From 5f5e2f1c447b292d9da7bb6ad9112e936dea6a0e Mon Sep 17 00:00:00 2001
From: Purva <ptand@uic.edu>
Date: Wed, 26 Nov 2025 02:08:47 -0600
Subject: [PATCH] done

---
 problem1.py | 31 +++++++++++++++++++++++++++++++
 problem2.py | 40 ++++++++++++++++++++++++++++++++++++++++
 problem3.py | 41 +++++++++++++++++++++++++++++++++++++++++
 3 files changed, 112 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..46f1f55b
--- /dev/null
+++ b/problem1.py
@@ -0,0 +1,31 @@
+"""
+We use a two pass approach to overcome the exponential brute force approach with i and j pointers, we calculate the product of elements to the left of number and to
+the right of the product, this way we can multiply the running product and get our answer
+TC is o(n) for traversing the whole array and SC is o(1) (resulant array doesnt count towards space)
+"""
+
+
+class Solution(object):
+    def productExceptSelf(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        #do a two pass approach, one left pass and one right pass
+
+        n = len(nums)
+        res = [0] * n
+        rp = 1
+        res[0] = 1
+        #left pass
+        for i in range(1,n):
+            rp = rp * nums[i-1]
+            res[i] = rp
+
+        rp = 1 #reset for right pass
+        for i in range(n-2, -1, -1):
+            rp = rp * nums[i + 1]
+            res[i] = res[i] * rp
+
+        return res
+
diff --git a/problem2.py b/problem2.py
new file mode 100644
index 00000000..11eac010
--- /dev/null
+++ b/problem2.py
@@ -0,0 +1,40 @@
+"""
+The idea is to see which conditions we hit at each pass and update the rows and cols accordingly
+the TC is o(m*n) and SC is O(1) since we build on result array and not on aux array
+"""
+class Solution(object):
+    def findDiagonalOrder(self, mat):
+        """
+        :type mat: List[List[int]]
+        :rtype: List[int]
+        """
+        m = len(mat)
+        n = len(mat[0]) #column
+        res = [0] * (m*n)
+        r, c = 0,0
+        flag = True
+
+        for i in range(m*n):
+            res[i] = mat[r][c]
+            if flag: #up right
+                if c == n - 1: #last element in the top right, increment row
+                    r += 1
+                    flag = False #go down
+                elif r == 0: #its in the first row, increment row then flag hanges
+                    flag = False
+                    c += 1 #shift the column
+                else: #else, update the conditions to move in a up-right direction (column-increase, row decrease)
+                    c += 1
+                    r -= 1
+            else:
+                if r == m - 1: #if it is the last row, increment column and switch flag 
+                    flag = True
+                    c += 1
+                elif c == 0: #if its the the first column, increment the row
+                    flag = True
+                    r += 1
+                else:
+                    r += 1
+                    c -= 1
+        return res
+
diff --git a/problem3.py b/problem3.py
new file mode 100644
index 00000000..e493f422
--- /dev/null
+++ b/problem3.py
@@ -0,0 +1,41 @@
+"""
+We use the boundaries method and follow the path that we would usually do while traversing on a notebook, and then update the pointers
+TC is o(m*n) and SC is o(1) since its directly stored in resultant space
+"""
+
+
+class Solution(object):
+    def spiralOrder(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: List[int]
+        """
+        m = len(matrix)
+        n = len(matrix[0])
+        top = 0
+        bottom = m - 1
+        right = n - 1
+        left = 0
+        res = []
+        while top <= bottom and left <= right:
+            for i in range(left, right + 1): #right + 1 so its included
+                res.append(matrix[top][i])
+            top += 1
+
+            for i in range(top, bottom + 1):
+                res.append(matrix[i][right])
+            right -= 1
+
+            if top <= bottom: #we have to update the pointers, so that nothing is left nor repeated
+                for i in range(right, left - 1, -1): #you dont want to repeat left row again, so left - 1
+                    res.append(matrix[bottom][i])
+                bottom -= 1
+
+            if left <= right:
+                for i in range(bottom, top - 1, -1):
+                    res.append(matrix[i][left])
+                left += 1
+        return res
+
+
+        
\ No newline at end of file