Completed Two-Pointers-1-3Sum

3Sum.java

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+// Time Complexity : O(nlogn)(sorting) + O(n)*O(n)(traversal*2sum computation)= O(nlogn) + O(n^2) = O(n^2)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this :no
+
+// Your code here along with comments explaining your approach
+/*
+Sort the given array first.Now, use left, right and i pointers to compute the sum and check if it equals
+0.If so, form the values of those pointers as a list and make sure to check duplicate values exist to
+avoid duplicate traversals and calculations.If sum is not 0, move left and right pointers accordingly.
+ */
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> result = new ArrayList<>();
+        Arrays.sort(nums);
+        int n = nums.length;
+
+        for(int i = 0 ; i < n ; i++) {
+            int left = i + 1 , right = n - 1;
+            if(i != 0 && nums[i] == nums[i - 1])
+                continue;
+            while(left < right) {
+                int sum = nums[left] + nums[right] + nums[i];
+                if(sum == 0) {
+                    List<Integer> list = Arrays.asList(nums[i], nums[left], nums[right]);
+                    result.add(list);
+                    left++;
+                    right--;
+                    while(left < right && nums[left] == nums[left - 1])
+                        left++;
+                    while(left < right && nums[right] == nums[right + 1])
+                        right--;
+                }
+                else if(sum < 0)
+                    left++;
+                else if(sum > 0)
+                    right--;
+            }
+        }
+        return result;
+    }
+}