - 1)$. Continue enumerating $l$ to get the final answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
 
 <!-- tabs:start -->