Completed TwoPointer1

Solutions.py

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+Problem1 (https://leetcode.com/problems/sort-colors/)
+Time Complexity : O(n)
+Space Complexity : O(1)
+
+
+## we need to collect 0 at the start
+## 1 in the mid 
+## 2 in the end
+## so we can have three different pointers, check if the value is 2 replace
+## check if the value is 0, replace from low , increment pointers
+## else just increment mid
+
+class Solution(object):
+    def sortColors(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: None Do not return anything, modify nums in-place instead.
+        """
+        low,high = 0,len(nums)-1
+        mid = 0
+
+        while mid<=high:
+            if nums[mid] == 2:
+                nums[mid],nums[high] = nums[high],nums[mid]
+                high-=1
+            elif nums[mid] == 0:
+                nums[mid],nums[low] = nums[low],nums[mid]
+                low+=1
+                mid +=1
+            else:
+                mid+=1
+
+
+
+
+
+Problem2 (https://leetcode.com/problems/3sum/)
+Time Complexity : O(n^2)
+Space Complexity : O(1)
+
+
+## sort the array so that we can manipulate left and right pointer
+## for removing the duplicates, we can move i until it remains the same
+## choose the l and r, check the sum, if sum ==0 add
+## move left if sum <0
+## move right if sum > 0 
+
+class Solution(object):
+    def threeSum(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        res = []
+        nums.sort()
+        for i in range(len(nums)-1):
+            if i>0 and nums[i] == nums[i-1]:
+                continue
+
+            l = i+1
+            r = len(nums)-1
+            while l<r:
+
+                sum_ = nums[i] + nums[l] + nums[r]
+                if sum_<0:
+                    l+=1
+                elif sum_ >0:
+                    r-=1
+                else:
+                    res.append([nums[i],nums[l],nums[r]])
+                    l+=1
+                    r-=1
+                    while(l < r and nums[l] == nums[l - 1]): 
+                        l+=1
+                    while(l < r and nums[r] == nums[r + 1]): 
+                        r-=1
+
+        return res
+
+
+
+Problem3 (https://leetcode.com/problems/container-with-most-water/)
+Time Complexity : O(n)
+Space Complexity : O(1)
+
+class Solution(object):
+    def maxArea(self, height):
+        """
+        :type height: List[int]
+        :rtype: int
+        """
+       ## initialize left and right
+       ## check for water stored
+       ## replace with max 
+       ## reduce or increase pointers
+
+        l=0
+        r=len(height)-1
+        cur_area = max_area =0  
+        while(l<r):
+
+            cur_area = min(height[r],height[l]) * (r-l)
+            max_area = max(max_area,cur_area)
+
+            if height[l] > height[r]:
+                r-=1
+            else:
+                l+=1
+
+        return max_area