From d86167a0bf061682377f09009b2d3dc4e6197661 Mon Sep 17 00:00:00 2001
From: Purva <ptand@uic.edu>
Date: Sun, 30 Nov 2025 09:57:12 -0600
Subject: [PATCH] done

---
 problem1.py | 21 +++++++++++++++++++++
 problem2.py | 35 +++++++++++++++++++++++++++++++++++
 problem3.py | 27 +++++++++++++++++++++++++++
 3 files changed, 83 insertions(+)
 create mode 100644 problem1.py
 create mode 100644 problem2.py
 create mode 100644 problem3.py

diff --git a/problem1.py b/problem1.py
new file mode 100644
index 00000000..6739c498
--- /dev/null
+++ b/problem1.py
@@ -0,0 +1,21 @@
+"""
+we use two pointers, i and left, such that i traverses the array an slow pointer remains on the last valid element, and later replaces another valid
+element found in future. TC is o(n) and sc is o(1)"""
+
+class Solution(object):
+    def removeDuplicates(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        slow = 1
+        count = 1
+        for i in range(1, len(nums)):
+            if nums[i] == nums[i-1]:
+                count += 1
+            else:
+                count = 1
+            if count <= 2:
+                nums[slow] = nums[i] #overwrite the number we want to keep at the slow pointer
+                slow += 1
+        return slow
\ No newline at end of file
diff --git a/problem2.py b/problem2.py
new file mode 100644
index 00000000..b17fe50d
--- /dev/null
+++ b/problem2.py
@@ -0,0 +1,35 @@
+"""
+We use two pointers from the end of the lists and maintain a last pointer as a third one to swap the elements greater than one at m and n position to the last
+place of the nums1 array. The remaining elements, if any are placed in the last element as a loop
+TC is o(m+n) and SC is o(!)"""
+
+
+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.
+        """
+        last = m + n - 1
+        while m > 0 and n > 0:
+            if nums1[m - 1] > nums2[n - 1]:
+                nums1[last] = nums1[m - 1]
+                m -= 1
+            else:
+                nums1[last] = nums2[n - 1]
+                n -= 1
+            last -= 1
+
+        #if there are still elements left in the nums 2 array
+        while n > 0:
+            nums1[last] = nums2[n-1]
+            n -= 1
+            last -= 1
+
+        
+
+
+        
\ No newline at end of file
diff --git a/problem3.py b/problem3.py
new file mode 100644
index 00000000..dcdfb3c1
--- /dev/null
+++ b/problem3.py
@@ -0,0 +1,27 @@
+"""
+we use boundary condtions and check that if we move left and down as two options, we can reach the answer in o(m+n) time, hence we traverse the array in such
+way that we move left and down
+TC is o(m+n) and SC is o(1)"""
+
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+        m = len(matrix) #rows
+        n = len(matrix[0])
+        r, c = 0, n - 1
+        while r < m and c >= 0:
+            if matrix[r][c] == target:
+                return True
+            elif matrix[r][c] > target:
+                c -= 1
+            else:
+                r += 1
+        return False
+
+
+