#3153

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Sum of Digit Differences of All Pairs

Sum of Digit Differences of All Pairs is a problem that involves counting digit differences in pairs from an array of integers.

array hash table math counting

Problem Statement

You are given an array of positive integers, nums, where all integers have the same number of digits. The digit difference between two integers is the count of different digits in the same position. Your task is to compute the sum of digit differences between all pairs of integers in nums.

The problem asks you to return this sum after checking every possible pair of integers. Each pair consists of two integers from the array, and you need to compare their digits to determine how many digits differ at corresponding positions.

Examples

Example 1

Input: nums = [13,23,12]

Output: 4

We have the following: - The digit difference between 1 3 and 2 3 is 1. - The digit difference between 1 3 and 1 2 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

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.

Solution Approach

Digit-by-Digit Comparison

The core approach involves comparing digits at each position separately. For each digit position, count how many pairs have different digits at that position. This reduces the problem into smaller, easier sub-problems.

Hash Lookup for Efficiency

To speed up the process, use hash tables to store the frequency of digits at each position. This allows you to compute digit differences efficiently by leveraging quick lookups instead of manually comparing all pairs.

Summing Up Differences

After calculating the differences for each digit position, sum up the results for all positions to get the final answer. This approach ensures that you consider all digit positions while avoiding unnecessary repetitions.

Complexity Analysis

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

The time complexity depends on the number of digits in the integers and the length of the array. Since you are processing each digit in the integers separately, the overall time complexity is linear with respect to the total number of digits in all integers. The space complexity depends on the hash tables used to store digit frequencies, which scale with the number of digits per integer.

What Interviewers Usually Probe

Look for an understanding of how to optimize comparing digits by position.
Evaluate the candidate's ability to reduce a complex problem into smaller sub-tasks.
Check if the candidate correctly utilizes hash tables to reduce time complexity.

Common Pitfalls or Variants

Common pitfalls

Candidates may try to compare each pair of integers directly, leading to a brute-force solution with poor time complexity.
Forgetting to optimize digit position comparisons could result in inefficient solutions.
Not summing up the results correctly across all digit positions could lead to an incorrect final answer.

Follow-up variants

What if the integers have varying numbers of digits?
What if the array contains non-positive integers?
Can the problem be optimized for space instead of time complexity?

FAQ

What is the key pattern in solving the Sum of Digit Differences of All Pairs problem?

The key pattern is to handle the digit positions separately and then sum up the results, using efficient methods like hash tables for better performance.

How can I avoid brute force in this problem?

By focusing on comparing digits at each position separately and using hash tables for efficient lookups, you avoid brute force comparisons of all pairs.

What is the time complexity of this problem?

The time complexity is linear with respect to the total number of digits across all integers in the array, due to the digit-by-digit comparison approach.

What are the potential pitfalls to watch out for?

Common pitfalls include failing to optimize digit comparisons, neglecting to use hash tables, or incorrectly summing the differences for each digit position.

Can this problem be optimized further for space?

The space complexity is driven by the hash tables used for digit frequency. While there’s limited opportunity for significant space optimization, consider whether a more memory-efficient approach could be applicable in specific cases.

terminal

Solution

Solution 1: Counting

First, we get the number of digits $m$ in the array. Then for each digit, we count the occurrence of each number at this digit in the array `nums`, denoted as `cnt`. Therefore, the sum of the digit differences of all number pairs at this digit is:

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class Solution:
    def sumDigitDifferences(self, nums: List[int]) -> int:
        n = len(nums)
        m = int(log10(nums[0])) + 1
        ans = 0
        for _ in range(m):
            cnt = Counter()
            for i, x in enumerate(nums):
                nums[i], y = divmod(x, 10)
                cnt[y] += 1
            ans += sum(v * (n - v) for v in cnt.values()) // 2
        return ans

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