Array - 2 Complete

program1.py

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+"""
+
+Time Complexity : o(2N)
+Space Complexity :O(N)
+Did this code successfully run on Leetcode :yes
+Any problem you faced while coding this :no
+
+
+Approach
+Here we are using indexs and temp array mutation to get the result 
+because of the mutation we are keeping the space complexity same
+
+basic idea is that we keep track of all the number that we have seen in the array but,
+converting the number at the index of the number found 
+as negative after we are do with one pass
+
+on the second pass we just see if a number is negative of not if now then that index or nunmber is missing
+
+"""
+
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+
+        result = list()
+
+        for i in range(len(nums)):
+            index = abs(nums[i]) -1
+            if nums[index] > 0:
+                nums[index] *= -1
+
+        for i in range(len(nums)):
+            if nums[i] > 0:
+                result.append(i+1)
+        for i in range(len(nums)):
+            if nums[i] < 0:
+                nums[i] *= -1
+        return result
+
+
+

program2.py

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+"""
+
+Time Complexity : o(2nm)
+Space Complexity :O(N)
+Did this code successfully run on Leetcode :yes
+Any problem you faced while coding this :no
+
+
+Approach
+we are travesing all the numbers one by one and checking for the rules
+and according to that we are changing the value, we are comparing all the 8 possible values
+
+and to do the change in place we are changing value to 2 and 3 as temp so was to keep track of all the value before and after mutation
+
+"""
+
+
+
+
+class Solution:
+    def countalive(self, board: List[List[int]], r: int, c: int) -> int:
+        m, n = len(board), len(board[0])
+        dirs = [(0,1),(0,-1),(-1,0),(1,0),(-1,1),(-1,-1),(1,1),(1,-1)]
+        count = 0
+        for dr, dc in dirs:
+            nr, nc = r + dr, c + dc
+            if 0 <= nr < m and 0 <= nc < n and (board[nr][nc] == 1 or board[nr][nc] == 2):
+                count += 1
+        return count
+
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        m, n = len(board), len(board[0])
+
+        # 1st pass: mark transitions
+        for i in range(m):
+            for j in range(n):
+                alive_neighbors = self.countalive(board, i, j)
+                if board[i][j] == 1 and (alive_neighbors < 2 or alive_neighbors > 3):
+                    board[i][j] = 2  
+                elif board[i][j] == 0 and alive_neighbors == 3:
+                    board[i][j] = 3  
+
+        # 2nd pass: finalize
+        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