diff --git a/findAll.py b/findAll.py
new file mode 100644
index 00000000..01b53ad0
--- /dev/null
+++ b/findAll.py
@@ -0,0 +1,26 @@
+# Time Complexity : O(n)
+# Space Complexity : O(1) (excluding output)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: Use index marking.
+# Iterate through the array and treat each value as an index (value - 1).
+# Mark the corresponding index as negative to indicate the number exists.
+# After marking, iterate again and collect indices that remain positive,
+# as they represent the missing numbers.
+
+
+
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        idx = -1
+        res = []
+        for i in nums:
+            idx = abs(i)
+            if nums[idx - 1] > 0:
+                nums[idx - 1] = nums[idx - 1] * -1 
+        
+        for i in range(len(nums)):
+            if nums[i] > 0:
+                res.append(i + 1)
+        return res
\ No newline at end of file
diff --git a/game.py b/game.py
new file mode 100644
index 00000000..09019dbd
--- /dev/null
+++ b/game.py
@@ -0,0 +1,45 @@
+# Time Complexity : O(m * n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach: In-place state encoding.
+# Traverse the board and count live neighbors for each cell.
+# Use temporary states:
+# 2 → was alive (1) → becomes dead (0)
+# 3 → was dead (0) → becomes alive (1)
+# This allows us to preserve original state while updating.
+# In the second pass, convert 2 → 0 and 3 → 1 to finalize the board.
+
+
+
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        dirs = [(-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)]
+        m, n = len(board), len(board[0])
+
+        def getCount(i, j):
+            count = 0
+            for dx, dy in dirs:
+                r, c = i + dx, j + dy
+                if 0 <= r < m and 0 <= c < n:
+                    if board[r][c] == 1 or board[r][c] == 2:
+                        count += 1
+            return count
+
+        for i in range(m):
+            for j in range(n):
+                cnt = getCount(i, j)
+                if board[i][j] == 0 and cnt == 3:
+                    board[i][j] = 3
+                elif board[i][j] == 1 and (cnt < 2 or cnt > 3):
+                    board[i][j] = 2
+
+        for i in range(m):
+            for j in range(n):
+                if board[i][j] == 2:
+                    board[i][j] = 0
+                elif board[i][j] == 3:
+                    board[i][j] = 1