Leetcode Problem - 3Sum
🧠 Problem
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.
🏷️ Tags
Array,Two Pointers,Sorting
📊 Difficulty
Medium
✅ Success Rate: 31%
📥 Submissions: 6,090,299
📈 Accepted: 1,886,780
❤️ Reactions
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💡 Hints
So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand! For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next numbery which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster? The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?
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