Two pointers done

Problem1.py

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+#Remove Duplicates from Sorted Array II
+
+# Time comlexity -> O(n)
+# Space complexity -> O1
+# Logic -> maitain 2 pointers, slow and fast, fast will take care of occurences, for 1st 2 ocurences it will update the slow with fast value 
+# once occurences move above 2 it's just on keeps moving ahead to find new number that needs to be moved in place of slow pointer's value
+
+from typing import List
+class Solution:
+    def removeDuplicates(self, nums: List[int]) -> int:
+        slow = 0
+        fast = 0
+        length = len(nums)
+        count=0
+        k = 2
+        for fast in range(0,length):
+            if (fast != 0 and nums[fast]==nums[fast-1]):
+                count+=1
+            else:
+                count = 1
+
+            if count <=k:
+                nums[slow]=nums[fast]
+                slow+=1
+
+        return slow
+

Problem2.py

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+#Merge Sorted Array
+# Time complexity O(m+n)
+# Space complexity -> O1
+# Logic -> Start from updating the biggest elemnt in the correct place this way when we overwrite the bigger items 
+# with the smaller items bigger items won't be lost as they will already be copied in the correct position 
+
+class Solution:
+    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
+        """
+        Do not return anything, modify nums1 in-place instead.
+        """
+        smallListPointer = n-1
+        bigListpointer = m-1
+        finalListPOinter = len(nums1)-1
+
+        while smallListPointer >=0 and bigListpointer>=0:
+            nums1item = nums1[bigListpointer]
+            nums2Item = nums2[smallListPointer]
+            if nums1item > nums2Item:
+                nums1[finalListPOinter] = nums1item
+                bigListpointer-=1
+            else:
+                nums1[finalListPOinter] = nums2Item
+                smallListPointer-=1
+            finalListPOinter-=1
+
+        if bigListpointer == -1:
+            while finalListPOinter >= 0:
+                nums1[finalListPOinter] = nums2[smallListPointer]
+                finalListPOinter-=1
+                smallListPointer-=1
+
+

Problem3.py

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+#Search a 2D Matrix II
+
+# Time complexity O(m+n) where m is rows and n is columns
+# Space complexity -> O1
+# Logic -> start from last row 1st column. based on the target comparison we can move up if target is lower than cuurrnt element if bigger move right
+# keep on moving until you find element or you cross the matrix boundary which means target is not there
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        startRow = len(matrix)-1
+        startColumn = 0
+        maxColumns = len(matrix[0]) 
+
+        while startRow>=0 and startColumn<maxColumns:
+            # if current number is target return true
+            currentItem = matrix[startRow][startColumn]
+            if currentItem == target:
+                return True
+
+            #if target is smaller than current matrix move up because all other items in row will ve big
+            if target < currentItem:
+                startRow-=1
+
+            #if target is greate than current matrix move right because all other items in column above will be smaller
+            if target > currentItem:
+                startColumn+=1
+
+        return False
+
+