Completed Two Pointer - 1

Problem1.java

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+1// Time Complexity :O(n)
+// 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
+/* we can count the number of 0s,1s and 2s and just set them in the array which works but that would be pass through the array twice
+I have proceeded with two point approach along with an additional pointer as marker for tracking the mid value.
+*/
+class Solution {
+    public void sortColors(int[] nums) {
+        //two pointer
+        int l=0, mid =0;//left for tracking 0s and mid for tracking 1s
+        int r = nums.length-1;// right for tracking 2s
+        while(mid<=r){
+            if(nums[mid]==2){
+                swap(nums, mid, r);//move the 2s to right
+                r--;//reduce right by 1 since we know 2 is in the end
+            }
+            else if(nums[mid]==0){
+                swap(nums, l, mid);//move 0 to the front
+                l++;//move left since we know 0 is in the front
+                mid++;
+            }
+            else{
+                mid++;//since the number is 1 do nothing as 1 is in mid
+            }
+        }
+    }
+
+    public void swap(int []nums, int x, int y){
+        int temp = nums[x];
+        nums[x]=nums[y];
+        nums[y]=temp;
+    }
+}

Problem2.java

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+// Time Complexity :O(n2)
+// 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
+/*
+we can implement by selecting a pivot element and sum other two elements which would bring total sum to 0
+the other two elements can be identified wither by 
+1. two sum problem - this will need a additional hashset of triplets to avoid duplicate triplets 
+time - O(n2) 
+space - O(n)
+2. binary search
+time - O(n2logn)
+3.we can use two pointers 
+time- O(n2)
+space - O(1)
+*/
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> result= new ArrayList<>();
+        Arrays.sort(nums);// sort because we want to implement two pointer
+        for(int i=0;i<nums.length;i++){
+            if(nums[i]>0) break;//we dont want to iterate if one element all the elements after this is >0 because sum is >0
+            if(i>0 && nums[i]==nums[i-1]) continue;//remove external duplicate triplets
+            int l=i+1;
+            int r=nums.length-1;
+            //two pointer for remaining elements after the pivot
+            while(l<r){
+                if(nums[l]+nums[r]+nums[i]==0){
+                    List<Integer> li= Arrays.asList(nums[l],nums[r],nums[i]);
+                    result.add(li);
+                    l++;
+                    r--;
+                    //remove interval duplicate elements
+                    while(l<r && nums[l]==nums[l-1]) l++;
+                    while(l<r && nums[r]==nums[r+1]) r--;
+                }
+                else if(nums[l]+nums[r]+nums[i]<0){
+                    l++;
+                }
+                else{
+                    r--;
+                }
+            }
+        }
+        return result;
+    }
+}

Problem3.java

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+// Time Complexity :O(n)
+// 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
+/* use two pointer approach to find tha maximum area of the rectangle 
+height being value and width being difference between indexes
+time-O(n)
+space- O(1)*/
+
+class Solution {
+    public int maxArea(int[] height) {
+        int l=0;
+        int m=0;
+        int a=0;
+        int r=height.length-1;
+        while(l<r){
+            //find maximum area
+            a=Math.max(Math.min(height[l], height[r])*(r-l),a);
+            if(height[l]<height[r]){
+                l++;
+            }
+            else{
+                r--;
+            }
+        }
+        return a;
+    }
+}