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question:You are given a 2D list of integers representing a grid where each cell has a certain amount of "gold". An entry in matrix `grid[i][j]` represents the amount of gold in the cell `(i, j)`. Your task is to design a function that summarizes how much maximum gold can be collected, starting from any cell and moving only in four possible directions (up, down, left, right) to adjacent cells, without visiting any cell more than once. **Function Signature:** ```python def getMaximumGold(grid: List[List[int]]) -> int: ``` **Example:** ```python grid = [ [1, 0, 7], [2, 0, 6], [3, 4, 5], [0, 3, 0] ] Output: 28 Explanation: One of the paths to get the maximum gold is 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7. ``` **Constraints:** - `1 <= grid.length, grid[i].length <= 15` - `0 <= grid[i][j] <= 100` - There are at most 25 cells with gold. **Note:** - You can start and stop collecting gold from any grid cell. - You may assume that there is always a path that allows at least one gold to be collected.

answer:from typing import List def getMaximumGold(grid: List[List[int]]) -> int: def dfs(x, y, visited): if x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or grid[x][y] == 0 or (x, y) in visited: return 0 visited.add((x, y)) gold = grid[x][y] max_gold = 0 for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]: max_gold = max(max_gold, dfs(x + dx, y + dy, visited)) visited.remove((x, y)) return gold + max_gold max_gold = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] != 0: max_gold = max(max_gold, dfs(i, j, set())) return max_gold

question:Design a min-stack that supports push, pop, top, and retrieving the minimum element in constant time. Example 1: Input: push(-2) push(0) push(-3) getMin() -> -3 pop() top() -> 0 getMin() -> -2 Your Task: You are required to implement a class `MinStack` with the following methods: - `void push(int x)`: Pushes an integer x onto the stack. - `void pop()`: Removes the element on the top of the stack. - `int top()`: Gets the top element of the stack. - `int getMin()`: Retrieves the minimum element in the stack. Expected Time Complexity: O(1) for all methods. Expected Auxiliary Space: O(n), where n is the number of elements in the stack. Constraints: 1 <= Number of operations <= 10^4 -10^5 <= value of elements <= 10^5

answer:class MinStack: def __init__(self): Initialize the stack and a helper stack to keep track of minimums. self.stack = [] self.min_stack = [] def push(self, x: int) -> None: Push element x onto the stack and update the min stack. self.stack.append(x) # If the min stack is empty or the current element is smaller or equal # to the element at the top of the min stack, push it onto the min stack. if not self.min_stack or x <= self.min_stack[-1]: self.min_stack.append(x) def pop(self) -> None: Removes the element on the top of the stack and updates the min stack. if self.stack: top_element = self.stack.pop() # If the top element is the same as the top of the min stack, pop from the min stack as well. if top_element == self.min_stack[-1]: self.min_stack.pop() def top(self) -> int: Get the top element of the stack. if self.stack: return self.stack[-1] return None # If the stack is empty def getMin(self) -> int: Retrieve the minimum element in the stack. if self.min_stack: return self.min_stack[-1] return None # If the min stack is empty

question:Given an integer array arr and an integer k, rearrange the array in such a way that the difference between any two consecutive elements in the rearranged array is at most k. Input The first line of input contains an integer n (1 ≤ n ≤ 100000) — the number of elements in the array. The second line contains n space-separated integers arr1, arr2, ..., arrn (1 ≤ arri ≤ 1000000) — the elements of the array. The third line contains a single integer k (1 ≤ k ≤ 1000000), the maximum allowed difference between consecutive elements. Output Print any valid rearrangement of the array such that the absolute difference between any two consecutive elements is at most k. If no such arrangement exists, print a single line containing "-1". Examples Input 6 1 2 9 12 15 16 3 Output 1 2 1 5 6 9 12 15 Input 3 10 3 20 15 Output 10 3 20 Note In the first example, the output array could be [1, 2, 5, 6, 9, 12, 15]. One possible valid output is [1, 4, 6, 8], but the numbers could be in any order that satisfies the conditions. In the second example, the output array could be [10, 3, 20] or [10, 20], which fulfill the conditions imposed by k.

answer:def rearrange_array(arr, k): Rearranges the array so that the difference between any two consecutive elements is at most k. Parameters: arr (list): List of integers. k (int): Maximum allowed difference between consecutive elements. Returns: list: Rearranged list that satisfies the condition or [-1] if not possible. arr.sort() for i in range(1, len(arr)): if abs(arr[i] - arr[i - 1]) > k: return [-1] return arr # Example usage: # arr = [1, 2, 9, 12, 15, 16] # k = 3 # print(rearrange_array(arr, k)) # Output: [1, 2, 9, 12, 15, 16] or any valid permutation

question:Alice and Bob are playing a modified version of a popular game called "Tower of Hanoi". In this game, instead of moving all disks from one rod to another, players need to move a subset of disks from the initial rod to a target rod in a restricted number of moves. The goal is to check whether it's possible to achieve the desired configuration within the given number of moves. # Problem Specification You are given: - Number of disks (`n`) - Number of moves allowed (`m`) - Initial configuration of the rods (`initial`) - Desired configuration of the rods (`target`) The rods are labeled as A, B, and C. Each rod can contain between 0 to n disks, represented by numbers from 1 to n, where 1 is the smallest disk and n is the largest. # Input One or multiple test cases will be given. For each test case: 1. An integer n, representing the number of disks (1 ≤ n ≤ 10). 2. An integer m, representing the number of moves allowed (1 ≤ m ≤ 30). 3. Three lines describing the initial configuration of the rods, one rod per line. 4. Three lines describing the target configuration of the rods. Each of the rod configuration lines contains a rod's label followed by zero or more integers, representing the disks placed on that rod, from top to bottom. The end of input is signified by a line with a single zero. # Output For each test case, output "YES" if it's possible to achieve the target configuration of the rods within m moves, otherwise output "NO". # Example Input ``` 3 3 A 1 2 3 B C A 3 B 2 C 1 3 7 A 1 2 3 B C A B C 1 2 3 0 ``` Output ``` NO YES ``` # Explanation 1. In the first test case, it's not possible to achieve the target configuration within 3 moves, hence the output is "NO". 2. In the second test case, it is possible to achieve the target configuration within 7 moves, hence the output is "YES". --- By adhering to these constraints, you should ensure that implementations can efficiently solve the problem within reasonable time limits.

answer:def is_hanoi_solvable(n, m, initial, target): def move_disks(num_disks, from_rod, to_rod, aux_rod): if num_disks == 0: return 0 if num_disks == 1: return 1 # one move # Move num_disks-1 disks from from_rod to aux_rod using to_rod moves1 = move_disks(num_disks - 1, from_rod, aux_rod, to_rod) # Move the nth disk from from_rod to to_rod moves2 = 1 # Move the num_disks-1 disks from aux_rod to to_rod using from_rod moves3 = move_disks(num_disks - 1, aux_rod, to_rod, from_rod) return moves1 + moves2 + moves3 def extract_state(config): state = {'A': [], 'B': [], 'C': []} for line in config.splitlines(): parts = line.split() rod = parts[0] disks = list(map(int, parts[1:])) state[rod] = disks return state initial_state = extract_state(initial) target_state = extract_state(target) def is_same_state(state1, state2): for rod in ['A', 'B', 'C']: if state1[rod] != state2[rod]: return False return True if is_same_state(initial_state, target_state): return "YES" total_moves = move_disks(n, 'A', 'C', 'B') if total_moves <= m: return "YES" else: return "NO" def solve_hanoi_games(input_data): lines = input_data.strip().split('n') index = 0 results = [] while index < len(lines): n = int(lines[index]) if n == 0: break m = int(lines[index + 1]) initial_config = 'n'.join(lines[index + 2: index + 5]) target_config = 'n'.join(lines[index + 5: index + 8]) result = is_hanoi_solvable(n, m, initial_config, target_config) results.append(result) index += 8 return 'n'.join(results)