Done Array-2

find-all-numbers-disappeared-in-array.py

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+''' Time Complexity : O(n) 
+    Space Complexity : O(1) ; 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+
+
+   Approach : Inplace marking method : traverse the array and mark the indexes negative for each element
+'''
+
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        result = []
+        n = len(nums)
+        for i in range(0,n):
+            idx = abs(nums[i]) - 1
+            if nums[idx] > 0:
+                nums[idx] *= -1
+        print(nums)
+        for i in range(n):
+            if nums[i] < 0:
+                nums[i] *= -1
+            else:
+                result.append(i+1)
+        return result
+

game-of-life.py

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+''' 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 : Temporary mark hthe change 1 -> 0 : 2 and 0 -> 1 : 3
+'''
+
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        def getLiveCount(board, i , j):
+                    # 
+            count = 0
+            dirs = [(i,j+1),(i,j-1),(i+1,j),(i-1,j),(i -1,j+1),(i+1,j+1),(i+1,j-1),(i-1,j-1)]
+            for r, c in dirs:
+                if (0<=r<rows and 0<=c<cols and (board[r][c] == 1 or board[r][c] == 2)):
+                    count += 1
+            return count
+
+        rows, cols = len(board), len(board[0])
+        for i in range(rows):
+            for j in range(cols):
+                count = getLiveCount(board, i , j)
+                if board[i][j] == 0 and count == 3:
+                    board[i][j] = 3
+                elif (board[i][j] == 1 and (count < 2 or count > 3)):
+                    board[i][j] = 2
+        for i in range(rows):
+            for j in range(cols):
+                if board[i][j] == 2:
+                    board[i][j] = 0
+                if board[i][j] == 3:
+                    board[i][j] = 1