diff --git a/problem31.py b/problem31.py
new file mode 100644
index 00000000..f2612025
--- /dev/null
+++ b/problem31.py
@@ -0,0 +1,75 @@
+#sorting + linear search
+
+from typing import List
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        #brute force will be O(n2)
+        #sorting + linear search
+        nums.sort()
+        result = []
+        i = 0
+
+        for expected in range(1, len(nums)+1):
+            if i == len(nums) or nums[i] > expected: #to handle cases like [1,1]
+                result.append(expected)
+            while i < len(nums) and nums[i] == expected: #loop till next unqiue element
+                i += 1
+
+        return result        
+        #nums[i] won't be less than expected because we are sorting it and the no.are in the range from 1 to n. where n is length of nums.      
+    
+# Time Complexity: O(nlogn + n) = O(nlogn)
+# Space complexity: O(1)
+
+#Treating the array as a hashset
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        #using the array as set itself
+
+        result = []
+
+        for i in range(len(nums)):
+            idx = abs(nums[i])-1 #the no.s actual index in the array from 1 to N, 4 will be 3
+            #1 will be 0 and so on.
+            if nums[idx] > 0:
+                nums[idx] *= -1 #mark the number at that idx negative
+            
+        for i in range(len(nums)):
+            if nums[i] > 0:
+                result.append(i+1)
+
+        return result
+
+# Time Complexity: O(N)
+# Space Complexity: O(1)
+
+#if negative numbers are involved example:
+# [0,-1,3,-2,4,-2,-1,-3]
+# range -3 to 4 = 4-(-3)+1 = 8
+# offset everything by range/2 = 4
+#this only works when the range is equal to the length of the array.
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        #using the array as set itself
+
+        result = []
+
+        shift = 8//2 #(range//2)
+
+        for i in range(len(nums)):
+            nums[i] = nums[i]+shift #this should be range/2
+
+        for i in range(len(nums)):
+            idx = abs(nums[i])-1 #the no.s actual index in the array from 1 to N, 4 will be 3
+            #1 will be 0 and so on.
+            if nums[idx] > 0:
+                nums[idx] *= -1 #mark the number at that idx negative
+            
+        for i in range(len(nums)):
+            if nums[i] > 0:
+                result.append(i+1-shift)
+
+        return result
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diff --git a/problem32.py b/problem32.py
new file mode 100644
index 00000000..0134ca18
--- /dev/null
+++ b/problem32.py
@@ -0,0 +1,37 @@
+#find min and max in an array
+#if we compare each and every element with the min and max values, then its 2N comparison so instead of that take pairs, 
+#there will be n//2 pairs and each pair will have 3 comparisons worse case so thats 1.5N comparisons.
+#so think in pairs here
+from typing import List
+
+class Solution:
+    def findMinandMax(self, nums: List[int]):
+
+        if not nums:
+             return [-1, -1]
+        
+        n = len(nums)
+
+        max_val = float('-inf')
+        min_val = float('inf')
+
+        i = 0
+
+        while i <= n-2:
+                if nums[i] > nums[i+1]:
+                    max_val = max(max_val, nums[i])
+                    min_val = min(min_val, nums[i+1])
+                else:
+                    max_val = max(max_val, nums[i+1])
+                    min_val = min(min_val, nums[i])
+                
+                i += 2
+        
+        if n % 2 != 0: #handle the last element if the length is odd and not even
+             max_val = max(max_val, nums[-1])
+             min_val = min(min_val, nums[-1])
+
+        return [min_val, max_val]
+    
+# Time Complexity: O(N)
+# Space Complexity: O(1)
diff --git a/problem33.py b/problem33.py
new file mode 100644
index 00000000..f4e9b5cc
--- /dev/null
+++ b/problem33.py
@@ -0,0 +1,51 @@
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        #we need to store the state change within the board itself without any external         variables.
+        # 1 -> 0 use 2
+        # 0 -> 1 use 3
+        # 2, 3 indicates original states were 1 and 0 respectively.
+
+        m = len(board)
+        n = len(board[0])
+
+        def helper(board, i, j):
+                #right   #left.   #up.   #bottom. #diagonals
+            directions = [(0,1), (0,-1), (-1,0), (1,0), (1,1), (1,-1), (-1,1), (-1,-1)]
+
+            count = 0
+
+            for dx, dy in directions:
+                r,c = i+dx, j+dy
+                if 0 <= r < m and 0 <= c < n:
+                    if board[r][c] == 1 or board[r][c] == 2:#we just need to count alive cells
+                        count += 1
+
+            return count
+
+        #use temp state change value so that we won't lose track of original value as well
+        for i in range(m):
+            for j in range(n):
+                count = helper(board, i ,j) #we just need count of alive neighbors
+                if board[i][j] and (count < 2 or count > 3):
+                    board[i][j] = 2
+                elif not board[i][j] and count == 3:
+                    board[i][j] = 3
+
+        
+        for i in range(m):
+            for j in range(n):
+                if board[i][j] == 2:
+                    board[i][j] = 0 #new state
+                elif board[i][j] == 3:
+                    board[i][j] = 1 #new state
+
+# Time Complexity: O(m*n)
+# Space Complexity: O(1)
+
+# 1 -> 0 if neighbors < 2 alive
+# 1 -> 1 if neighbors 2 or 3 alive
+# 1 -> 0 if neighbors > 3 alive
+# 0 -> 1 if neighbors = 3 alive
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