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committedOct 1, 2021
feat: add solutions to lcof2 problem: No.073.Koko Eating Bananas
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‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/README.md

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<!-- 这里可写通用的实现逻辑 -->
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二分查找。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minEatingSpeed(self, piles: List[int], h: int) -> int:
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left, right = 1, max(piles)
72+
while left < right:
73+
mid = (left + right) >> 1
74+
s = sum((pile + mid - 1) // mid for pile in piles)
75+
if s <= h:
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right = mid
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else:
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left = mid + 1
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return left
6880
```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
87+
class Solution {
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public int minEatingSpeed(int[] piles, int h) {
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int mx = 0;
90+
for (int pile : piles) {
91+
mx = Math.max(mx, pile);
92+
}
93+
int left = 1, right = mx;
94+
while (left < right) {
95+
int mid = (left + right) >>> 1;
96+
int s = 0;
97+
for (int pile : piles) {
98+
s += (pile + mid - 1) / mid;
99+
}
100+
if (s <= h) {
101+
right = mid;
102+
} else {
103+
left = mid + 1;
104+
}
105+
}
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return left;
107+
}
108+
}
109+
```
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### **C++**
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```cpp
114+
class Solution {
115+
public:
116+
int minEatingSpeed(vector<int>& piles, int h) {
117+
int left = 1, right = *max_element(piles.begin(), piles.end());
118+
while (left < right)
119+
{
120+
int mid = left + right >> 1;
121+
int s = 0;
122+
for (int pile : piles) s += (pile + mid - 1) / mid;
123+
if (s <= h) right = mid;
124+
else left = mid + 1;
125+
}
126+
return left;
127+
}
128+
};
129+
```
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### **Go**
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```go
134+
func minEatingSpeed(piles []int, h int) int {
135+
mx := 0
136+
for _, pile := range piles {
137+
mx = max(mx, pile)
138+
}
139+
left, right := 1, mx
140+
for left < right {
141+
mid := (left + right) >> 1
142+
s := 0
143+
for _, pile := range piles {
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s += (pile + mid - 1) / mid
145+
}
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if s <= h {
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right = mid
148+
} else {
149+
left = mid + 1
150+
}
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}
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return left
153+
}
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func max(a, b int) int {
156+
if a > b {
157+
return a
158+
}
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return b
160+
}
161+
```
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163+
### **C#**
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```cs
166+
public class Solution {
167+
public int MinEatingSpeed(int[] piles, int h) {
168+
int left = 1, right = piles.Max();
169+
while (left < right)
170+
{
171+
int mid = (left + right) >> 1;
172+
int s = 0;
173+
foreach (int pile in piles)
174+
{
175+
s += (pile + mid - 1) / mid;
176+
}
177+
if (s <= h)
178+
{
179+
right = mid;
180+
}
181+
else
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{
183+
left = mid + 1;
184+
}
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}
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return left;
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}
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}
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```
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### **...**

‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/Solution.cpp

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class Solution {
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public:
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int minEatingSpeed(vector<int>& piles, int h) {
4+
int left = 1, right = *max_element(piles.begin(), piles.end());
5+
while (left < right)
6+
{
7+
int mid = left + right >> 1;
8+
int s = 0;
9+
for (int pile : piles) s += (pile + mid - 1) / mid;
10+
if (s <= h) right = mid;
11+
else left = mid + 1;
12+
}
13+
return left;
14+
}
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};

‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/Solution.cs

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public class Solution {
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public int MinEatingSpeed(int[] piles, int h) {
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int left = 1, right = piles.Max();
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while (left < right)
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{
6+
int mid = (left + right) >> 1;
7+
int s = 0;
8+
foreach (int pile in piles)
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{
10+
s += (pile + mid - 1) / mid;
11+
}
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if (s <= h)
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{
14+
right = mid;
15+
}
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else
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{
18+
left = mid + 1;
19+
}
20+
}
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return left;
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}
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}

‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/Solution.go

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func minEatingSpeed(piles []int, h int) int {
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mx := 0
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for _, pile := range piles {
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mx = max(mx, pile)
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}
6+
left, right := 1, mx
7+
for left < right {
8+
mid := (left + right) >> 1
9+
s := 0
10+
for _, pile := range piles {
11+
s += (pile + mid - 1) / mid
12+
}
13+
if s <= h {
14+
right = mid
15+
} else {
16+
left = mid + 1
17+
}
18+
}
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return left
20+
}
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func max(a, b int) int {
23+
if a > b {
24+
return a
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}
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return b
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}

‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/Solution.java

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class Solution {
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public int minEatingSpeed(int[] piles, int h) {
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int mx = 0;
4+
for (int pile : piles) {
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mx = Math.max(mx, pile);
6+
}
7+
int left = 1, right = mx;
8+
while (left < right) {
9+
int mid = (left + right) >>> 1;
10+
int s = 0;
11+
for (int pile : piles) {
12+
s += (pile + mid - 1) / mid;
13+
}
14+
if (s <= h) {
15+
right = mid;
16+
} else {
17+
left = mid + 1;
18+
}
19+
}
20+
return left;
21+
}
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}

‎lcof2/剑指 Offer II 073. 狒狒吃香蕉/Solution.py

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class Solution:
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def minEatingSpeed(self, piles: List[int], h: int) -> int:
3+
left, right = 1, max(piles)
4+
while left < right:
5+
mid = (left + right) >> 1
6+
s = sum([(pile + mid - 1) // mid for pile in piles])
7+
if s <= h:
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right = mid
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else:
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left = mid + 1
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return left

‎solution/0800-0899/0875.Koko Eating Bananas/README.md

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left, right = 1, max(piles)
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while left < right:
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mid = (left + right) >> 1
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s = sum([(pile + mid - 1) // mid for pile in piles])
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s = sum((pile + mid - 1) // mid for pile in piles)
7070
if s <= h:
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right = mid
7272
else:
@@ -109,22 +109,14 @@ class Solution {
109109
class Solution {
110110
public:
111111
int minEatingSpeed(vector<int>& piles, int h) {
112-
int mx = 0;
113-
for (auto pile : piles) {
114-
mx = max(mx, pile);
115-
}
116-
int left = 1, right = mx;
117-
while (left < right) {
112+
int left = 1, right = *max_element(piles.begin(), piles.end());
113+
while (left < right)
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{
118115
int mid = left + right >> 1;
119116
int s = 0;
120-
for (auto pile : piles) {
121-
s += (pile + mid - 1) / mid;
122-
}
123-
if (s <= h) {
124-
right = mid;
125-
} else {
126-
left = mid + 1;
127-
}
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for (int pile : piles) s += (pile + mid - 1) / mid;
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if (s <= h) right = mid;
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else left = mid + 1;
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}
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return left;
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}

‎solution/0800-0899/0875.Koko Eating Bananas/README_EN.md

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class Solution {
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public:
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int minEatingSpeed(vector<int>& piles, int h) {
102-
int mx = 0;
103-
for (auto pile : piles) {
104-
mx = max(mx, pile);
105-
}
106-
int left = 1, right = mx;
107-
while (left < right) {
102+
int left = 1, right = *max_element(piles.begin(), piles.end());
103+
while (left < right)
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{
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int mid = left + right >> 1;
109106
int s = 0;
110-
for (auto pile : piles) {
111-
s += (pile + mid - 1) / mid;
112-
}
113-
if (s <= h) {
114-
right = mid;
115-
} else {
116-
left = mid + 1;
117-
}
107+
for (int pile : piles) s += (pile + mid - 1) / mid;
108+
if (s <= h) right = mid;
109+
else left = mid + 1;
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}
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return left;
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}

‎solution/0800-0899/0875.Koko Eating Bananas/Solution.cpp

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class Solution {
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public:
33
int minEatingSpeed(vector<int>& piles, int h) {
4-
int mx = 0;
5-
for (auto pile : piles) {
6-
mx = max(mx, pile);
7-
}
8-
int left = 1, right = mx;
9-
while (left < right) {
4+
int left = 1, right = *max_element(piles.begin(), piles.end());
5+
while (left < right)
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{
107
int mid = left + right >> 1;
118
int s = 0;
12-
for (auto pile : piles) {
13-
s += (pile + mid - 1) / mid;
14-
}
15-
if (s <= h) {
16-
right = mid;
17-
} else {
18-
left = mid + 1;
19-
}
9+
for (int pile : piles) s += (pile + mid - 1) / mid;
10+
if (s <= h) right = mid;
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else left = mid + 1;
2012
}
2113
return left;
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}

‎solution/2000-2099/2020.Number of Accounts That Did Not Stream/README.md

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# [2020. Number of Accounts That Did Not Stream](https://leetcode-cn.com/problems/number-of-accounts-that-did-not-stream)
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[English Version](/solution/2000-2099/2020.Number%20of%20Accounts%20That%20Did%20Not%20Stream/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Subscriptions</code></p>
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<pre>
12+
+-------------+------+
13+
| Column Name | Type |
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+-------------+------+
15+
| account_id | int |
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| start_date | date |
17+
| end_date | date |
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+-------------+------+
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account_id is the primary key column for this table.
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Each row of this table indicates the start and end dates of an account&#39;s subscription.
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Note that always start_date &lt; end_date.
22+
</pre>
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<p>&nbsp;</p>
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<p>Table: <code>Streams</code></p>
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<pre>
29+
+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| session_id | int |
33+
| account_id | int |
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| stream_date | date |
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+-------------+------+
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session_id is the primary key column for this table.
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account_id is a foreign key from the Subscriptions table.
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Each row of this table contains information about the account and the date associated with a stream session.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to report the number of accounts that bought a subscription in <code>2021</code> but did not have any stream session.</p>
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<p>The query result format is in the following example.</p>
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47+
<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
49+
50+
<pre>
51+
<strong>Input:</strong>
52+
Subscription table:
53+
+------------+------------+------------+
54+
| account_id | start_date | end_date |
55+
+------------+------------+------------+
56+
| 9 | 2020-02-18 | 2021-10-30 |
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| 3 | 2021-09-21 | 2021-11-13 |
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| 11 | 2020-02-28 | 2020-08-18 |
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| 13 | 2021-04-20 | 2021-09-22 |
60+
| 4 | 2020-10-26 | 2021-05-08 |
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| 5 | 2020-09-11 | 2021-01-17 |
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+------------+------------+------------+
63+
Streams table:
64+
+------------+------------+-------------+
65+
| session_id | account_id | stream_date |
66+
+------------+------------+-------------+
67+
| 14 | 9 | 2020-05-16 |
68+
| 16 | 3 | 2021-10-27 |
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| 18 | 11 | 2020-04-29 |
70+
| 17 | 13 | 2021-08-08 |
71+
| 19 | 4 | 2020-12-31 |
72+
| 13 | 5 | 2021-01-05 |
73+
+------------+------------+-------------+
74+
<strong>Output:</strong>
75+
+----------------+
76+
| accounts_count |
77+
+----------------+
78+
| 2 |
79+
+----------------+
80+
<strong>Explanation:</strong> Users 4 and 9 did not stream in 2021.
81+
User 11 did not subscribe in 2021.
82+
</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
89+
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
93+
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```sql
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```
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<!-- tabs:end -->

‎solution/2000-2099/2020.Number of Accounts That Did Not Stream/README_EN.md

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# [2020. Number of Accounts That Did Not Stream](https://leetcode.com/problems/number-of-accounts-that-did-not-stream)
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[中文文档](/solution/2000-2099/2020.Number%20of%20Accounts%20That%20Did%20Not%20Stream/README.md)
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## Description
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7+
<p>Table: <code>Subscriptions</code></p>
8+
9+
<pre>
10+
+-------------+------+
11+
| Column Name | Type |
12+
+-------------+------+
13+
| account_id | int |
14+
| start_date | date |
15+
| end_date | date |
16+
+-------------+------+
17+
account_id is the primary key column for this table.
18+
Each row of this table indicates the start and end dates of an account&#39;s subscription.
19+
Note that always start_date &lt; end_date.
20+
</pre>
21+
22+
<p>&nbsp;</p>
23+
24+
<p>Table: <code>Streams</code></p>
25+
26+
<pre>
27+
+-------------+------+
28+
| Column Name | Type |
29+
+-------------+------+
30+
| session_id | int |
31+
| account_id | int |
32+
| stream_date | date |
33+
+-------------+------+
34+
session_id is the primary key column for this table.
35+
account_id is a foreign key from the Subscriptions table.
36+
Each row of this table contains information about the account and the date associated with a stream session.
37+
</pre>
38+
39+
<p>&nbsp;</p>
40+
41+
<p>Write an SQL query to report the number of accounts that bought a subscription in <code>2021</code> but did not have any stream session.</p>
42+
43+
<p>The query result format is in the following example.</p>
44+
45+
<p>&nbsp;</p>
46+
<p><strong>Example 1:</strong></p>
47+
48+
<pre>
49+
<strong>Input:</strong>
50+
Subscription table:
51+
+------------+------------+------------+
52+
| account_id | start_date | end_date |
53+
+------------+------------+------------+
54+
| 9 | 2020-02-18 | 2021-10-30 |
55+
| 3 | 2021-09-21 | 2021-11-13 |
56+
| 11 | 2020-02-28 | 2020-08-18 |
57+
| 13 | 2021-04-20 | 2021-09-22 |
58+
| 4 | 2020-10-26 | 2021-05-08 |
59+
| 5 | 2020-09-11 | 2021-01-17 |
60+
+------------+------------+------------+
61+
Streams table:
62+
+------------+------------+-------------+
63+
| session_id | account_id | stream_date |
64+
+------------+------------+-------------+
65+
| 14 | 9 | 2020-05-16 |
66+
| 16 | 3 | 2021-10-27 |
67+
| 18 | 11 | 2020-04-29 |
68+
| 17 | 13 | 2021-08-08 |
69+
| 19 | 4 | 2020-12-31 |
70+
| 13 | 5 | 2021-01-05 |
71+
+------------+------------+-------------+
72+
<strong>Output:</strong>
73+
+----------------+
74+
| accounts_count |
75+
+----------------+
76+
| 2 |
77+
+----------------+
78+
<strong>Explanation:</strong> Users 4 and 9 did not stream in 2021.
79+
User 11 did not subscribe in 2021.
80+
</pre>
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## Solutions
83+
84+
<!-- tabs:start -->
85+
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### **SQL**
87+
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```sql
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```
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<!-- tabs:end -->

‎solution/2000-2099/2021.Brightest Position on Street/README.md

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# [2021. Brightest Position on Street](https://leetcode-cn.com/problems/brightest-position-on-street)
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[English Version](/solution/2000-2099/2021.Brightest%20Position%20on%20Street/README_EN.md)
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## 题目描述
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7+
<!-- 这里写题目描述 -->
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<p>A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array <code>lights</code>. Each <code>lights[i] = [position<sub>i</sub>, range<sub>i</sub>]</code> indicates that there is a street lamp at position <code>position<sub>i</sub></code> that lights up the area from <code>[position<sub>i</sub> - range<sub>i</sub>, position<sub>i</sub> + range<sub>i</sub>]</code> (<strong>inclusive</strong>).</p>
10+
11+
<p>The <strong>brightness</strong> of a position <code>p</code> is defined as the number of street lamp that light up the position <code>p</code>.</p>
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13+
<p>Given <code>lights</code>, return <em>the <strong>brightest</strong> position on the</em><em> street. If there are multiple brightest positions, return the <strong>smallest</strong> one.</em></p>
14+
15+
<p>&nbsp;</p>
16+
<p><strong>Example 1:</strong></p>
17+
<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2021.Brightest%20Position%20on%20Street/images/image-20210928155140-1.png" style="width: 700px; height: 165px;" />
18+
<pre>
19+
<strong>Input:</strong> lights = [[-3,2],[1,2],[3,3]]
20+
<strong>Output:</strong> -1
21+
<strong>Explanation:</strong>
22+
The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
23+
The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
24+
The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].
25+
26+
Position -1 has a brightness of 2, illuminated by the first and second street light.
27+
Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
28+
Out of all these positions, -1 is the smallest, so return it.
29+
</pre>
30+
31+
<p><strong>Example 2:</strong></p>
32+
33+
<pre>
34+
<strong>Input:</strong> lights = [[1,0],[0,1]]
35+
<strong>Output:</strong> 1
36+
<strong>Explanation:</strong>
37+
The first street lamp lights up the area from [1 - 0, 1 + 0] = [1, 1].
38+
The second street lamp lights up the area from [0 - 1, 0 + 1] = [-1, 1].
39+
40+
Position 1 has a brightness of 2, illuminated by the first and second street light.
41+
Return 1 because it is the brightest position on the street.
42+
</pre>
43+
44+
<p><strong>Example 3:</strong></p>
45+
46+
<pre>
47+
<strong>Input:</strong> lights = [[1,2]]
48+
<strong>Output:</strong> -1
49+
<strong>Explanation:</strong>
50+
The first street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
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Positions -1, 0, 1, 2, and 3 have a brightness of 1, illuminated by the first street light.
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Out of all these positions, -1 is the smallest, so return it.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= lights.length &lt;= 10<sup>5</sup></code></li>
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<li><code>lights[i].length == 2</code></li>
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<li><code>-10<sup>8</sup> &lt;= position<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
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<li><code>0 &lt;= range<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2000-2099/2021.Brightest Position on Street/README_EN.md

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# [2021. Brightest Position on Street](https://leetcode.com/problems/brightest-position-on-street)
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[中文文档](/solution/2000-2099/2021.Brightest%20Position%20on%20Street/README.md)
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## Description
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<p>A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array <code>lights</code>. Each <code>lights[i] = [position<sub>i</sub>, range<sub>i</sub>]</code> indicates that there is a street lamp at position <code>position<sub>i</sub></code> that lights up the area from <code>[position<sub>i</sub> - range<sub>i</sub>, position<sub>i</sub> + range<sub>i</sub>]</code> (<strong>inclusive</strong>).</p>
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<p>The <strong>brightness</strong> of a position <code>p</code> is defined as the number of street lamp that light up the position <code>p</code>.</p>
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<p>Given <code>lights</code>, return <em>the <strong>brightest</strong> position on the</em><em> street. If there are multiple brightest positions, return the <strong>smallest</strong> one.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2021.Brightest%20Position%20on%20Street/images/image-20210928155140-1.png" style="width: 700px; height: 165px;" />
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<pre>
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<strong>Input:</strong> lights = [[-3,2],[1,2],[3,3]]
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<strong>Output:</strong> -1
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<strong>Explanation:</strong>
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The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
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The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
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The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].
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Position -1 has a brightness of 2, illuminated by the first and second street light.
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Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
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Out of all these positions, -1 is the smallest, so return it.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> lights = [[1,0],[0,1]]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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The first street lamp lights up the area from [1 - 0, 1 + 0] = [1, 1].
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The second street lamp lights up the area from [0 - 1, 0 + 1] = [-1, 1].
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Position 1 has a brightness of 2, illuminated by the first and second street light.
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Return 1 because it is the brightest position on the street.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> lights = [[1,2]]
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<strong>Output:</strong> -1
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<strong>Explanation:</strong>
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The first street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
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Positions -1, 0, 1, 2, and 3 have a brightness of 1, illuminated by the first street light.
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Out of all these positions, -1 is the smallest, so return it.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= lights.length &lt;= 10<sup>5</sup></code></li>
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<li><code>lights[i].length == 2</code></li>
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<li><code>-10<sup>8</sup> &lt;= position<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
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<li><code>0 &lt;= range<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2000-2099/2021.Brightest Position on Street/images/image-20210928155140-1.png

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‎solution/summary.md

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- [2017.Grid Game](/solution/2000-2099/2017.Grid%20Game/README.md)
20612061
- [2018.Check if Word Can Be Placed In Crossword](/solution/2000-2099/2018.Check%20if%20Word%20Can%20Be%20Placed%20In%20Crossword/README.md)
20622062
- [2019.The Score of Students Solving Math Expression](/solution/2000-2099/2019.The%20Score%20of%20Students%20Solving%20Math%20Expression/README.md)
2063+
- [2020.Number of Accounts That Did Not Stream](/solution/2000-2099/2020.Number%20of%20Accounts%20That%20Did%20Not%20Stream/README.md)
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- [2021.Brightest Position on Street](/solution/2000-2099/2021.Brightest%20Position%20on%20Street/README.md)

‎solution/summary_en.md

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- [2017.Grid Game](/solution/2000-2099/2017.Grid%20Game/README_EN.md)
20612061
- [2018.Check if Word Can Be Placed In Crossword](/solution/2000-2099/2018.Check%20if%20Word%20Can%20Be%20Placed%20In%20Crossword/README_EN.md)
20622062
- [2019.The Score of Students Solving Math Expression](/solution/2000-2099/2019.The%20Score%20of%20Students%20Solving%20Math%20Expression/README_EN.md)
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- [2020.Number of Accounts That Did Not Stream](/solution/2000-2099/2020.Number%20of%20Accounts%20That%20Did%20Not%20Stream/README_EN.md)
2064+
- [2021.Brightest Position on Street](/solution/2000-2099/2021.Brightest%20Position%20on%20Street/README_EN.md)

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