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question:You are given N disks of different diameters and three rods, A, B, and C. Initially, all the disks are stacked on rod A in decreasing order of their diameters, with the largest disk at the bottom and the smallest at the top. The goal is to move all the disks to rod C using rod B as an auxiliary rod, i.e., through a series of legal moves. Only one disk can be moved at a time, and a larger disk cannot be placed on top of a smaller disk. Write a function `hanoi(N, from_rod, to_rod, aux_rod)` that prints out the sequence of moves required to solve the Towers of Hanoi puzzle for N disks. Constraints: - Your function should properly handle up to 20 disks. - Each function call should print the moves in the format "Move disk X from rod {from_rod} to rod {to_rod}". Function Signature: ```python def hanoi(N: int, from_rod: str, to_rod: str, aux_rod: str) -> None: pass ``` Example: ```python hanoi(2, 'A', 'C', 'B') ``` Output: ``` Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C ``` Explanation: - The smallest disk (disk 1) is moved from rod A to rod B. - The larger disk (disk 2) is moved from rod A to rod C. - Finally, the smallest disk (disk 1) is moved from rod B to rod C, completing the puzzle for 2 disks.

answer:def hanoi(N, from_rod, to_rod, aux_rod): Print the sequence of moves required to solve the Towers of Hanoi puzzle for N disks. Parameters: N (int): Number of disks. from_rod (str): The rod from which disks need to be moved initially. to_rod (str): The destination rod to which disks need to be moved ultimately. aux_rod (str): The rod used as an auxiliary for intermediate moves. if N == 1: print(f"Move disk 1 from rod {from_rod} to rod {to_rod}") return hanoi(N - 1, from_rod, aux_rod, to_rod) print(f"Move disk {N} from rod {from_rod} to rod {to_rod}") hanoi(N - 1, aux_rod, to_rod, from_rod) # Example usage: # hanoi(2, 'A', 'C', 'B')

question:# Task You are given a list of ordered pairs, where each pair represents a connection between two cities. Each city is represented by an integer. Determine if these given connections form a single connected component. In other words, starting from any city, you should be able to travel to any other city using the given connections. If all the cities form a single connected component, return True. Otherwise, return False. # Example For pairs `[(0, 1), (1, 2), (2, 3)]`, the output should be `True`. ``` Here, 0 is connected to 1, 1 is connected to 2, and 2 is connected to 3. Therefore, all cities are reachable from any other city. ``` For pairs `[(0, 1), (1, 2), (3, 4)]`, the output should be `False`. ``` Here, there is no connection between cities 2 and 3, so not all cities are interconnected. ``` # Function Signature ```python def is_single_connected_component(pairs: List[Tuple[int, int]]) -> bool: pass ```

answer:from typing import List, Tuple from collections import defaultdict, deque def is_single_connected_component(pairs: List[Tuple[int, int]]) -> bool: if not pairs: return True # No pairs means there's no connection required to check, it's trivially connected # Create an adjacency list adjacency_list = defaultdict(list) for a, b in pairs: adjacency_list[a].append(b) adjacency_list[b].append(a) # We need to find all unique cities unique_cities = set() for a, b in pairs: unique_cities.add(a) unique_cities.add(b) # Perform BFS or DFS to check if all nodes are connected visited = set() def bfs(start_node): queue = deque([start_node]) visited.add(start_node) while queue: node = queue.popleft() for neighbor in adjacency_list[node]: if neighbor not in visited: visited.add(neighbor) queue.append(neighbor) # Start BFS/DFS from one of the cities start_city = next(iter(unique_cities)) bfs(start_city) # Check if we visited all the cities return visited == unique_cities

question:Given a rectangular chocolate bar of dimensions n x m squares, you need to determine if it is possible to break it into exactly k squares by making some number of straight breaks between the squares. Each break involves breaking the chocolate bar along one of the lines between its squares. You can only break the chocolate bar into two rectangular pieces at a time. Input There are three integers in a single line: n, m, and k (1 ≤ n, m ≤ 100, 1 ≤ k ≤ n * m). Output Print YES if it's possible to break the chocolate bar into exactly k squares, otherwise print NO. Examples Input 4 2 6 Output YES Input 3 3 8 Output NO Note For the first example, we can break the chocolate bar as follows: - Make a break after 2 squares to get a piece of 2 x 2 (4 squares) and another piece of 2 x 2 (4 squares). - Then make a break in one of the 2 x 2 pieces after 2 squares to get 2 pieces of 2 squares each. Thus, we have 4 squares + 2 squares = 6 squares in total, so the answer is YES. For the second example, whatever breaks we make, we cannot get exactly 8 squares out of a 3 x 3 chocolate bar. Thus, the answer is NO.

answer:def can_break_chocolate(n, m, k): Function to determine if it is possible to break a n x m chocolate bar into exactly k squares. if k > n * m: return "NO" # We can break k squares if we can break either # an n x b or a x m bar with k = i * b or k = i * a if k % n == 0 or k % m == 0: return "YES" return "NO"

question:You are given a list of integers. The task is to find and print the element that occurs the maximum number of times. If multiple elements have the same maximum frequency, print the smallest one among them. Function Description Complete the `mostFrequentElement` function in the editor below. `mostFrequentElement` has the following parameter: - `int arr[n]`: an array of integers Returns - `int`: the element that occurs the most frequently. If more than one element has the same maximal frequency, return the smallest one. Input Format The first line contains an integer n, the size of the array arr. The second line contains n space-separated integers arr[i]. Constraints - 1 le n le 10^5 - -10^4 le arr[i] le 10^4 Sample Input ``` 7 1 3 2 3 2 3 1 ``` Sample Output ``` 3 ``` Explanation The element `3` appears three times, which is more frequent than the other elements. Therefore, the return value is `3`. Among elements with the same frequency (like `1` and `3`), `3` is the most frequent, so it is selected. If multiple elements tied in frequency, the smallest one would be selected.

answer:def mostFrequentElement(arr): Returns the element that occurs the most frequently. If more than one element has the same maximal frequency, returns the smallest one among them. from collections import Counter count = Counter(arr) max_frequency = max(count.values()) most_frequent_elements = [key for key, value in count.items() if value == max_frequency] return min(most_frequent_elements)