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 \(O(1)\).

Python3JavaC++GoTypeScript

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)

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;
    }
};

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);
}

3079. Find the Sum of Encrypted Integers

Description

Solutions

Solution 1: Simulation

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