leetcode 15 - 3sum | Optimal Approach in JAVA скачать в хорошем качестве

leetcode 15 - 3sum | Optimal Approach in JAVA

#15leetcode #java #education Problem it solves The code solves the 3Sum problem: Given an array of integers, it finds all unique triplets (a, b, c) such that a + b + c = 0. How it works Sort the array Sorting helps to handle duplicates easily and allows the use of the two-pointer approach. Fix one number at a time Use a loop with index i. For each nums[i], try to find two other numbers (nums[j] and nums[k]) that make the total sum zero. Skip duplicates for the fixed number If the current number is the same as the previous one, skip it to avoid repeating the same triplet. Two-pointer search for the other two numbers Initialize j = i+1 (next element after i) and k = n-1 (last element). Move j forward or k backward depending on whether the sum is too small or too large: If sum less than 0 → move j forward (to increase the sum). If sum greater than 0 → move k backward (to decrease the sum). If sum == 0 → we found a valid triplet. Add the triplet to the result Store the triplet [nums[i], nums[j], nums[k]] in the answer list. Skip duplicates for j and k After adding a triplet, move j and k while skipping over duplicate values to avoid repeating the same triplet. Continue until all possibilities are checked The process repeats for all values of i. Return the list of triplets At the end, we get all unique triplets that sum up to zero. 👉 In short: The algorithm sorts the array, fixes one number at a time, and uses a two-pointer technique to find pairs that, together with the fixed number, make the sum zero. Duplicate values are skipped to ensure only unique triplets are returned. Time Complexity: O(n²) Space Complexity: O(1) auxiliary (O(n²) if you count output storage).