Brute force checks all triplets in . Can we do better? Here's the idea: if you fix one element, you're left with a two-sum problem. For each element nums[], find two elements that sum to nums[]. Two Sum with a hash map is . So fixing one element and doing Two Sum gives . But there's an even cleaner approach using sorting and two pointers.
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$ curl repovive.com/roadmaps/maang-interview-prep/two-pointers-sliding-window/3sum-hint
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