From e88750fa49465cc3262a0edb97a6964d7a3562b4 Mon Sep 17 00:00:00 2001
From: ankitakulkarnigit <kulkarniankita555@gmail.com>
Date: Thu, 23 Oct 2025 23:38:34 -0700
Subject: [PATCH] Arrays-2 Completed

---
 Sample.java |  7 -----
 Sample.py   | 84 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 84 insertions(+), 7 deletions(-)
 delete mode 100644 Sample.java
 create mode 100644 Sample.py

diff --git a/Sample.java b/Sample.java
deleted file mode 100644
index f5c45b5f..00000000
--- a/Sample.java
+++ /dev/null
@@ -1,7 +0,0 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
-// Any problem you faced while coding this :
-
-
-// Your code here along with comments explaining your approach
\ No newline at end of file
diff --git a/Sample.py b/Sample.py
new file mode 100644
index 00000000..a346d71c
--- /dev/null
+++ b/Sample.py
@@ -0,0 +1,84 @@
+# Time Complexity : O(n) + O(n) = O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+
+# Your code here along with comments explaining your approach
+# Mark the visited indices values as Negative. Whichever number is not negative, that indices are the disappeared number
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        res = []
+        for i in range(len(nums)):
+            idx_val = abs(nums[i]) - 1
+            if nums[idx_val] > 0:
+                nums[idx_val] = -nums[idx_val]
+        for i in range(len(nums)):
+            if nums[i] < 0:
+                nums[i] = abs(nums[i])
+            else:
+                res.append(i+1)
+        return res
+
+
+
+# Time Complexity : O(n^2) + O(n) = O(n^2)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : 
+
+
+# Your code here along with comments explaining your approach
+# Check all directions to count alive cells around the current cell
+# When turning previously alive to dead, turn them to 2
+# When turning previously dead to alive, turn them to 3
+# This preseves the original state as well
+# Finally turn 2s to 0 and 3s to 1
+
+
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        
+        1 -> 1 if livecount > 2 nc
+        1 -> 0  if livecount < 2 up
+        1 -> 0 if livecount > 3 op
+        0 -> 1 if livecount == 3 rebirth
+        """
+        m = len(board)
+        n = len(board[0])
+        for i in range(m):
+            for j in range(n):
+                countAlives = self.countAlive(board,i,j)
+                if board[i][j] == 1 and (countAlives < 2 or countAlives > 3):
+                    board[i][j] = 2
+                if board[i][j] == 0 and countAlives == 3:
+                    board[i][j] = 3
+
+        for i in range(m):
+            for j in range(n):
+                if board[i][j] == 2:
+                    board[i][j] = 0
+                if board[i][j] == 3:
+                    board[i][j] = 1
+
+
+    def countAlive(self,board,i,j):
+        alive = 0
+        m = len(board)
+        n = len(board[0])
+        # left,right,top,down,top-left,top-right,bottom-left,bottom-right
+        dirs = [(0,-1), (0,1), (-1,0), (1,0), (-1,-1), (-1,1), (1,-1), (1,1)]
+        for dir in dirs:
+            r = i+dir[0]
+            c = j+dir[1]
+            if (r >= 0 and c >= 0 and r < m and c < n and (board[r][c] == 1 or board[r][c] == 2)):
+                alive += 1
+
+        return alive
+            
+
+
+        
\ No newline at end of file