Completed Two-Pointers-2

merge_sorted_array.py

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+# https://leetcode.com/problems/merge-sorted-array/
+
+# Using two pointers started updenting the elements from second array into the first array. While doing that, compare the last element of the first array with last element of second array and update accordinlgy
+
+# Time Complexity- O((m+n) Space Complexity- O(1)
+class Solution:
+    def merge(self, nums1: list[int], m: int, nums2: list[int], n: int) -> None:
+        # finding the last index of the merge array
+        last_index= m + n - 1
+
+        # merge from last index
+        while m > 0 and n > 0:
+            # checking whether the last element of first array is greater than the last element of second array 
+            if nums1[m - 1] > nums2[n - 1]:
+                # update the last index of the nums1 with greater element
+                nums1[last_index] = nums1[m - 1]
+                # decrement the index 
+                m -= 1
+            # when the last element of second array is greater than the last element of first array
+            else:
+                nums1[last_index] = nums2[n - 1]
+                # decrement the index
+                n -= 1
+            # decrement the last index 
+            last_index -= 1
+
+        # update the remaining elements from second array into the first array
+        # check if n is greater than 0 after decrementing
+        while n > 0:
+            # update with the elements
+            nums1[last_index] = nums2[n-1]
+            # decrement the pointer
+            n = n - 1
+            last_index = last_index - 1
+
+solution = Solution()
+print(solution.merge([1,2,3,0,0,0],3,[2,5,6],3))
+

remove_duplicate.py

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+# https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
+
+# Time Complexity- O(n) Space Complexity- O(1)
+class Solution:
+    def removeDuplicates(self, nums: list[int]) -> int:
+        k = 2
+        old = 0
+        new = 0
+        count = 0
+        while new < len(nums):
+            # check whether the previous number is same as the current
+            if new != 0 and nums[new] == nums[new - 1]:
+                # increase the count
+                count += 1
+            else:
+                # if the first occurance
+                count = 1
+            # check the count is in given limit 
+            if count <= k:
+                # update the list with old 
+                nums[old] = nums[new]
+                # increment the old 
+                old += 1
+            # increment the new
+            new += 1
+        # return the old 
+        return old
+
+
+solution = Solution()
+print(solution.removeDuplicates([1,1,1,2,2,3]))
+# nums = [1,1,1,2,2,3]
+# output = 5

search_a_2d_matrix.py

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+# https://leetcode.com/problems/search-a-2d-matrix-ii/
+
+# To find the target element, start checking from the top left corner of the matrix. If the target is less than the current element decrease the column as the matrix is sorted. and if it's greater than the current element increase the row.
+# Time Complexity- O(m + n) Space Complexity- O(1)
+class Solution:
+    def searchMatrix(self, matrix: list[list[int]], target: int) -> bool:
+        m = len(matrix)
+        n = len(matrix[0])
+        # starting from the top-right corner
+        r, c = 0 , n - 1
+
+        while r < m and c >= 0:
+            # if the target is present at this coordinate
+            if matrix[r][c] == target:
+                return True
+            # if the target is less than the element
+            elif matrix[r][c] > target:
+                # decrement the column
+                c -= 1
+            else:
+                # increment the row 
+                r += 1
+
+        return False
+
+solution = Solution()
+print(solution.searchMatrix([[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], 19))
+
+# target = 5
+# output = true