medium two pointers sorted ArraysasList great tryyy

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Hint
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.

 

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
 

Constraints:

3 <= nums.length <= 3000
-105 <= nums[i] <= 105

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Hint 1
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!
Hint 2
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 number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?
Hint 3
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?

Author: Gauranshgoel123

three-sum.java

@@ -0,0 +1,30 @@
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> res=new ArrayList<>();
+        Arrays.sort(nums);
+        for(int i=0;i<nums.length;i++){
+            if(i>0 && nums[i]==nums[i-1]){
+                continue;
+            }
+            int l=i+1;
+            int r=nums.length-1;
+            while(l<r){
+                if(nums[i]+nums[l]+nums[r]==0){
+                    res.add(Arrays.asList(nums[i],nums[r],nums[l]));
+                    l++;
+                    //to avoid duplicates
+                    while(l<r && nums[l]==nums[l-1]){
+                        l++;
+                    }
+                }
+                else if(nums[i]+nums[l]+nums[r]>0){
+                    r--;
+                }
+                else{
+                    l++;
+                }
+            }
+        }
+        return res;
+    }
+}