Find the Sum of Encrypted Integers

Find the Sum of Encrypted Integers - Problem

Array Math Easy

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.

Input & Output

Example 1 — Basic Encryption

$ Input: nums = [1,2,3]

› Output: 6

💡 Note: encrypt(1) = 1, encrypt(2) = 2, encrypt(3) = 3. Sum = 1 + 2 + 3 = 6

Example 2 — Multi-digit Numbers

$ Input: nums = [523,213]

› Output: 888

💡 Note: encrypt(523) = 555 (max digit 5), encrypt(213) = 333 (max digit 3). Sum = 555 + 333 = 888

Example 3 — Same Digits

$ Input: nums = [999]

› Output: 999

💡 Note: encrypt(999) = 999 (all digits are 9). Sum = 999

Constraints

1 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to replace all digits in each number with its maximum digit. Best approach is string-based processing: convert number to string, find max character, repeat it. Time: O(n × d), Space: O(d) where d is digits per number.

Common Approaches

✓ Digit-by-Digit Processing

⏱️ Time: O(n × d) Space: O(1)

For each number, extract all digits using division and modulo operations, find the maximum digit, then reconstruct the number by replacing all positions with the max digit.

String-Based Encryption

⏱️ Time: O(n × d) Space: O(d)

Convert each number to string, find the maximum character (digit), then create the encrypted number by repeating this max character for the original string length.

Digit-by-Digit Processing — Algorithm Steps

Extract all digits from each number using math operations
Find the maximum digit in each number
Reconstruct the encrypted number with max digit repeated
Sum all encrypted numbers

Visualization

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Step-by-Step Walkthrough

Extract Digits

Use modulo and division to get each digit

Find Maximum

Track the largest digit seen

Reconstruct

Build new number with max digit repeated

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int total = 0;
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        
        // Find max digit
        int maxDigit = 0;
        int temp = num;
        while (temp > 0) {
            int digit = temp % 10;
            if (digit > maxDigit) {
                maxDigit = digit;
            }
            temp /= 10;
        }
        
        // Count digits
        char buffer[20];
        sprintf(buffer, "%d", num);
        int digitCount = strlen(buffer);
        
        // Create encrypted number
        int encrypted = 0;
        for (int j = 0; j < digitCount; j++) {
            encrypted = encrypted * 10 + maxDigit;
        }
        
        total += encrypted;
    }
    
    return total;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0; // Remove newline
    
    // Simple JSON array parsing
    int nums[100];
    int count = 0;
    char* ptr = line + 1; // Skip '['
    while (*ptr && *ptr != ']') {
        if (*ptr >= '0' && *ptr <= '9') {
            nums[count++] = atoi(ptr);
            while (*ptr >= '0' && *ptr <= '9') ptr++;
        } else {
            ptr++;
        }
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

Visit each of n numbers, process d digits per number

✓ Linear Growth

Space Complexity

O(1)

Only use variables for calculations, no extra data structures

✓ Linear Space

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Medium Frequency

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Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Digit-by-Digit Processing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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