#3079

Easy

LeetCode Problem Workspace

Find the Sum of Encrypted Integers

Compute the sum of encrypted integers by replacing each digit with the largest digit, combining array traversal with digit math.

array math

Problem Statement

Given an array of positive integers nums, define an encryption function where encrypt(x) replaces each digit in x with the largest digit present in x. For instance, encrypt(523) becomes 555 and encrypt(213) becomes 333.

Return the total sum of all encrypted numbers. For example, given nums = [1,2,3], the encrypted elements are [1,2,3] and the sum is 6. Constraints: 1 <= nums.length <= 50, 1 <= nums[i] <= 1000.

Examples

Example 1

Input: nums = [1,2,3]

Output: 6

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

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

Solution Approach

Iterate and Encrypt Each Number

Loop through each number in the array, extract its digits, determine the maximum digit, and construct the encrypted number by repeating that digit according to the original number's length.

Sum Encrypted Numbers

Maintain a running total of the encrypted numbers as they are generated. Add each encrypted number to a sum variable to get the final answer efficiently.

Optimize Digit Extraction

Use integer division and modulus operations to extract digits instead of converting to strings for slightly better performance. This reduces overhead while maintaining clarity for array plus math pattern.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * d) where n is the array length and d is the max number of digits in nums[i]. Space complexity is O(1) extra space if summing in place, otherwise O(n) for storing encrypted numbers.

What Interviewers Usually Probe

Check if the candidate correctly identifies the largest digit in each number.
Observe if the candidate efficiently loops through digits without unnecessary conversions.
Notice if the candidate maintains the correct sum while preserving number lengths in encryption.

Common Pitfalls or Variants

Common pitfalls

Assuming the encrypted number is simply the max digit rather than repeating it for the length of the original number.
Using string conversion without handling numbers with multiple digits correctly.
Forgetting to sum the encrypted numbers and returning only the last encrypted value.

Follow-up variants

Compute the sum of encrypted integers where only odd digits are replaced by the largest odd digit in each number.
Encrypt numbers by replacing each digit with the smallest digit and return their sum.
Given a 2D array of integers, encrypt each element as described and sum all values.

FAQ

What does 'Find the Sum of Encrypted Integers' require exactly?

It requires replacing each digit in every number with the largest digit in that number and summing all results.

Can I convert numbers to strings to extract digits?

Yes, but be cautious to repeat the max digit correctly according to the original number's length.

What is the time complexity for this array plus math problem?

Time complexity is O(n * d) where n is the number of elements and d is the maximum digits per number.

Are there constraints on the array size or number values?

Yes, 1 <= nums.length <= 50 and 1 <= nums[i] <= 1000.

How can I optimize summing encrypted integers?

Compute each encrypted number on the fly and add it directly to a sum variable to avoid extra storage.

terminal

Solution

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:

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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)

Continue Practicing

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