Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit 5a549bf

Browse files
feat: update solutions to lc problems (doocs#3624)
1 parent 865f078 commit 5a549bf

File tree

6 files changed

+36
-70
lines changed

6 files changed

+36
-70
lines changed

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

Lines changed: 8 additions & 31 deletions
Original file line numberDiff line numberDiff line change
@@ -25,20 +25,20 @@ tags:
2525
<p><strong>示例 1:</strong></p>
2626

2727
<pre><strong>输入:</strong>n = 234
28-
<strong>输出:</strong>15
28+
<strong>输出:</strong>15
2929
<strong>解释:</strong>
30-
各位数之积 = 2 * 3 * 4 = 24
31-
各位数之和 = 2 + 3 + 4 = 9
30+
各位数之积 = 2 * 3 * 4 = 24
31+
各位数之和 = 2 + 3 + 4 = 9
3232
结果 = 24 - 9 = 15
3333
</pre>
3434

3535
<p><strong>示例 2:</strong></p>
3636

3737
<pre><strong>输入:</strong>n = 4421
3838
<strong>输出:</strong>21
39-
<strong>解释:
40-
</strong>各位数之积 = 4 * 4 * 2 * 1 = 32
41-
各位数之和 = 4 +たす 4 +たす 2 +たす 1 =わ 11
39+
<strong>解释:
40+
</strong>各位数之积 = 4 * 4 * 2 * 1 = 32
41+
各位数之和 = 4 +たす 4 +たす 2 +たす 1 =わ 11
4242
结果 = 32 - 11 = 21
4343
</pre>
4444

@@ -73,12 +73,8 @@ tags:
7373
```python
7474
class Solution:
7575
def subtractProductAndSum(self, n: int) -> int:
76-
x, y = 1, 0
77-
while n:
78-
n, v = divmod(n, 10)
79-
x *= v
80-
y += v
81-
return x - y
76+
nums = list(map(int, str(n)))
77+
return prod(nums) - sum(nums)
8278
```
8379

8480
#### Java
@@ -196,23 +192,4 @@ int subtractProductAndSum(int n) {
196192
197193
<!-- solution:end -->
198194
199-
<!-- solution:start -->
200-
201-
### 方法二
202-
203-
<!-- tabs:start -->
204-
205-
#### Python3
206-
207-
```python
208-
class Solution:
209-
def subtractProductAndSum(self, n: int) -> int:
210-
nums = list(map(int, str(n)))
211-
return prod(nums) - sum(nums)
212-
```
213-
214-
<!-- tabs:end -->
215-
216-
<!-- solution:end -->
217-
218195
<!-- problem:end -->

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

Lines changed: 9 additions & 32 deletions
Original file line numberDiff line numberDiff line change
@@ -25,10 +25,10 @@ Given an integer number <code>n</code>, return the difference between the produc
2525

2626
<pre>
2727
<strong>Input:</strong> n = 234
28-
<strong>Output:</strong> 15
29-
<b>Explanation:</b>
30-
Product of digits = 2 * 3 * 4 = 24
31-
Sum of digits = 2 + 3 + 4 = 9
28+
<strong>Output:</strong> 15
29+
<b>Explanation:</b>
30+
Product of digits = 2 * 3 * 4 = 24
31+
Sum of digits = 2 + 3 + 4 = 9
3232
Result = 24 - 9 = 15
3333
</pre>
3434

@@ -37,9 +37,9 @@ Result = 24 - 9 = 15
3737
<pre>
3838
<strong>Input:</strong> n = 4421
3939
<strong>Output:</strong> 21
40-
<b>Explanation:
41-
</b>Product of digits = 4 * 4 * 2 * 1 = 32
42-
Sum of digits = 4 +たす 4 +たす 2 +たす 1 =わ 11
40+
<b>Explanation:
41+
</b>Product of digits = 4 * 4 * 2 * 1 = 32
42+
Sum of digits = 4 +たす 4 +たす 2 +たす 1 =わ 11
4343
Result = 32 - 11 = 21
4444
</pre>
4545

@@ -73,12 +73,8 @@ The time complexity is $O(\log n),ドル where $n$ is the given integer. The space co
7373
```python
7474
class Solution:
7575
def subtractProductAndSum(self, n: int) -> int:
76-
x, y = 1, 0
77-
while n:
78-
n, v = divmod(n, 10)
79-
x *= v
80-
y += v
81-
return x - y
76+
nums = list(map(int, str(n)))
77+
return prod(nums) - sum(nums)
8278
```
8379

8480
#### Java
@@ -196,23 +192,4 @@ int subtractProductAndSum(int n) {
196192
197193
<!-- solution:end -->
198194
199-
<!-- solution:start -->
200-
201-
### Solution 2
202-
203-
<!-- tabs:start -->
204-
205-
#### Python3
206-
207-
```python
208-
class Solution:
209-
def subtractProductAndSum(self, n: int) -> int:
210-
nums = list(map(int, str(n)))
211-
return prod(nums) - sum(nums)
212-
```
213-
214-
<!-- tabs:end -->
215-
216-
<!-- solution:end -->
217-
218195
<!-- problem:end -->

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

Lines changed: 0 additions & 4 deletions
This file was deleted.

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

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -66,7 +66,7 @@ tags:
6666

6767
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,ドル其中 $s[i]$ 表示数组前 $i$ 个元素之和。
6868

69-
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,ドル此时数组中大于 `value` 的元素个数为 $n - i,ドル因此数组中小于等于 `value` 的元素个数为 $i,ドル此时数组中小于等于 `value` 的元素之和为 $s[i],ドル数组中大于 `value` 的元素之和为 $(n - i) \times value,ドル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`。
69+
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,ドル此时数组中大于 `value` 的元素个数为 $n - i,ドル因此数组中小于等于 `value` 的元素个数为 $i,ドル此时数组中小于等于 `value` 的元素之和为 $s[i],ドル数组中大于 `value` 的元素之和为 $(n - i) \times \textit{value},ドル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`。
7070

7171
枚举完所有 `value` 后,即可得到最终答案 `ans`。
7272

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

Lines changed: 9 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -63,7 +63,15 @@ tags:
6363

6464
<!-- solution:start -->
6565

66-
### Solution 1
66+
### Solution 1: Sorting + Prefix Sum + Binary Search + Enumeration
67+
68+
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,ドル where $s[i]$ represents the sum of the first $i$ elements of the array.
69+
70+
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,ドル 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],ドル 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`.
71+
72+
After enumerating all `value` values, we can get the final answer `ans`.
73+
74+
Time complexity $O(n \times \log n),ドル space complexity $O(n)$. Where $n$ is the length of the array `arr`.
6775

6876
<!-- tabs:start -->
6977

‎solution/1300-1399/1301.Number of Paths with Max Score/README_EN.md‎

Lines changed: 9 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -66,7 +66,15 @@ tags:
6666

6767
<!-- solution:start -->
6868

69-
### Solution 1
69+
### Solution 1: Dynamic Programming
70+
71+
We define $f[i][j]$ to represent the maximum score from the starting point $(n - 1, n - 1)$ to $(i, j),ドル 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,ドル and $g[i][j]$ are all 0ドル$.
72+
73+
For the current position $(i, j),ドル it can be transferred from three positions: $(i + 1, j),ドル $(i, j + 1),ドル 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],ドル then we update $f[i][j] = f[x][y]$ and $g[i][j] = g[x][y]$. If $f[x][y] = f[i][j],ドル 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 $f[i][j] = f[i][j] + board[i][j]$.
74+
75+
Finally, if $f[0][0] \lt 0,ドル 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ドル^9 + 7$.
76+
77+
Time complexity $O(n^2),ドル space complexity $O(n^2)$. Where $n$ is the side length of the array.
7078

7179
<!-- tabs:start -->
7280

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /