solved 3sum

Author: samcho0608

3sum/samcho0608.java

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+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+// link: https://leetcode.com/problems/3sum/
+// difficulty: Medium
+class Solution {
+    // Problem:
+    // * return: all triplets of elements in nums such that the sum == 0 (indices must differ)
+    // Solution:
+    // * Time Complexity: O(N^2)
+    // * Space Complexity: O(1)
+    public List<List<Integer>> threeSum(int[] nums) {
+        // sort for simplicity
+        // Time Complexity: O(N log N)
+        Arrays.sort(nums);
+
+        List<List<Integer>> answers = new ArrayList<>();
+
+        // if there are only positive or only negative numbers, there exist no solution
+        if(nums[0] > 0 || nums[nums.length-1] < 0) return answers;
+
+
+
+        // Time Complexity: O(N^2)
+        // * for loop * inner while loop = O(N) * O(N) = O(N^2)
+
+        // nums[i]: first of the triplet
+        // * skip positive because the other two will also be positive
+        for(int i = 0; i < nums.length && nums[i] <= 0; i++) {
+            // skip if same first of the triplet met
+            if(i != 0 && nums[i] == nums[i-1]) continue;
+
+            int numI = nums[i];
+
+            int left = i + 1;
+            int right = nums.length - 1;
+
+            // Time Complexity: O(N)
+            while(left < right) {
+                int numLeft = nums[left];
+                int numRight = nums[right];
+
+                int sum = numI + numLeft + numRight;
+
+                if(sum == 0) {
+                    answers.add(Arrays.asList(numI, numLeft, numRight));
+
+                    // skip same lefts and rights two prevent duplicate
+                    // * must update both left and right
+                    //    * e.g. if only left is moved, newNumLeft = 0 - (numI + numRight) and that can only be numLeft that's already visited
+                    // * Time Complexity: O(1) because there is no actual operation other than skipping
+                    while(left < nums.length && nums[left] == numLeft) left++;
+
+                    while(right > left && nums[right] == numRight) right--;
+                } else if(sum > 0) {
+                    right--;
+                } else {
+                    left++;
+                }
+            }
+        }
+
+        return answers;
+    }
+}