Done Two-Pointers-2

merge-sorted-array.py

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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
+
+
+   Approach : Maintain 2 pointers from end of array, and one right pointer at end of 1st array
+               swap the largest element at end of array and move the right pointer inside
+'''
+
+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.
+        """
+        p1 = m-1
+        p2 = n-1
+        r = (m+n)-1
+        print(p1,p2,r)
+        while p1 >= 0 and p2 >= 0:
+            if nums1[p1] >= nums2[p2]:
+                nums1[r] = nums1[p1]
+                r -= 1
+                p1 -= 1
+            else:
+                nums1[r] = nums2[p2]
+                r -= 1
+                p2 -= 1
+        while p2 >= 0:
+            nums1[r] = nums2[p2]
+            r -= 1
+            p2 -= 1

remove-duplicates-from-sorted-array-ii.py

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+''' Time Complexity : O(2n) 
+    Space Complexity : O(1) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+
+   Approach : Maintain two pointers, one to overwirte and one to scan, update the array where count <= k
+'''
+class Solution:
+    def removeDuplicates(self, nums: List[int]) -> int:
+        k = 2
+        p1 = 0
+        i = 0
+        count = 0
+        while i < len(nums):
+            if i != 0 and nums[i] == nums[i-1]:
+                count += 1
+            else:
+                count = 1
+
+            if count <= k:
+                nums[p1] = nums[i]
+                p1 += 1
+
+            i += 1
+        return p1
+
+
+
+

search-a-2d-matrix-ii.py

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+''' Time Complexity : O(m+n) 
+    Space Complexity : O(1) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+
+   Approach : Start from Top-Right corner, if target is less then go to left, else go to downwards
+'''
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        rows,cols = len(matrix), len(matrix[0])
+        r, c = 0, cols-1
+        while 0<=r<rows and 0<=c<cols:
+            if target == matrix[r][c]:
+                return True
+            elif target < matrix[r][c]:
+                c -= 1
+            else:
+                r += 1
+        return False