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Commit 3ef2fee

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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
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14 files changed

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

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## 解法
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### 方法一
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### 方法一:模拟
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我们直接模拟加密的过程,定义一个函数 $encrypt(x),ドル将一个整数 $x$ 中每一个数位都用 $x$ 中的最大数位替换。函数的实现如下:
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我们可以通过不断地对 $x$ 取模和整除 10ドル$ 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。
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时间复杂度 $O(n \times \log M),ドル其中 $n$ 是数组的长度,而 $M$ 是数组中元素的最大值。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def sumOfEncryptedInt(self, nums: List[int]) -> int:
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def encrypt(x: int) -> int:
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mx = p = 0
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while x:
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x, v = divmod(x, 10)
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mx = max(mx, v)
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p = p * 10 + 1
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return mx * p
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return sum(encrypt(x) for x in nums)
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```
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```java
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class Solution {
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public int sumOfEncryptedInt(int[] nums) {
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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private int encrypt(int x) {
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int mx = 0, p = 0;
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for (; x > 0; x /= 10) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int sumOfEncryptedInt(vector<int>& nums) {
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auto encrypt = [&](int x) {
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int mx = 0, p = 0;
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for (; x; x /= 10) {
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mx = max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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};
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```
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```go
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func sumOfEncryptedInt(nums []int) (ans int) {
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encrypt := func(x int) int {
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mx, p := 0, 0
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for ; x > 0; x /= 10 {
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mx = max(mx, x%10)
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p = p*10 + 1
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}
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return mx * p
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}
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for _, x := range nums {
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ans += encrypt(x)
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}
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return
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}
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```
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```ts
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function sumOfEncryptedInt(nums: number[]): number {
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const encrypt = (x: number): number => {
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let [mx, p] = [0, 0];
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for (; x > 0; x = Math.floor(x / 10)) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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return nums.reduce((acc, x) => acc + encrypt(x), 0);
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}
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```
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<!-- tabs:end -->

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

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## Solutions
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### Solution 1
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### Solution 1: Simulation
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We directly simulate the encryption process by defining a function $encrypt(x),ドル which replaces each digit in an integer $x$ with the maximum digit in $x$. The implementation of the function is as follows:
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We can obtain each digit of $x$ by continuously taking the modulus and integer division of $x$ by 10ドル,ドル 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,ドル i.e., $p = 1, 11, 111, \cdots$. Finally, return $mx \times p$.
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The time complexity is $O(n \times \log M),ドル where $n$ is the length of the array, and $M$ is the maximum value in the array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def sumOfEncryptedInt(self, nums: List[int]) -> int:
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def encrypt(x: int) -> int:
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mx = p = 0
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while x:
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x, v = divmod(x, 10)
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mx = max(mx, v)
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p = p * 10 + 1
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return mx * p
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return sum(encrypt(x) for x in nums)
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```
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```java
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class Solution {
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public int sumOfEncryptedInt(int[] nums) {
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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private int encrypt(int x) {
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int mx = 0, p = 0;
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for (; x > 0; x /= 10) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int sumOfEncryptedInt(vector<int>& nums) {
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auto encrypt = [&](int x) {
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int mx = 0, p = 0;
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for (; x; x /= 10) {
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mx = max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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};
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```
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```go
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func sumOfEncryptedInt(nums []int) (ans int) {
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encrypt := func(x int) int {
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mx, p := 0, 0
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for ; x > 0; x /= 10 {
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mx = max(mx, x%10)
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p = p*10 + 1
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}
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return mx * p
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}
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for _, x := range nums {
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ans += encrypt(x)
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}
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return
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}
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```
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```ts
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function sumOfEncryptedInt(nums: number[]): number {
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const encrypt = (x: number): number => {
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let [mx, p] = [0, 0];
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for (; x > 0; x = Math.floor(x / 10)) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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return nums.reduce((acc, x) => acc + encrypt(x), 0);
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int sumOfEncryptedInt(vector<int>& nums) {
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auto encrypt = [&](int x) {
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int mx = 0, p = 0;
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for (; x; x /= 10) {
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mx = max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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};
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func sumOfEncryptedInt(nums []int) (ans int) {
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encrypt := func(x int) int {
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mx, p := 0, 0
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for ; x > 0; x /= 10 {
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mx = max(mx, x%10)
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p = p*10 + 1
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}
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return mx * p
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}
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for _, x := range nums {
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ans += encrypt(x)
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}
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return
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}
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@@ -0,0 +1,18 @@
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class Solution {
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public int sumOfEncryptedInt(int[] nums) {
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int ans = 0;
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for (int x : nums) {
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ans += encrypt(x);
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}
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return ans;
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}
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private int encrypt(int x) {
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int mx = 0, p = 0;
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for (; x > 0; x /= 10) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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}
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}
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class Solution:
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def sumOfEncryptedInt(self, nums: List[int]) -> int:
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def encrypt(x: int) -> int:
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mx = p = 0
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while x:
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x, v = divmod(x, 10)
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mx = max(mx, v)
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p = p * 10 + 1
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return mx * p
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return sum(encrypt(x) for x in nums)
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
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function sumOfEncryptedInt(nums: number[]): number {
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const encrypt = (x: number): number => {
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let [mx, p] = [0, 0];
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for (; x > 0; x = Math.floor(x / 10)) {
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mx = Math.max(mx, x % 10);
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p = p * 10 + 1;
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}
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return mx * p;
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};
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return nums.reduce((acc, x) => acc + encrypt(x), 0);
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}

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