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This is a medium difficulty Coding & Algorithms question, commonly asked during Technical Screen rounds at Akuna Capital.

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This question is commonly asked for Data Scientist candidates at Akuna Capital during technical interviews.

Maximize profitable pairs | Akuna Capital Coding Question

Quick Overview

This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Maximize profitable pairs states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

Maximize profitable pairs

Company: Akuna Capital

Role: Data Scientist

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Technical Screen

You are given an array profits of n integers representing net profit per item and an integer threshold T. You may form disjoint pairs (i, j). What is the maximum number of pairs such that profits[i] + profits[j] >= T? Describe an O(n log n) algorithm using sorting and a left/right pointer strategy, prove correctness, analyze complexity, and discuss edge cases (odd n, negative values, very large or very small T). Provide pseudocode.

Quick Answer: This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Maximize profitable pairs states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.

You are given an array `profits` of `n` integers representing the net profit per item, and an integer threshold `T`. You may form disjoint pairs `(i, j)` (each index used in at most one pair). Return the maximum number of pairs you can form such that `profits[i] + profits[j] >= T`. Use an O(n log n) approach: sort the array, then use a left/right two-pointer scan. If the smallest and largest available values already sum to at least `T`, pair them and move both pointers inward; otherwise the smallest value cannot reach `T` even with the largest partner, so discard it (advance the left pointer). Example 1: Input: profits = [1, 2, 3, 4, 5], T = 6 Output: 2 Explanation: Pair (5,1) -> sum 6, and pair (4,2) -> sum 6. The 3 is left unpaired. 2 pairs. Example 2: Input: profits = [1, 1, 1, 1], T = 10 Output: 0 Explanation: No two values sum to at least 10. Example 3: Input: profits = [10, -10, 3, 3], T = 0 Output: 2 Explanation: Pair (10,-10) -> 0, pair (3,3) -> 6. 2 pairs.

Constraints

0 <= n <= 10^5
-10^9 <= profits[i] <= 10^9
-10^9 <= T <= 10^9
Each index may be used in at most one pair (pairs are disjoint).

Examples

Input: ([1, 2, 3, 4, 5], 6)

Expected Output: 2

Explanation: Pairs (5,1) and (4,2) each sum to 6; 3 is left over.

Input: ([5, 5, 5, 5], 10)

Expected Output: 2

Explanation: Two pairs of (5,5), each summing to 10.

Hints

Sort the array first. After sorting, the optimal strategy can be decided greedily from the two ends.
Use two pointers: left at the smallest value, right at the largest. If their sum >= T, this is a valid pair — count it and move both inward.
If profits[left] + profits[right] < T, then profits[left] cannot meet the threshold with ANY available partner (right is the largest left). Discard left (advance it) and try again.
Greedy exchange argument: pairing the current smallest unpairable-or-pairable value with the largest never loses an achievable pairing, so this maximizes the count.