From 3cf3734ecacbdd9b68831f53f06b28585ba32f74 Mon Sep 17 00:00:00 2001
From: pranjay01 <pranjay.sagar01@gmail.com>
Date: Tue, 3 Mar 2026 22:06:43 -0800
Subject: [PATCH] Done Two pointers 1

---
 Problem1.py | 33 +++++++++++++++++++++++++++++++++
 Problem2.py | 42 ++++++++++++++++++++++++++++++++++++++++++
 Problem3.py | 26 ++++++++++++++++++++++++++
 3 files changed, 101 insertions(+)
 create mode 100644 Problem1.py
 create mode 100644 Problem2.py
 create mode 100644 Problem3.py

diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..5a1c7c4e
--- /dev/null
+++ b/Problem1.py
@@ -0,0 +1,33 @@
+#Sort Colors
+
+#O(n) time complexity
+#O(1) space complecity
+#Logic -> Have 3 ointers, left, mid and right. if mid is 2, replace with right pointer and move right pointer by 1 on left. 
+# If mid is 1 move mid pointer by 1 to right. If mid is 0 replace it with left and move both left and pointers 
+class Solution:
+    def sortColors(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        leftPointer = 0
+        midPointer = 0
+        rightPointer = len(nums)-1
+
+        while midPointer <=rightPointer:
+            numleft = nums[leftPointer]
+            numRight = nums[rightPointer]
+            numMid = nums[midPointer]
+            if numMid == 2:
+                nums[midPointer] = numRight
+                nums[rightPointer] = numMid
+                rightPointer-=1
+            elif numMid == 1:
+                midPointer+=1
+            else:
+                nums[midPointer] = numleft
+                nums[leftPointer] = numMid
+                leftPointer+=1
+                midPointer+=1
+
+
+
diff --git a/Problem2.py b/Problem2.py
new file mode 100644
index 00000000..e0f16969
--- /dev/null
+++ b/Problem2.py
@@ -0,0 +1,42 @@
+#3Sum
+# Time complexity -> On^2)
+# Space complexity -> O1
+# Sort the nums, the for each element use 2sum logic to find the numbers which will be equal to -1*element basically then -vs of the element will become
+# the target for the 2sum
+
+# from typing import List
+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        result=[]
+        nums.sort()
+        for i in range(0,len(nums)):
+            num1 = nums[i]
+            if i>0 and nums[i]==nums[i-1]:
+                continue
+            self.twoSum(nums,i+1,num1*-1,result)
+        return result
+        
+    
+    def twoSum(self,nums,startingIndex,target,result):
+        # subResult = set()
+        leftPointer = startingIndex
+        rightPointer = len(nums)-1
+        while leftPointer < rightPointer:
+            sum = nums[leftPointer] + nums[rightPointer]
+
+            if sum == target:
+                result.append([-1*target, nums[leftPointer],  nums[rightPointer]])
+                while leftPointer < rightPointer and nums[leftPointer] == nums[leftPointer+1]:
+                    leftPointer+=1
+                while leftPointer < rightPointer and nums[rightPointer] == nums[rightPointer-1]:
+                    rightPointer-=1
+                leftPointer+=1
+                rightPointer-=1
+
+            elif sum < target:
+                leftPointer+=1
+            else:
+                rightPointer-=1
+        
+        # return subResult
+
diff --git a/Problem3.py b/Problem3.py
new file mode 100644
index 00000000..52110354
--- /dev/null
+++ b/Problem3.py
@@ -0,0 +1,26 @@
+# Container With Most Water
+#time complexity -> On
+#space complexity -> O1
+#logic, start from both ends, get the total volume, then whichever wall is smaller move the pointers inside from that wall as 
+# small wall is our limiting factor, idea is to find the biggest wall that can generate the most volume
+
+from typing import List
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        left = 0
+        right = len(height)-1
+        result = 0 
+        while left < right:
+            leftWall = height[left]
+            rightWall = height[right]
+            result = max(result,min(leftWall,rightWall)*(right-left))
+
+            if leftWall<rightWall:
+                left+=1
+            else:
+                right-=1
+        return result
+
+sol = Solution()
+
+print(sol.maxArea([3,2,3]))
\ No newline at end of file