solution two pointers 2

mergesorted.py

@@ -0,0 +1,25 @@
+# // Time Complexity : O(m+n)
+# // Space Complexity : O(1)
+# // Did this code successfully run on Leetcode : Yes
+
+class Solution(object):
+    def merge(self, nums1, m, nums2, n):
+        """
+        :type nums1: List[int]
+        :type m: int
+        :type nums2: List[int]
+        :type n: int
+        :rtype: None Do not return anything, modify nums1 in-place instead.
+        """
+        p1=m-1
+        p2=n-1
+
+        for i in range(m+n-1,-1,-1):
+            if p2<0:
+                break
+            if p1>=0 and nums1[p1]>nums2[p2]:
+                nums1[i]=nums1[p1]
+                p1-=1
+            else:
+                nums1[i]=nums2[p2]
+                p2-=1 

removeduplicates.py

@@ -0,0 +1,28 @@
+# // Time Complexity : O(n)
+# // Space Complexity : O(1)
+# // Did this code successfully run on Leetcode : Yes
+
+class Solution(object):
+    def removeDuplicates(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        p1=0
+        p2=1
+        count=1
+        while (p2<len(nums)):
+            if nums[p2]==nums[p1]:
+                if count>=2:
+                    p2+=1
+                    continue
+                else:
+                    count+=1
+            else:
+                count=1            
+            p1+=1
+            nums[p1]=nums[p2]
+            p2+=1
+        return p1+1
+
+

search2dmatrix.py

@@ -0,0 +1,29 @@
+# // Time Complexity : O(m+n)
+# // Space Complexity : O(1)
+# // Did this code successfully run on Leetcode : Yes
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+        m=len(matrix)
+        n=len(matrix[0])
+
+        diag =  max(m,n)
+        row=0
+        col=n-1
+        while 0<=row<m and 0<=col<n:
+            if target<matrix[row][col]:
+                col-=1
+            elif target>matrix[row][col]:
+                row+=1
+            else:
+                return True
+
+        return False
+
+
+