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Leetcode 3079. Find the Sum of Encrypted Integers

Problem Statement

You are given an array encryptedNums which represents a series of integers that have been encrypted. Your task is to decrypt these integers and calculate their sum. The encryption rule is as follows:

• Every digit d in the encrypted integers was originally the digit 9 - d.

For example, the integer 284 was originally 715 before encryption (2 -> 7, 8 -> 1, 4 -> 5).

Write a function findSumOfDecryptedNums that takes in the array encryptedNums and returns the sum of the decrypted integers.

Examples

Input: encryptedNums = [284, 619]
Output: 1334

Explanation:
- Decrypting 284: 7(9-2), 1(9-8), 5(9-4) -> 715
- Decrypting 619: 3(9-6), 8(9-1), 0(9-9) -> 380
- Sum is 715 + 380 = 1095

Clarifying Questions

1. Input Format: Are encryptedNums guaranteed to be positive integers?
2. Output Format: Should the output be a single integer (the sum of decrypted integers)?
3. Constraints on the size of encryptedNums: What are the maximum possible values for the length and elements of encryptedNums?
4. Leading Zeros: How should leading zeros after decryption be handled?

Strategy

1. Decrypt Each Number: We will create a helper function decryptNumber to decrypt each number based on the rule d = 9 - encryptedDigit.
2. Sum Decrypted Numbers: Sum all the decrypted numbers and return the result.

Code

public class SumOfEncryptedIntegers {

    public static int findSumOfDecryptedNums(int[] encryptedNums) {
        int sum = 0;
        for (int encryptedNum : encryptedNums) {
            sum += decryptNumber(encryptedNum);
        }
        return sum;
    }

    private static int decryptNumber(int encryptedNum) {
        int decryptedNum = 0;
        int placeValue = 1;
        
        while (encryptedNum > 0) {
            int digit = encryptedNum % 10;
            int decryptedDigit = 9 - digit;
            
            decryptedNum += decryptedDigit * placeValue;
            placeValue *= 10;
            
            encryptedNum /= 10;
        }
        
        return decryptedNum;
    }

    public static void main(String[] args) {
        int[] encryptedNums = {284, 619};
        System.out.println(findSumOfDecryptedNums(encryptedNums));  // Output: 1095
    }
}

Time Complexity

• Decrypting Each Number: The decryption process involves processing each digit in the encrypted number, so it is O(d) where d is the number of digits in the number.
• Summing Decrypted Numbers: Summing n decrypted numbers involves O(n) operations.
• Overall Complexity: If n is the number of elements in encryptedNums and d is the typical number of digits in each number, the overall complexity is O(n * d).

This solution efficiently decrypts each number and sums them up in linear time relative to the size of the input and the size of the numbers.

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