Introduction

In this blog post, we will explore how to solve the LeetCode problem "3153" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array nums consisting of positive integers where all integers have the same number of digits. The digit difference between two integers is the count of different digits that are in the same position in the two integers. Return the sum of the digit differences between all pairs of integers in nums. Example 1: Input: nums = [13,23,12] Output: 4 Explanation: We have the following: - The digit difference between 13 and 23 is 1. - The digit difference between 13 and 12 is 1. - The digit difference between 23 and 12 is 2. So the total sum of digit differences between all pairs of integers is 1 + 1 + 2 = 4. Example 2: Input: nums = [10,10,10,10] Output: 0 Explanation: All the integers in the array are the same. So the total sum of digit differences between all pairs of integers will be 0. Constraints: 2 <= nums.length <= 105 1 <= nums[i] < 109 All integers in nums have the same number of digits.

Explanation

Here's an efficient solution to calculate the sum of digit differences between all pairs of integers in the given array:

• Key Idea: Iterate through each digit position of the numbers. For each position, count the frequency of each digit (0-9) that appears in that position across all numbers. Then, for each digit, multiply its count by the sum of counts of all other digits in that position. This gives the digit differences involving that digit at that position. Summing this over all digits at all positions provides the total digit difference sum.
• Optimization: We can optimize the counting process using an array to store digit frequencies.

• Efficiency: This approach avoids direct pairwise comparisons, making it more efficient for large arrays.

• Runtime Complexity: O(N D), where N is the number of integers in nums and D is the number of digits in each integer. Storage Complexity: O(1), specifically O(10 D) which simplifies to O(D) where D is the number of digits in each integer, since we store digit counts in fixed-size arrays.

Code

    def digit_difference_sum(nums):
    n = len(nums)
    num_digits = len(str(nums[0]))
    total_difference = 0

    for digit_index in range(num_digits):
        digit_counts = [0] * 10
        for num in nums:
            digit = (num // (10**(num_digits - 1 - digit_index))) % 10
            digit_counts[digit] += 1

        for i in range(10):
            total_difference += digit_counts[i] * (n - digit_counts[i])

    return total_difference // 2  # Divide by 2 since each pair is counted twice