Sum of Digit Differences of All Pairs
A medium-tier problem at 42% community acceptance, tagged with Array, Hash Table, Math. Reported in interviews at Turing and 0 others.
You've got an array of positive integers and need to compute the sum of digit differences across all pairs. Turing asks this one, and it shows up with a 42% acceptance rate, which means most people either misread the problem or burn time on a brute-force approach that times out. The trick is realizing you don't need to iterate all pairs. You can calculate the contribution of each digit position independently using a hash table or frequency map, turning an O(n^2) nightmare into something linear. If you hit this live and blank on the optimization, StealthCoder surfaces the correct approach in seconds, invisible to the proctor.
Companies that ask "Sum of Digit Differences of All Pairs"
Sum of Digit Differences of All Pairs is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE.
Get StealthCoderThe naive approach: iterate every pair (i, j), compute absolute differences of digits, sum them all up. That's O(n^2 * d) where d is the number of digits. It fails on large inputs. The real solution exploits the independence of digit positions. For each position (ones, tens, hundreds, etc.), count how many numbers have each digit 0-9 at that position, then calculate how much each digit contributes to the total. A hash table or counter tracks frequency. This converts the problem into counting and arithmetic, cutting the work dramatically. Common pitfall: forgetting to handle the absolute value correctly, or not realizing you can process each digit place separately. The acceptance rate tells you most candidates don't see the counting angle fast enough.
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Sum of Digit Differences of All Pairs recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Sum of Digit Differences of All Pairs interview FAQ
Why does brute force fail on this problem?+
Brute force iterates all pairs and computes digit differences for each one, yielding O(n^2) complexity. With n up to 10^5, that's 10 billion operations. You need to count digit frequencies by position instead, which is O(n * d). The problem is designed to reject the obvious loop.
How do I use a hash table to solve this efficiently?+
For each digit position (ones, tens, hundreds), create a frequency map of which digits appear there. Then for each digit d that appears k times, calculate how it contributes to the sum: it pairs with (n - k) numbers that have different digits at that position. Sum across all positions and digits. This avoids iterating pairs.
What's the trick with digit positions?+
Each digit position is independent. The ones place of one number only interacts with the ones place of another, never with tens or hundreds. By processing each position separately and counting frequencies, you reduce a pairwise problem to a frequency-counting problem. That's where the speedup lives.
Is this problem still asked at top companies?+
Turing reports asking it. It's not a household FAANG problem, but it appears in specialized interview rounds focused on optimization and counting tricks. A 42% acceptance rate suggests it's recent or deliberately hard. It's worth knowing the digit-position decomposition pattern.
What's the difference between this and other counting problems?+
Most counting problems fix the outcome and count how many ways to achieve it. This one fixes the computation (digit difference) and counts contributions by position. The mental shift from 'iterate all pairs' to 'process digit by digit' is non-obvious. That's why acceptance is below 50%.
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