How do you find a triplet where a squared equals b squared plus c squared?
This problem requires finding three numbers in an array that satisfy the Pythagorean theorem condition. It tests your skills in array manipulation and optimization techniques.
Why Interviewers Ask This
This question checks if you can optimize a brute-force O(N^3) solution to something more efficient like O(N^2). Interviewers look for your ability to transform mathematical conditions into algorithmic steps. They also evaluate your understanding of hashing or two-pointer techniques to reduce search space. It simulates scenarios where finding specific patterns in large datasets is crucial.
How to Answer This Question
Begin by sorting the array to enable the two-pointer technique. Iterate through each element considering it as the potential hypotenuse 'a'. Use two pointers, one starting just after 'a' and one at the end, to find 'b' and 'c'. Calculate sums and adjust pointers based on whether the sum is too small or too large. Alternatively, mention using a hash set for O(N^2) solutions without sorting. Clearly articulate the trade-offs between sorting and hashing approaches.
Key Points to Cover
- Square elements to simplify equation
- Apply two-pointer technique after sorting
- Optimize from O(N^3) to O(N^2)
- Handle edge cases like duplicates
Sample Answer
First, I would square all elements in the array to simplify the check. Then, I could iterate through each element treating it as 'a'. For the remaining elements, I would use a hash set to store squares of 'b' and check if 'a_squared - b_squared' exists in the set. A more optimized approach involves sorting the array first. Fix 'a' and use two pointers for 'b' and 'c'. If the sum of squares is less than 'a', move the left pointer up; if greater, move the right pointer down. This reduces the complexity to O(N^2) and avoids the overhead of hashing collisions.
Common Mistakes to Avoid
- Ignoring negative numbers or zero
- Forgetting to sort before using two pointers
- Not handling duplicate triplets correctly
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