Answer: Airbnb Associate Developer Interview Question 2024 August | Two Pointers & DP |

Solution 1: 3Sum — Two Pointers Approach To avoid O(N3) brute force search, we sort the array and iterate through it, using a **Two-Pointer** search for each element. This reduces the time complexity to O(N2). For each element nums[i], if it is positive, we can break early since sorting guarantees no three positive elements can sum to 0. We set two pointers: left = i + 1 and right = nums.length - 1. If the sum is too small, we ...