# 3079. Find the Sum of Encrypted Integers

## Description

You are given an integer array nums containing positive integers. We define a function encrypt such that encrypt(x) replaces every digit in x with the largest digit in x. For example, encrypt(523) = 555 and encrypt(213) = 333.

Return the sum of encrypted elements.

Example 1:

Input: nums = [1,2,3]

Output: 6

Explanation: The encrypted elements are [1,2,3]. The sum of encrypted elements is 1 + 2 + 3 == 6.

Example 2:

Input: nums = [10,21,31]

Output: 66

Explanation: The encrypted elements are [11,22,33]. The sum of encrypted elements is 11 + 22 + 33 == 66.

Constraints:

• 1 <= nums.length <= 50
• 1 <= nums[i] <= 1000

## Solutions

### Solution 1: Simulation

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:

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 $mx \times p$.

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 $O(1)$.

• class Solution {
public int sumOfEncryptedInt(int[] nums) {
int ans = 0;
for (int x : nums) {
ans += encrypt(x);
}
return ans;
}

private int encrypt(int x) {
int mx = 0, p = 0;
for (; x > 0; x /= 10) {
mx = Math.max(mx, x % 10);
p = p * 10 + 1;
}
return mx * p;
}
}

• class Solution {
public:
int sumOfEncryptedInt(vector<int>& nums) {
auto encrypt = [&](int x) {
int mx = 0, p = 0;
for (; x; x /= 10) {
mx = max(mx, x % 10);
p = p * 10 + 1;
}
return mx * p;
};
int ans = 0;
for (int x : nums) {
ans += encrypt(x);
}
return ans;
}
};

• class Solution:
def sumOfEncryptedInt(self, nums: List[int]) -> int:
def encrypt(x: int) -> int:
mx = p = 0
while x:
x, v = divmod(x, 10)
mx = max(mx, v)
p = p * 10 + 1
return mx * p

return sum(encrypt(x) for x in nums)


• func sumOfEncryptedInt(nums []int) (ans int) {
encrypt := func(x int) int {
mx, p := 0, 0
for ; x > 0; x /= 10 {
mx = max(mx, x%10)
p = p*10 + 1
}
return mx * p
}
for _, x := range nums {
ans += encrypt(x)
}
return
}

• function sumOfEncryptedInt(nums: number[]): number {
const encrypt = (x: number): number => {
let [mx, p] = [0, 0];
for (; x > 0; x = Math.floor(x / 10)) {
mx = Math.max(mx, x % 10);
p = p * 10 + 1;
}
return mx * p;
};
return nums.reduce((acc, x) => acc + encrypt(x), 0);
}