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+# Problem 1 : https://leetcode.com/problems/product-of-array-except-self/
+# Time complexity : O(n)
+# Space Complexity : O(1)
+
+class Solution(object):
+    def productExceptSelf(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        arr = [1]* len(nums)
+
+        ##running product
+        ## creating a running product from right
+        ## and then going from left to right and multiplying to the same product
+        ## because the product will be all the elements on the left and right
+        ## finally returning the arr
+        rp =1
+        for i in range(len(nums)):
+            arr[i] = arr[i] * rp
+            rp = rp *nums[i]
+            
+        rp =1
+        for i in range(len(nums)-1,-1,-1):
+            arr[i] = arr[i] * rp
+            rp = rp * nums[i]
+              
+        return arr
+
+
+# Problem 2: https://leetcode.com/problems/diagonal-traverse/
+# Time complexity : O(n)
+# Space Complexity : O(1)
+
+class Solution(object):
+    def findDiagonalOrder(self, matrix):
+        """
+        :type mat: List[List[int]]
+        :rtype: List[int]
+        """
+        rows = len(matrix)
+        cols = len(matrix[0])
+
+        ## we can either go up or down, introducing a flag to store the condition
+        ## and then identifying what will be the next r.c
+        ## 
+
+        r,c = 0,0
+        direction = True ##up
+        res = []
+        while r<rows and c<cols:
+            res.append(matrix[r][c])
+            ## find out next cell, in case direction is up
+            if direction:
+                ## to check if row becomes out of bound
+                if r ==0 and c!= cols-1 :
+                    c+=1
+                    direction = False
+                ## if col becomes out of bound    
+                elif c == cols-1:
+                    r+=1
+                    direction = False
+                ## if none of of the above condition, then just do this for next element
+                else:
+                    r-=1
+                    c+=1            
+            else:
+                ## base case to check out of bounds
+                if c ==0 and r!= rows-1 :
+                    r+=1
+                    direction = True
+                ## base case to check row out of bound
+                elif r == rows-1:
+                    c+=1
+                    direction = True
+                ## if none of the condition just get the next element
+                else:
+                    c-=1
+                    r+=1
+        return res 
+                
+# Problem 3 : https://leetcode.com/problems/spiral-matrix/description/
+# Time complexity : O(n)
+# Space Complexity : O(1)
+
+
+class Solution(object):
+    def spiralOrder(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: List[int]
+        """
+        rows = len(matrix)
+        cols = len(matrix[0])
+
+        ## four pointer for squeezing the array
+        ## we can squeeze the array for next iteration
+        top,left =0,0
+        bottom = rows-1
+        right = cols-1
+
+        res = []
+        
+        while left<=right and top<=bottom:
+            ## printing rows
+            for j in range(left,right+1):
+                res.append(matrix[top][j])
+            top+=1
+            ## printing cols from last onwards
+            ## checking the while loop base condition because it is changing
+            if top<=bottom:
+                for j in range(top,bottom+1):
+                    res.append(matrix[j][right])
+                right -=1
+            ## print rows from bottom
+            ## checking the while loop base condition because it is changing
+            if top<=bottom:
+                for j in range(right,left-1,-1):
+                    res.append(matrix[bottom][j])
+                bottom-=1
+            ## printing cols from starting
+            ## checking the while loop base condition because it is changing
+            if left<=right:
+                for j in range (bottom,top-1,-1):
+                    res.append(matrix[j][left])
+                left +=1
+        
+        return res