feat: update solutions to lc problems (doocs#3624)

Author: yanglbme

solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md

@@ -25,20 +25,20 @@ tags:
 <p><strong>示例 1：</strong></p>
 
 <pre><strong>输入：</strong>n = 234
-<strong>输出：</strong>15 
+<strong>输出：</strong>15
 <strong>解释：</strong>
-各位数之积 = 2 * 3 * 4 = 24 
-各位数之和 = 2 + 3 + 4 = 9 
+各位数之积 = 2 * 3 * 4 = 24
+各位数之和 = 2 + 3 + 4 = 9
 结果 = 24 - 9 = 15
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
 <pre><strong>输入：</strong>n = 4421
 <strong>输出：</strong>21
-<strong>解释： 
-</strong>各位数之积 = 4 * 4 * 2 * 1 = 32 
-各位数之和 = 4 + 4 + 2 + 1 = 11 
+<strong>解释：
+</strong>各位数之积 = 4 * 4 * 2 * 1 = 32
+各位数之和 = 4 + 4 + 2 + 1 = 11
 结果 = 32 - 11 = 21
 </pre>
 
@@ -73,12 +73,8 @@ tags:
 ```python
 class Solution:
     def subtractProductAndSum(self, n: int) -> int:
-        x, y = 1, 0
-        while n:
-            n, v = divmod(n, 10)
-            x *= v
-            y += v
-        return x - y
+        nums = list(map(int, str(n)))
+        return prod(nums) - sum(nums)
 ```
 
 #### Java
@@ -196,23 +192,4 @@ int subtractProductAndSum(int n) {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def subtractProductAndSum(self, n: int) -> int:
-        nums = list(map(int, str(n)))
-        return prod(nums) - sum(nums)
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md

@@ -25,10 +25,10 @@ Given an integer number <code>n</code>, return the difference between the produc
 
 <pre>
 <strong>Input:</strong> n = 234
-<strong>Output:</strong> 15 
-<b>Explanation:</b> 
-Product of digits = 2 * 3 * 4 = 24 
-Sum of digits = 2 + 3 + 4 = 9 
+<strong>Output:</strong> 15
+<b>Explanation:</b>
+Product of digits = 2 * 3 * 4 = 24
+Sum of digits = 2 + 3 + 4 = 9
 Result = 24 - 9 = 15
 </pre>
 
@@ -37,9 +37,9 @@ Result = 24 - 9 = 15
 <pre>
 <strong>Input:</strong> n = 4421
 <strong>Output:</strong> 21
-<b>Explanation: 
-</b>Product of digits = 4 * 4 * 2 * 1 = 32 
-Sum of digits = 4 + 4 + 2 + 1 = 11 
+<b>Explanation:
+</b>Product of digits = 4 * 4 * 2 * 1 = 32
+Sum of digits = 4 + 4 + 2 + 1 = 11
 Result = 32 - 11 = 21
 </pre>
 
@@ -73,12 +73,8 @@ The time complexity is $O(\log n)$, where $n$ is the given integer. The space co
 ```python
 class Solution:
     def subtractProductAndSum(self, n: int) -> int:
-        x, y = 1, 0
-        while n:
-            n, v = divmod(n, 10)
-            x *= v
-            y += v
-        return x - y
+        nums = list(map(int, str(n)))
+        return prod(nums) - sum(nums)
 ```
 
 #### Java
@@ -196,23 +192,4 @@ int subtractProductAndSum(int n) {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def subtractProductAndSum(self, n: int) -> int:
-        nums = list(map(int, str(n)))
-        return prod(nums) - sum(nums)
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/Solution2.py

@@ -66,7 +66,7 @@ tags:
 
 我们注意到，题目中要把所有大于 `value` 的值变成 `value`，并且求和，因此我们可以考虑先对数组 `arr` 进行排序，然后求出前缀和数组 $s$，其中 $s[i]$ 表示数组前 $i$ 个元素之和。
 
-接下来，我们可以从小到大枚举所有 `value` 值，对于每个 `value`，我们可以通过二分查找找到数组中第一个大于 `value` 的元素的下标 $i$，此时数组中大于 `value` 的元素个数为 $n - i$，因此数组中小于等于 `value` 的元素个数为 $i$，此时数组中小于等于 `value` 的元素之和为 $s[i]$，数组中大于 `value` 的元素之和为 $(n - i) \times value$，因此数组中所有元素之和为 $s[i] + (n - i) \times value$。如果 $s[i] + (n - i) \times value$ 与 `target` 的差的绝对值小于当前的最小差值 `diff`，则更新 `diff` 和 `ans`。
+接下来，我们可以从小到大枚举所有 `value` 值，对于每个 `value`，我们可以通过二分查找找到数组中第一个大于 `value` 的元素的下标 $i$，此时数组中大于 `value` 的元素个数为 $n - i$，因此数组中小于等于 `value` 的元素个数为 $i$，此时数组中小于等于 `value` 的元素之和为 $s[i]$，数组中大于 `value` 的元素之和为 $(n - i) \times \textit{value}$，因此数组中所有元素之和为 $s[i] + (n - i) \times \textit{value}$。如果 $s[i] + (n - i) \times \textit{value}$ 与 `target` 的差的绝对值小于当前的最小差值 `diff`，则更新 `diff` 和 `ans`。
 
 枚举完所有 `value` 后，即可得到最终答案 `ans`。
 

solution/1300-1399/1300.Sum of Mutated Array Closest to Target/README.md

@@ -63,7 +63,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting + Prefix Sum + Binary Search + Enumeration
+
+We notice that the problem requires changing all values greater than `value` to `value` and then summing them up. Therefore, we can consider sorting the array `arr` first, and then calculating the prefix sum array $s$, where $s[i]$ represents the sum of the first $i$ elements of the array.
+
+Next, we can enumerate all `value` values from smallest to largest. For each `value`, we can use binary search to find the index $i$ of the first element in the array that is greater than `value`. At this point, the number of elements in the array greater than `value` is $n - i$, so the number of elements in the array less than or equal to `value` is $i$. The sum of the elements in the array less than or equal to `value` is $s[i]$, and the sum of the elements in the array greater than `value` is $(n - i) \times value$. Therefore, the sum of all elements in the array is $s[i] + (n - i) \times \textit{value}$. If the absolute difference between $s[i] + (n - i) \times \textit{value}$ and `target` is less than the current minimum difference `diff`, update `diff` and `ans`.
+
+After enumerating all `value` values, we can get the final answer `ans`.
+
+Time complexity $O(n \times \log n)$, space complexity $O(n)$. Where $n$ is the length of the array `arr`.
 
 <!-- tabs:start -->
 

solution/1300-1399/1300.Sum of Mutated Array Closest to Target/README_EN.md

@@ -66,7 +66,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i][j]$ to represent the maximum score from the starting point $(n - 1, n - 1)$ to $(i, j)$, and $g[i][j]$ to represent the number of ways to achieve the maximum score from the starting point $(n - 1, n - 1)$ to $(i, j)$. Initially, $f[n - 1][n - 1] = 0$ and $g[n - 1][n - 1] = 1$. The other positions of $f[i][j]$ are all $-1$, and $g[i][j]$ are all $0$.
+
+For the current position $(i, j)$, it can be transferred from three positions: $(i + 1, j)$, $(i, j + 1)$, and $(i + 1, j + 1)$. Therefore, we can enumerate these three positions to update the values of $f[i][j]$ and $g[i][j]$. If the current position $(i, j)$ has an obstacle, or the current position is the starting point, or other positions are out of bounds, no update is performed. Otherwise, if another position $(x, y)$ satisfies $f[x][y] \gt f[i][j]$, then we update $f[i][j] = f[x][y]$ and $g[i][j] = g[x][y]$. If $f[x][y] = f[i][j]$, then we update $g[i][j] = g[i][j] + g[x][y]$. Finally, if the current position $(i, j)$ is reachable and is a number, we update $f[i][j] = f[i][j] + board[i][j]$.
+
+Finally, if $f[0][0] \lt 0$, it means there is no path to reach the endpoint, return $[0, 0]$. Otherwise, return $[f[0][0], g[0][0]]$. Note that the result needs to be taken modulo $10^9 + 7$.
+
+Time complexity $O(n^2)$, space complexity $O(n^2)$. Where $n$ is the side length of the array.
 
 <!-- tabs:start -->