Skip to content

Commit 3ef2fee

Browse files
yanglbmeyanglbme
authored
feat: add solutions to lc problems: No.3079,3080 (doocs#2454)
* No.3079.Find the Sum of Encrypted Integers * No.3080.Mark Elements on Array by Performing Queries
1 parent 6ba88a9 commit 3ef2fee

File tree

14 files changed

+683
-16
lines changed

14 files changed

+683
-16
lines changed

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/README.md

+81-4
Original file line numberDiff line numberDiff line change
@@ -45,24 +45,101 @@
4545

4646
## 解法
4747

48-
### 方法一
48+
### 方法一：模拟
49+
50+
我们直接模拟加密的过程，定义一个函数 $encrypt(x)$，将一个整数 $x$ 中每一个数位都用 $x$ 中的最大数位替换。函数的实现如下：
51+
52+
我们可以通过不断地对 $x$ 取模和整除 $10$ 来得到 $x$ 的每一位数，找到最大的数位，记为 $mx$。在循环的过程中，我们还可以用一个变量 $p$ 来记录 $mx$ 的基础底数，即 $p = 1, 11, 111, \cdots$。最后返回 $mx \times p$ 即可。
53+
54+
时间复杂度 $O(n \times \log M)$，其中 $n$ 是数组的长度，而 $M$ 是数组中元素的最大值。空间复杂度 $O(1)$。
4955

5056
<!-- tabs:start -->
5157

5258
```python
53-
59+
class Solution:
60+
def sumOfEncryptedInt(self, nums: List[int]) -> int:
61+
def encrypt(x: int) -> int:
62+
mx = p = 0
63+
while x:
64+
x, v = divmod(x, 10)
65+
mx = max(mx, v)
66+
p = p * 10 + 1
67+
return mx * p
68+
69+
return sum(encrypt(x) for x in nums)
5470
```
5571

5672
```java
57-
73+
class Solution {
74+
public int sumOfEncryptedInt(int[] nums) {
75+
int ans = 0;
76+
for (int x : nums) {
77+
ans += encrypt(x);
78+
}
79+
return ans;
80+
}
81+
82+
private int encrypt(int x) {
83+
int mx = 0, p = 0;
84+
for (; x > 0; x /= 10) {
85+
mx = Math.max(mx, x % 10);
86+
p = p * 10 + 1;
87+
}
88+
return mx * p;
89+
}
90+
}
5891
```
5992

6093
```cpp
61-
94+
class Solution {
95+
public:
96+
int sumOfEncryptedInt(vector<int>& nums) {
97+
auto encrypt = [&](int x) {
98+
int mx = 0, p = 0;
99+
for (; x; x /= 10) {
100+
mx = max(mx, x % 10);
101+
p = p * 10 + 1;
102+
}
103+
return mx * p;
104+
};
105+
int ans = 0;
106+
for (int x : nums) {
107+
ans += encrypt(x);
108+
}
109+
return ans;
110+
}
111+
};
62112
```
63113
64114
```go
115+
func sumOfEncryptedInt(nums []int) (ans int) {
116+
encrypt := func(x int) int {
117+
mx, p := 0, 0
118+
for ; x > 0; x /= 10 {
119+
mx = max(mx, x%10)
120+
p = p*10 + 1
121+
}
122+
return mx * p
123+
}
124+
for _, x := range nums {
125+
ans += encrypt(x)
126+
}
127+
return
128+
}
129+
```
65130

131+
```ts
132+
function sumOfEncryptedInt(nums: number[]): number {
133+
const encrypt = (x: number): number => {
134+
let [mx, p] = [0, 0];
135+
for (; x > 0; x = Math.floor(x / 10)) {
136+
mx = Math.max(mx, x % 10);
137+
p = p * 10 + 1;
138+
}
139+
return mx * p;
140+
};
141+
return nums.reduce((acc, x) => acc + encrypt(x), 0);
142+
}
66143
```
67144

68145
<!-- tabs:end -->

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/README_EN.md

+81-4
Original file line numberDiff line numberDiff line change
@@ -41,24 +41,101 @@
4141

4242
## Solutions
4343

44-
### Solution 1
44+
### Solution 1: Simulation
45+
46+
We directly simulate the encryption process by defining a function $encrypt(x)$, which replaces each digit in an integer $x$ with the maximum digit in $x$. The implementation of the function is as follows:
47+
48+
We can obtain each digit of $x$ by continuously taking the modulus and integer division of $x$ by $10$, and find the maximum digit, denoted as $mx$. During the loop, we can also use a variable $p$ to record the base number of $mx$, i.e., $p = 1, 11, 111, \cdots$. Finally, return $mx \times p$.
49+
50+
The time complexity is $O(n \times \log M)$, where $n$ is the length of the array, and $M$ is the maximum value in the array. The space complexity is $O(1)$.
4551

4652
<!-- tabs:start -->
4753

4854
```python
49-
55+
class Solution:
56+
def sumOfEncryptedInt(self, nums: List[int]) -> int:
57+
def encrypt(x: int) -> int:
58+
mx = p = 0
59+
while x:
60+
x, v = divmod(x, 10)
61+
mx = max(mx, v)
62+
p = p * 10 + 1
63+
return mx * p
64+
65+
return sum(encrypt(x) for x in nums)
5066
```
5167

5268
```java
53-
69+
class Solution {
70+
public int sumOfEncryptedInt(int[] nums) {
71+
int ans = 0;
72+
for (int x : nums) {
73+
ans += encrypt(x);
74+
}
75+
return ans;
76+
}
77+
78+
private int encrypt(int x) {
79+
int mx = 0, p = 0;
80+
for (; x > 0; x /= 10) {
81+
mx = Math.max(mx, x % 10);
82+
p = p * 10 + 1;
83+
}
84+
return mx * p;
85+
}
86+
}
5487
```
5588

5689
```cpp
57-
90+
class Solution {
91+
public:
92+
int sumOfEncryptedInt(vector<int>& nums) {
93+
auto encrypt = [&](int x) {
94+
int mx = 0, p = 0;
95+
for (; x; x /= 10) {
96+
mx = max(mx, x % 10);
97+
p = p * 10 + 1;
98+
}
99+
return mx * p;
100+
};
101+
int ans = 0;
102+
for (int x : nums) {
103+
ans += encrypt(x);
104+
}
105+
return ans;
106+
}
107+
};
58108
```
59109
60110
```go
111+
func sumOfEncryptedInt(nums []int) (ans int) {
112+
encrypt := func(x int) int {
113+
mx, p := 0, 0
114+
for ; x > 0; x /= 10 {
115+
mx = max(mx, x%10)
116+
p = p*10 + 1
117+
}
118+
return mx * p
119+
}
120+
for _, x := range nums {
121+
ans += encrypt(x)
122+
}
123+
return
124+
}
125+
```
61126

127+
```ts
128+
function sumOfEncryptedInt(nums: number[]): number {
129+
const encrypt = (x: number): number => {
130+
let [mx, p] = [0, 0];
131+
for (; x > 0; x = Math.floor(x / 10)) {
132+
mx = Math.max(mx, x % 10);
133+
p = p * 10 + 1;
134+
}
135+
return mx * p;
136+
};
137+
return nums.reduce((acc, x) => acc + encrypt(x), 0);
138+
}
62139
```
63140

64141
<!-- tabs:end -->

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/Solution.cpp

+18
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
class Solution {
2+
public:
3+
int sumOfEncryptedInt(vector<int>& nums) {
4+
auto encrypt = [&](int x) {
5+
int mx = 0, p = 0;
6+
for (; x; x /= 10) {
7+
mx = max(mx, x % 10);
8+
p = p * 10 + 1;
9+
}
10+
return mx * p;
11+
};
12+
int ans = 0;
13+
for (int x : nums) {
14+
ans += encrypt(x);
15+
}
16+
return ans;
17+
}
18+
};

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/Solution.go

+14
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
1+
func sumOfEncryptedInt(nums []int) (ans int) {
2+
encrypt := func(x int) int {
3+
mx, p := 0, 0
4+
for ; x > 0; x /= 10 {
5+
mx = max(mx, x%10)
6+
p = p*10 + 1
7+
}
8+
return mx * p
9+
}
10+
for _, x := range nums {
11+
ans += encrypt(x)
12+
}
13+
return
14+
}

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/Solution.java

+18
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
class Solution {
2+
public int sumOfEncryptedInt(int[] nums) {
3+
int ans = 0;
4+
for (int x : nums) {
5+
ans += encrypt(x);
6+
}
7+
return ans;
8+
}
9+
10+
private int encrypt(int x) {
11+
int mx = 0, p = 0;
12+
for (; x > 0; x /= 10) {
13+
mx = Math.max(mx, x % 10);
14+
p = p * 10 + 1;
15+
}
16+
return mx * p;
17+
}
18+
}

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/Solution.py

+11
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
class Solution:
2+
def sumOfEncryptedInt(self, nums: List[int]) -> int:
3+
def encrypt(x: int) -> int:
4+
mx = p = 0
5+
while x:
6+
x, v = divmod(x, 10)
7+
mx = max(mx, v)
8+
p = p * 10 + 1
9+
return mx * p
10+
11+
return sum(encrypt(x) for x in nums)

‎solution/3000-3099/3079.Find the Sum of Encrypted Integers/Solution.ts

+11
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
function sumOfEncryptedInt(nums: number[]): number {
2+
const encrypt = (x: number): number => {
3+
let [mx, p] = [0, 0];
4+
for (; x > 0; x = Math.floor(x / 10)) {
5+
mx = Math.max(mx, x % 10);
6+
p = p * 10 + 1;
7+
}
8+
return mx * p;
9+
};
10+
return nums.reduce((acc, x) => acc + encrypt(x), 0);
11+
}

0 commit comments

Comments
 (0)