Sum of Digit Differences of All Pairs

Sum of Digit Differences of All Pairs

What is this problem about?

The "Sum of Digit Differences of All Pairs" problem asks you to calculate the total number of differences between corresponding digits for all possible pairs of numbers in an array. All numbers in the array are assumed to have the same number of digits. For example, if you have the numbers 123 and 145, the differences are at the second position (2 vs 4) and the third position (3 vs 5), making for a total of 2 differences. You need to sum these differences for every pair (nums[i], nums[j]) where i < j.

Why is this asked in interviews?

This problem is commonly used in technical interviews at Turing and Google to test a candidate's ability to optimize counting problems. A brute-force approach (comparing every pair of numbers) would take O(N² * D) time, where D is the number of digits. The key to this problem is realizing that digits at each position (ones, tens, hundreds, etc.) are independent. It evaluates if a candidate can transition from pair-based thinking to position-based frequency counting.

Algorithmic pattern used

The primary pattern used is "Digit Frequency Counting" or "Contribution Technique." Instead of comparing pairs, we look at each digit position (say, the 0-th index) and count how many times each digit (0-9) appears at that position across all numbers. For a given position, if digit 'd' appears count[d] times, it will differ from every other number that doesn't have 'd' at that position. The contribution to the total difference is calculated using the formula: Total_Numbers - count[d].

Example explanation

Suppose the array is [12, 13, 23].

• At the tens position: Digits are {1, 1, 2}.
  • '1' appears twice, '2' appears once.
  • Differences: '1' vs '2' (twice). Total = 2.
• At the units position: Digits are {2, 3, 3}.
  • '2' appears once, '3' appears twice.
  • Differences: '2' vs '3' (twice). Total = 2. Grand Total Sum of Differences = 2 + 2 = 4. This is much faster than checking (12, 13), (12, 23), and (13, 23) individually.

Common mistakes candidates make

1. O(N²) Brute Force: Failing to optimize and trying to iterate through every pair of numbers.
2. Miscounting Position Differences: Forgetting that each pair is only counted once. If using frequency counts, ensure you don't double-count the differences.
3. Ignoring the same-length constraint: Not realizing that the numbers have a fixed number of digits, which simplifies the iteration logic significantly.

Interview preparation tip

When a problem asks for a sum across "all pairs," always try to see if you can calculate the "contribution" of each element or each property (like a digit position) independently. Moving from O(N²) to O(N) is the most common goal in these types of interview questions.