From 00df512ca7954677538df3c6a0135e7f6255a701 Mon Sep 17 00:00:00 2001
From: agk-s30 <anikesh.kamath@hotmail.com>
Date: Thu, 13 Aug 2026 19:52:50 -0700
Subject: [PATCH 1/3] Add merge function for merging sorted arrays

Implement merge function to combine two sorted arrays in-place.
---
 Problem_2.py | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)
 create mode 100644 Problem_2.py

diff --git a/Problem_2.py b/Problem_2.py
new file mode 100644
index 00000000..378b0e98
--- /dev/null
+++ b/Problem_2.py
@@ -0,0 +1,23 @@
+# https://leetcode.com/problems/merge-sorted-array/description/
+
+# TC: O(n + m)
+# SP: O(1)
+# Explanation: Place two points at the end of each nums array, and another pointer at the nums1;
+# keep comparing nums1 vs nums2 and insert the higher value, and in the end you will get the fully sorted array in nums1
+
+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
+        for p in range(n + m - 1, -1, -1):
+            if p2 < 0:
+                break
+            if p1 >= 0 and nums1[p1] > nums2[p2]:
+                nums1[p] = nums1[p1]
+                p1 -= 1
+            else:
+                nums1[p] = nums2[p2]
+                p2 -= 1

From f8ed5d4b8c8deeca15f07266fe3a549c7eb8ab48 Mon Sep 17 00:00:00 2001
From: agk-s30 <anikesh.kamath@hotmail.com>
Date: Tue, 18 Aug 2026 21:33:25 -0700
Subject: [PATCH 2/3] Add solution for removing duplicates from sorted array

Implement a solution to remove duplicates from a sorted array, allowing at most two occurrences of each element.
---
 Problem_1.py | 29 +++++++++++++++++++++++++++++
 1 file changed, 29 insertions(+)
 create mode 100644 Problem_1.py

diff --git a/Problem_1.py b/Problem_1.py
new file mode 100644
index 00000000..b606624b
--- /dev/null
+++ b/Problem_1.py
@@ -0,0 +1,29 @@
+# https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/description/
+
+# Time complexity: O(n) 
+# Space complexity: O(1)
+# Explanation: Maintain two points, one to check count and one to act as the current index being written to. Keep iterating the elements and then return the 2nd pointer value.
+
+class Solution:
+    def removeDuplicates(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+        
+        i = 1
+        j = 1
+        count = 1
+
+        while i < len(nums):
+            if nums[i] == nums[i-1]:
+                count += 1
+                if count > 2:
+                    i += 1
+                    continue
+            else:
+                count = 1
+            nums[j] = nums[i]
+            j += 1
+            i += 1
+        
+        del nums[j:]
+        return j + 1

From 9de7da18b69a30df9a042c34ea4cbc1e6aec2d3a Mon Sep 17 00:00:00 2001
From: agk-s30 <anikesh.kamath@hotmail.com>
Date: Tue, 18 Aug 2026 21:35:04 -0700
Subject: [PATCH 3/3] Add searchMatrix function for 2D matrix search

Implement searchMatrix method to find target in 2D matrix.
---
 Problem_3.py | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)
 create mode 100644 Problem_3.py

diff --git a/Problem_3.py b/Problem_3.py
new file mode 100644
index 00000000..2573d6cd
--- /dev/null
+++ b/Problem_3.py
@@ -0,0 +1,20 @@
+# https://leetcode.com/problems/search-a-2d-matrix-ii/description/
+
+# Time complexity: O(n) 
+# Space complexity: O(1)
+# Explanation: Start from the bottom left. If the curr value is greater than target, shift up a row. If smaller, then shift right. Finally return target.
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        m, n = len(matrix), len(matrix[0])
+        i, j = m - 1, 0
+
+        while i >=0 and j < n:
+            if matrix[i][j] > target:
+                i -= 1
+            elif matrix[i][j] < target:
+                j += 1
+            else:
+                return True
+        
+        return False