diff --git a/diagonal_traverse.py b/diagonal_traverse.py
new file mode 100644
index 00000000..f2e57676
--- /dev/null
+++ b/diagonal_traverse.py
@@ -0,0 +1,55 @@
+class Solution:
+    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
+        rowLen = len(mat)
+        if rowLen == 0:
+            return []
+        colLen = len(mat[0])
+        
+        res = []              # Result list to store elements in diagonal order
+        row = 0               # Start at the top-left corner
+        col = 0
+        directionUp = True    # Tracks the current direction of traversal (up-right or down-left)
+        
+        # Continue until we’ve processed all elements
+        while row < rowLen and col < colLen:
+            
+            if directionUp:
+                # --- Moving UP-RIGHT direction ---
+                # Keep going up (row--) and right (col++) while staying within matrix bounds
+                while row > 0 and col < (colLen - 1):
+                    res.append(mat[row][col])
+                    row -= 1
+                    col += 1
+                
+                # Append the last element before switching direction
+                res.append(mat[row][col])
+                
+                # If we reached the last column, we can only move down to next row
+                if col == (colLen - 1):
+                    row += 1
+                else:
+                    # Otherwise, move right to the next column
+                    col += 1
+            
+            else:
+                # --- Moving DOWN-LEFT direction ---
+                # Keep going down (row++) and left (col--) while staying within matrix bounds
+                while col > 0 and row < (rowLen - 1):
+                    res.append(mat[row][col])
+                    row += 1
+                    col -= 1
+                
+                # Append the last element before switching direction
+                res.append(mat[row][col])
+                
+                # If we reached the last row, we can only move right to next column
+                if row == (rowLen - 1):
+                    col += 1
+                else:
+                    # Otherwise, move down to the next row
+                    row += 1
+            
+            # Flip direction for the next diagonal
+            directionUp = not directionUp
+        
+        return res
diff --git a/product_array_except_self.py b/product_array_except_self.py
new file mode 100644
index 00000000..a2662d4b
--- /dev/null
+++ b/product_array_except_self.py
@@ -0,0 +1,48 @@
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        
+        # Step 1: Initialize result array with 1s.
+        # 'res[i]' will eventually hold the product of all elements
+        # to the LEFT of index i (initially all 1s as placeholders).
+        res = [1] * n
+
+        # Step 2: Compute prefix products (products of all elements to the left of i)
+        # Example: nums = [1, 2, 3, 4]
+        # After this loop, res = [1, 1, 2, 6]
+        # Explanation:
+        # res[1] = res[0] * nums[0] = 1 * 1 = 1
+        # res[2] = res[1] * nums[1] = 1 * 2 = 2
+        # res[3] = res[2] * nums[2] = 2 * 3 = 6
+        for i in range(1, n):
+            res[i] = res[i - 1] * nums[i - 1]
+
+        # Step 3: Initialize variable 'product' to hold suffix product
+        # (product of elements to the right of the current index)
+        # Start with the last element since nothing is to its right.
+        product = nums[-1]
+
+        # Step 4: Traverse array backward (from second last element to first)
+        # Multiply each res[i] (which currently has prefix product)
+        # by 'product' (which represents suffix product).
+        # Then, update 'product' by multiplying it with nums[i].
+        #
+        # Example trace (nums = [1, 2, 3, 4]):
+        # Initially product = 4
+        #
+        # i = 2 → res[2] = res[2] * 4 = 2 * 4 = 8
+        #          product = product * nums[2] = 4 * 3 = 12
+        #
+        # i = 1 → res[1] = res[1] * 12 = 1 * 12 = 12
+        #          product = product * nums[1] = 12 * 2 = 24
+        #
+        # i = 0 → res[0] = res[0] * 24 = 1 * 24 = 24
+        #          product = product * nums[0] = 24 * 1 = 24
+        #
+        # Final res = [24, 12, 8, 6]
+        for i in range(n - 2, -1, -1):
+            res[i] = res[i] * product
+            product = product * nums[i]
+
+        # Step 5: Return the final result array
+        return res
\ No newline at end of file
diff --git a/spiral_matrix.py b/spiral_matrix.py
new file mode 100644
index 00000000..e530e6a3
--- /dev/null
+++ b/spiral_matrix.py
@@ -0,0 +1,32 @@
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        rows = len(matrix)
+        if rows == 0:
+            return []
+        cols = len(matrix[0])
+        return self.helper(matrix, 0, rows - 1, 0, cols - 1)
+    
+    def helper(self, matrix: List[List[int]], t, b, l, r) -> List[int]:
+        if t > b or l > r:
+            return []
+        # Single cell left
+        if t == b and l == r:
+            return [matrix[t][l]]
+        
+        res = []
+        # top row
+        for i in range(l, r + 1):
+            res.append(matrix[t][i])
+        # right column
+        for i in range(t + 1, b + 1):
+            res.append(matrix[i][r])
+        # bottom row (if more than one row)
+        if t < b:
+            for i in range(r - 1, l - 1, -1):
+                res.append(matrix[b][i])
+        # left column (if more than one column)
+        if l < r:
+            for i in range(b - 1, t, -1):
+                res.append(matrix[i][l])
+        
+        return res + self.helper(matrix, t + 1, b - 1, l + 1, r - 1)