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+Problem1 (https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)
+Time Complexity : O(n)
+Space Complexity : O(1)
+
+
+## trying to do in place here again
+## we can iterate over the array and start looking for the element, 
+## Once we look at the element go the index meaning if the element is 6 go to index 6 
+## and convert it to negative. in the second pass look for any element that is +ve indicating 
+## those are not present in arr, add them in list and return
+class Solution(object):
+    def findDisappearedNumbers(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        for i in range(len(nums)):
+            
+            if nums[abs(nums[i])-1] > 0:
+                nums[abs(nums[i])-1] *= -1
+        res = []
+        for i in range(len(nums)):
+            if nums[i] >0:
+                res.append(i+1)
+        return res
+
+
+Problem 2 https://www.geeksforgeeks.org/maximum-and-minimum-in-an-array/
+Time Complexity: O(n)
+Space Complexity: O(1)
+
+
+
+class Solution:
+
+    def find(self, nums):
+        n = len(nums)
+        i = 0
+        ## first pass
+        if n % 2 == 0:
+            if nums[0] < nums[1]:
+                min_val = nums[0]
+                max_val = nums[1]
+            else:
+                min_val = nums[1]
+                max_val = nums[0]
+            i = 2
+        else:
+            min_val = max_val = nums[0]
+            i = 1
+        ## for all the pairs
+        while i < n - 1:
+            if nums[i] < nums[i + 1]:
+                min_val = min(min_val, nums[i])
+                max_val = max(max_val, nums[i + 1])
+            else:
+                min_val = min(min_val, nums[i + 1])
+                max_val = max(max_val, nums[i])
+            i += 2
+
+        return [min_val, max_val]
+
+
+Problem 3: https://leetcode.com/problems/game-of-life/
+Time Complexity: O(mn)
+Space Complexity: O()
+
+class Solution(object):
+    def gameOfLife(self, matrix):
+        """
+        :type board: List[List[int]]
+        :rtype: None Do not return anything, modify board in-place instead.
+        """
+        ## we want to do in place and hence cannot change the state right away
+        ## we can start assigning different states and then change them in second pass
+        ## assigning 2
+        ## less than 2 1->0
+        ## more than 3 1->0
+        ## it will remain 1 no change required
+        ## ==2 or ==3 1->1
+        ## assigning 3
+        ## ==3 0->1
+        def count(i,j):
+            count =0 
+            for dr,dc in [(-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)]:
+                row = i+dr
+                col = j+dc
+                if 0<=row<rows and 0<=col<cols :
+                    if matrix[row][col] ==1 or  matrix[row][col] ==2:
+                        count +=1
+            return count
+
+        rows = len(matrix)
+        cols = len(matrix[0])
+
+        for i in range(rows):
+            for j in range(cols):
+                c = count(i,j)
+                val = matrix[i][j]
+                if val ==1 and (c<2 or c>3):
+                    matrix[i][j] = 2
+                if val ==0 and c==3:
+                    matrix[i][j] =3
+        
+        for i in range(rows):
+            for j in range(cols):
+                if matrix[i][j] ==2:
+                    matrix[i][j] =0
+                elif matrix[i][j] ==3:
+                    matrix[i][j] =1
+
+        return matrix
+
+                
+