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Commit 04070a3

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feat: add solutions to lc problem: No.2237 (doocs#3938)
No.2237.Count Positions on Street With Required Brightness
1 parent 16116c4 commit 04070a3

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8 files changed

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‎solution/2200-2299/2237.Count Positions on Street With Required Brightness/README.md‎

Lines changed: 41 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -77,7 +77,15 @@ tags:
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### 方法一:差分数组
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80-
时间复杂度 $O(n)$。
80+
对一段连续的区间 $[i, j]$ 同时加上一个值 $v,ドル可以通过差分数组来实现。
81+
82+
我们定义一个长度为 $n + 1$ 的数组 $\textit{d},ドル接下来对于每个路灯,我们计算出它的左边界 $i = \max(0, p - r)$ 和右边界 $j = \min(n - 1, p + r),ドル然后将 $\textit{d}[i]$ 加上 1ドル,ドル将 $\textit{d}[j + 1]$ 减去 1ドル$。
83+
84+
然后,我们对 $\textit{d}$ 进行前缀和运算,对于每个位置 $i,ドル如果 $\textit{d}[i]$ 的前缀和大于等于 $\textit{requirement}[i],ドル则说明该位置满足要求,将答案加一。
85+
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最后返回答案即可。
87+
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为路灯数量。
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<!-- tabs:start -->
8391

@@ -88,7 +96,7 @@ class Solution:
8896
def meetRequirement(
8997
self, n: int, lights: List[List[int]], requirement: List[int]
9098
) -> int:
91-
d = [0] * 100010
99+
d = [0] * (n +1)
92100
for p, r in lights:
93101
i, j = max(0, p - r), min(n - 1, p + r)
94102
d[i] += 1
@@ -101,7 +109,7 @@ class Solution:
101109
```java
102110
class Solution {
103111
public int meetRequirement(int n, int[][] lights, int[] requirement) {
104-
int[] d = new int[100010];
112+
int[] d = new int[n +1];
105113
for (int[] e : lights) {
106114
int i = Math.max(0, e[0] - e[1]);
107115
int j = Math.min(n - 1, e[0] + e[1]);
@@ -127,16 +135,18 @@ class Solution {
127135
class Solution {
128136
public:
129137
int meetRequirement(int n, vector<vector<int>>& lights, vector<int>& requirement) {
130-
vector<int> d(100010);
131-
for (auto& e : lights) {
138+
vector<int> d(n + 1);
139+
for (const auto& e : lights) {
132140
int i = max(0, e[0] - e[1]), j = min(n - 1, e[0] + e[1]);
133141
++d[i];
134142
--d[j + 1];
135143
}
136144
int s = 0, ans = 0;
137145
for (int i = 0; i < n; ++i) {
138146
s += d[i];
139-
if (s >= requirement[i]) ++ans;
147+
if (s >= requirement[i]) {
148+
++ans;
149+
}
140150
}
141151
return ans;
142152
}
@@ -146,21 +156,42 @@ public:
146156
#### Go
147157
148158
```go
149-
func meetRequirement(n int, lights [][]int, requirement []int) int {
150-
d := make([]int, 100010)
159+
func meetRequirement(n int, lights [][]int, requirement []int) (ans int) {
160+
d := make([]int, n+1)
151161
for _, e := range lights {
152162
i, j := max(0, e[0]-e[1]), min(n-1, e[0]+e[1])
153163
d[i]++
154164
d[j+1]--
155165
}
156-
var s, ans int
166+
s := 0
157167
for i, r := range requirement {
158168
s += d[i]
159169
if s >= r {
160170
ans++
161171
}
162172
}
163-
return ans
173+
return
174+
}
175+
```
176+
177+
#### TypeScript
178+
179+
```ts
180+
function meetRequirement(n: number, lights: number[][], requirement: number[]): number {
181+
const d: number[] = Array(n + 1).fill(0);
182+
for (const [p, r] of lights) {
183+
const [i, j] = [Math.max(0, p - r), Math.min(n - 1, p + r)];
184+
++d[i];
185+
--d[j + 1];
186+
}
187+
let [ans, s] = [0, 0];
188+
for (let i = 0; i < n; ++i) {
189+
s += d[i];
190+
if (s >= requirement[i]) {
191+
++ans;
192+
}
193+
}
194+
return ans;
164195
}
165196
```
166197

‎solution/2200-2299/2237.Count Positions on Street With Required Brightness/README_EN.md‎

Lines changed: 43 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -73,7 +73,17 @@ Positions 0, 1, 2, and 4 meet the requirement so we return 4.
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<!-- solution:start -->
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76-
### Solution 1
76+
### Solution 1: Difference Array
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78+
To add a value $v$ to a continuous interval $[i, j]$ simultaneously, we can use a difference array.
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80+
We define an array $\textit{d}$ of length $n + 1$. For each streetlight, we calculate its left boundary $i = \max(0, p - r)$ and right boundary $j = \min(n - 1, p + r),ドル then add 1ドル$ to $\textit{d}[i]$ and subtract 1ドル$ from $\textit{d}[j + 1]$.
81+
82+
Next, we perform a prefix sum operation on $\textit{d}$. For each position $i,ドル if the prefix sum of $\textit{d}[i]$ is greater than or equal to $\textit{requirement}[i],ドル it means that the position meets the requirement, and we increment the answer by one.
83+
84+
Finally, return the answer.
85+
86+
The time complexity is $O(n),ドル and the space complexity is $O(n),ドル where $n$ is the number of streetlights.
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7888
<!-- tabs:start -->
7989

@@ -84,7 +94,7 @@ class Solution:
8494
def meetRequirement(
8595
self, n: int, lights: List[List[int]], requirement: List[int]
8696
) -> int:
87-
d = [0] * 100010
97+
d = [0] * (n +1)
8898
for p, r in lights:
8999
i, j = max(0, p - r), min(n - 1, p + r)
90100
d[i] += 1
@@ -97,7 +107,7 @@ class Solution:
97107
```java
98108
class Solution {
99109
public int meetRequirement(int n, int[][] lights, int[] requirement) {
100-
int[] d = new int[100010];
110+
int[] d = new int[n +1];
101111
for (int[] e : lights) {
102112
int i = Math.max(0, e[0] - e[1]);
103113
int j = Math.min(n - 1, e[0] + e[1]);
@@ -123,16 +133,18 @@ class Solution {
123133
class Solution {
124134
public:
125135
int meetRequirement(int n, vector<vector<int>>& lights, vector<int>& requirement) {
126-
vector<int> d(100010);
127-
for (auto& e : lights) {
136+
vector<int> d(n + 1);
137+
for (const auto& e : lights) {
128138
int i = max(0, e[0] - e[1]), j = min(n - 1, e[0] + e[1]);
129139
++d[i];
130140
--d[j + 1];
131141
}
132142
int s = 0, ans = 0;
133143
for (int i = 0; i < n; ++i) {
134144
s += d[i];
135-
if (s >= requirement[i]) ++ans;
145+
if (s >= requirement[i]) {
146+
++ans;
147+
}
136148
}
137149
return ans;
138150
}
@@ -142,21 +154,42 @@ public:
142154
#### Go
143155
144156
```go
145-
func meetRequirement(n int, lights [][]int, requirement []int) int {
146-
d := make([]int, 100010)
157+
func meetRequirement(n int, lights [][]int, requirement []int) (ans int) {
158+
d := make([]int, n+1)
147159
for _, e := range lights {
148160
i, j := max(0, e[0]-e[1]), min(n-1, e[0]+e[1])
149161
d[i]++
150162
d[j+1]--
151163
}
152-
var s, ans int
164+
s := 0
153165
for i, r := range requirement {
154166
s += d[i]
155167
if s >= r {
156168
ans++
157169
}
158170
}
159-
return ans
171+
return
172+
}
173+
```
174+
175+
#### TypeScript
176+
177+
```ts
178+
function meetRequirement(n: number, lights: number[][], requirement: number[]): number {
179+
const d: number[] = Array(n + 1).fill(0);
180+
for (const [p, r] of lights) {
181+
const [i, j] = [Math.max(0, p - r), Math.min(n - 1, p + r)];
182+
++d[i];
183+
--d[j + 1];
184+
}
185+
let [ans, s] = [0, 0];
186+
for (let i = 0; i < n; ++i) {
187+
s += d[i];
188+
if (s >= requirement[i]) {
189+
++ans;
190+
}
191+
}
192+
return ans;
160193
}
161194
```
162195

Lines changed: 6 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,17 +1,19 @@
11
class Solution {
22
public:
33
int meetRequirement(int n, vector<vector<int>>& lights, vector<int>& requirement) {
4-
vector<int> d(100010);
5-
for (auto& e : lights) {
4+
vector<int> d(n + 1);
5+
for (constauto& e : lights) {
66
int i = max(0, e[0] - e[1]), j = min(n - 1, e[0] + e[1]);
77
++d[i];
88
--d[j + 1];
99
}
1010
int s = 0, ans = 0;
1111
for (int i = 0; i < n; ++i) {
1212
s += d[i];
13-
if (s >= requirement[i]) ++ans;
13+
if (s >= requirement[i]) {
14+
++ans;
15+
}
1416
}
1517
return ans;
1618
}
17-
};
19+
};
Lines changed: 5 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -1,16 +1,16 @@
1-
func meetRequirement(n int, lights [][]int, requirement []int) int {
2-
d := make([]int, 100010)
1+
func meetRequirement(n int, lights [][]int, requirement []int) (ansint) {
2+
d := make([]int, n+1)
33
for _, e := range lights {
44
i, j := max(0, e[0]-e[1]), min(n-1, e[0]+e[1])
55
d[i]++
66
d[j+1]--
77
}
8-
vars, ansint
8+
s:=0
99
for i, r := range requirement {
1010
s += d[i]
1111
if s >= r {
1212
ans++
1313
}
1414
}
15-
returnans
16-
}
15+
return
16+
}

‎solution/2200-2299/2237.Count Positions on Street With Required Brightness/Solution.java‎

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -1,6 +1,6 @@
11
class Solution {
22
public int meetRequirement(int n, int[][] lights, int[] requirement) {
3-
int[] d = new int[100010];
3+
int[] d = new int[n + 1];
44
for (int[] e : lights) {
55
int i = Math.max(0, e[0] - e[1]);
66
int j = Math.min(n - 1, e[0] + e[1]);
@@ -17,4 +17,4 @@ public int meetRequirement(int n, int[][] lights, int[] requirement) {
1717
}
1818
return ans;
1919
}
20-
}
20+
}

‎solution/2200-2299/2237.Count Positions on Street With Required Brightness/Solution.py‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -2,7 +2,7 @@ class Solution:
22
def meetRequirement(
33
self, n: int, lights: List[List[int]], requirement: List[int]
44
) -> int:
5-
d = [0] * 100010
5+
d = [0] * (n+1)
66
for p, r in lights:
77
i, j = max(0, p - r), min(n - 1, p + r)
88
d[i] += 1
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
function meetRequirement(n: number, lights: number[][], requirement: number[]): number {
2+
const d: number[] = Array(n + 1).fill(0);
3+
for (const [p, r] of lights) {
4+
const [i, j] = [Math.max(0, p - r), Math.min(n - 1, p + r)];
5+
++d[i];
6+
--d[j + 1];
7+
}
8+
let [ans, s] = [0, 0];
9+
for (let i = 0; i < n; ++i) {
10+
s += d[i];
11+
if (s >= requirement[i]) {
12+
++ans;
13+
}
14+
}
15+
return ans;
16+
}

‎solution/2200-2299/2239.Find Closest Number to Zero/README_EN.md‎

Lines changed: 7 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -56,7 +56,13 @@ Thus, the closest number to 0 in the array is 1.
5656

5757
<!-- solution:start -->
5858

59-
### Solution 1
59+
### Solution 1: One Pass
60+
61+
We define a variable $d$ to record the current minimum distance, initially $d=\infty$. Then we traverse the array, for each element $x,ドル we calculate $y=|x|$. If $y \lt d$ or $y=d$ and $x \gt ans,ドル we update the answer $ans=x$ and $d=y$.
62+
63+
After the traversal, return the answer.
64+
65+
Time complexity is $O(n),ドル where $n$ is the length of the array. Space complexity is $O(1)$.
6066

6167
<!-- tabs:start -->
6268

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