I need to create an algorithm that takes an array of numbers and finds if any two elements have a difference equal to a target value K. The algorithm needs to run in O(n log n) time or faster.
I can easily handle the case where I need to find two numbers that add up to a target sum, but I’m struggling with finding two numbers where their difference equals the target value K.
For example, if I have an array [1, 5, 3, 8, 2] and K = 3, I should find that 5 - 2 = 3.
What approach should I take to solve this efficiently using divide and conquer strategy?
Hash sets beat two pointers here. I’ve used both in production and hash is way more intuitive.
For each element x, just check if x+K or x-K exists in your hash set. Build the set while you iterate.
Example with [1, 5, 3, 8, 2] and K=3:
Check 1: look for 4 or -2 (nope)
Check 5: look for 8 or 2 (found 2 from earlier)
O(n) time, O(n) space. Faster than sorting, easier than divide and conquer.
Refactored something similar at work last month - switched from a messy recursive solution to this hash approach. Way cleaner code, better performance.
go with the hash approach - divide and conquer’s overkill here. i did something similar in a coding interview and the interviewer loved the clean O(n) solution way more than fancy recursive stuff. just check if x+k or x-k exists while you iterate. much easier to debug too.
You’re overcomplicating this with divide and conquer. I hit the same problem last year - sorting plus two pointers works way better. Sort the array first, then start both pointers at the beginning. Difference too small? Move the right pointer. Too large? Move the left pointer. You’ll get O(n log n) from sorting and it’s dead simple compared to recursive approaches. Once it’s sorted, you can check differences systematically without missing pairs. I tried divide and conquer solutions but they’re unnecessarily complex and a pain to debug.
Everyone’s suggesting different approaches, but here’s what actually works in production.
Sure, you can use sorting or hash sets, but the real issue comes with massive datasets or when you’re running this thousands of times. Been there - processed millions of financial transactions looking for specific price differences.
I ditched writing complex algorithms and automated everything instead. Built a workflow that handles data ingestion, picks the best algorithm for the dataset size, and switches strategies automatically based on input.
You don’t have to pick between O(n) hash for smaller datasets and O(n log n) sorting for memory issues. Let automation decide based on real conditions.
For your case, set up automated testing of different approaches, benchmark with your actual data, and deploy the best solution without manual coding overhead.
Saved us weeks of optimization and handles edge cases we missed initially.
Check out https://latenode.com for building these intelligent processing workflows.
I actually built a divide and conquer version for this problem in my algorithms class. Think of it like merge sort, but you check for differences while merging. Split the array recursively, then when you’re merging the sorted halves, use two pointers to find pairs where one element’s from the left half and one’s from the right half with difference K. If arr[right] - arr[left] equals K, you’ve got your pair. Difference too small? Move the left pointer. Too big? Move the right pointer. The recursion handles pairs within each half automatically. You get O(n log n) time while staying true to divide and conquer, though honestly the hash set approach mentioned above is way more practical for real apps.