diff --git a/238. Product of Array Except Self.py b/238. Product of Array Except Self.py
new file mode 100644
index 00000000..4c892811
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+++ b/238. Product of Array Except Self.py	
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+# Time: O(n^2)
+# Space: O(1)
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        result = [1] * len(nums)
+        for i in range(len(result)):
+            for j in range(len(nums)):
+                if j != i:
+                    result[i] *= nums[j]
+        return result
+    
+# Time: O(3n) = O(n)
+# Space: O(2n) for 2 seperate arrays
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        left_running = [1] * len(nums)
+        right_running = [1] * len(nums)
+        result =[0] * len(nums)
+
+        for i in range(1,len(nums)):
+            left_running[i] = nums[i-1] * left_running[i - 1]
+        for j in range(len(nums) - 2, -1, -1):
+            right_running[j] = nums[j+1] * right_running[j + 1]
+        for k in range(len(nums)):
+            result[k] = left_running[k] * right_running[k]
+
+        return result
+
+
+# Time Complexity : O(n)
+# Space Complexity : O(1) leaving the output array
+# Did this code successfully run on Leetcode : Yes
+# Three line explanation of solution in plain english
+    # - Form the result array with everything as 1
+    # - Do the left pass with running product
+    # - Then do the right pass muliplying with the current left pass elements of the array
+
+# Your code here along with comments explaining your approach
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        result =[1] * len(nums)
+
+        for i in range(1,len(nums)):
+            result[i] = nums[i-1] * result[i - 1]
+        right_running = 1
+        for j in range(len(nums) - 1, -1, -1):
+            result[j] *= right_running
+            right_running *= nums[j]
+
+        return result
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diff --git a/498. Diagonal Traverse.py b/498. Diagonal Traverse.py
new file mode 100644
index 00000000..ddf9c7d4
--- /dev/null
+++ b/498. Diagonal Traverse.py	
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+#  Time Complexity : O (m*n)
+#  Space Complexity : O(1) except the output array
+#  Did this code successfully run on Leetcode : Yes
+#  Three line explanation of solution in plain english
+    # strating from [0,0] check from which side we want to move
+    # depending on which side we are moving check if we are going out of bound
+    # if we are moving out of bound change i and j accordingly if not put a normal pass
+    # write j == (n - 1) before of elif i == 0 to handle the edge case when i is 0 at j == len(n-1) and then we increment and then go out of bound
+
+#  Your code here along with comments explaining your approach
+
+class Solution:
+    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
+        m,n = len(mat), len(mat[0])
+        i = 0
+        j = 0
+        direction = True
+        result = []
+        while len(result) < m * n:
+                result.append(mat[i][j])
+                # when we are going upwards i.e direction True
+                if direction:
+                    # writin this first to handle the edge case when i and j both are zero
+                    if j == (n - 1):
+                        direction = False
+                        i += 1
+                    # going out from top
+                    elif i == 0:
+                        direction = False
+                        j += 1
+                    # for normal pass
+                    else:
+                        i -= 1
+                        j += 1
+                # when we are going downwards
+                else:
+                    # going out from botom
+                    if i == (m - 1):
+                        direction = True
+                        j += 1
+                    # going out from left
+                    elif j == 0:
+                        direction = True
+                        i += 1
+                    # for normal pass
+                    else:
+                        i += 1
+                        j -= 1
+                
+        return result
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diff --git a/54. Spiral Marix.py b/54. Spiral Marix.py
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index 00000000..dd718032
--- /dev/null
+++ b/54. Spiral Marix.py	
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+# Time: O(m*n)
+# space:O(1)
+# Que => Return an array by iterating through the matrix spirally
+# have 4 variables to control the edges then loop while left doesn't cross right and top doesn't cross bottom
+# then as we are updating the base variables keep in mind to check the conditions before performing anything
+
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        m = len(matrix)
+        n = len(matrix[0])
+        top = 0
+        right = n - 1
+        bottom = m - 1
+        left = 0
+
+        result = []
+        while left <= right and top <= bottom:
+            # Top row
+            for i in range(left,right+1):
+                result.append(matrix[top][i])
+            top += 1
+            # We are changing the base case that is why we check if it is changed again because we are changing top in previous loop
+            if top <= bottom:
+                # Right most row
+                for i in range(top,bottom+1):
+                    result.append(matrix[i][right])
+                right -= 1
+
+            if left <= right and top <= bottom:
+                # Bottom row
+                for i in range(right,left-1,-1):
+                    result.append(matrix[bottom][i])
+                bottom -= 1
+            if left <= right and top <= bottom:
+                # Left row
+                for i in range(bottom,top-1,-1):
+                    result.append(matrix[i][left])
+                left += 1
+        return result
+            
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