Main

1260. Shift 2D Grid/1260. Shift 2D Grid.py

@@ -0,0 +1,47 @@
+"""
+Given a 2D grid of size m x n and an integer k. You need to shift the grid k times.
+
+In one shift operation:
+
+Element at grid[i][j] moves to grid[i][j + 1].
+Element at grid[i][n - 1] moves to grid[i + 1][0].
+Element at grid[m - 1][n - 1] moves to grid[0][0].
+Return the 2D grid after applying shift operation k times.
+
+
+
+Example 1:
+
+
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
+Output: [[9,1,2],[3,4,5],[6,7,8]]
+Example 2:
+
+
+Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
+Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
+Example 3:
+
+Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
+Output: [[1,2,3],[4,5,6],[7,8,9]]
+
+
+Constraints:
+
+m == grid.length
+n == grid[i].length
+1 <= m <= 50
+1 <= n <= 50
+-1000 <= grid[i][j] <= 1000
+0 <= k <= 100
+"""
+class Solution:
+    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
+        m,n = len(grid), len(grid[0])
+        k = k % (m*n)
+        temp = [grid[i][j] for i in range(m) for j in range(n)]
+        temp = temp[-k:] + temp[:-k]
+        result = []
+        for i in range(m):
+            result.append(temp[i*n:i*n+n])
+        return result

700. Search in a Binary Search Tree/700. Search in a Binary Search Tree.py

@@ -20,26 +20,24 @@
 
 Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.
 """
-
-
 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, val=0, left=None, right=None):
 #         self.val = val
 #         self.left = left
 #         self.right = right
 class Solution:
-    def searchBST(self, root: TreeNode, val: int) -> TreeNode:
-        if not root:
-            return None
+    def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        if not root: return None
         q = collections.deque()
         q.append(root)
         while q:
             node = q.popleft()
             if node.val == val:
                 return node
-            if node.left:
-                q.append(node.left)
-            if node.right:
+            elif node.val < val and node.right:
                 q.append(node.right)
+            else:
+                if node.left:
+                    q.append(node.left)
         return None