Monk goes to graph city to learn about graphs, and meets an undirected graph having \$ N \$ nodes, where each node has a value \$ i \$ such that \$ 1 \le i \le N \$. Each node of the graph is very curious and wants to know something about the nodes which are directly connected to it.
For each node, if we sort the nodes directly connected to it, according to their values in descending order, what is the value of the node at the \$ k \$ th position? The graph is 1-based.
Note: If two nodes have the same value, they are sorted according to their indices in ascending order.
My accepted solution:
import sys from collections import defaultdict node_value_dict = defaultdict(int) connected_graph_component = defaultdict(list) def connect_graph(): n, m, k = list(map(int, sys.stdin.readline().split())) node_values = list(map(int, sys.stdin.readline().split())) for i in range(1, len(node_values) + 1): node_value_dict[i] = node_values[i - 1] for m_i in range(m): n1, n2 = list(map(int, input().split())) connected_graph_component[n1].append(n2) connected_graph_component[n2].append(n1) for i in range(1, n + 1): if i in connected_graph_component: curr_co = connected_graph_component[i] sort_list = sorted(curr_co, key=lambda value: node_value_dict[ value], reverse=True) for i in range(len(sort_list) - 1): if node_value_dict[sort_list[i]] == node_value_dict[sort_list[i + 1]] and sort_list[i] < sort_list[i + 1]: sort_list[i], sort_list[ i + 1] = sort_list[i + 1], sort_list[i] print(sort_list[k - 1]) else: print(-1) connect_graph()
With a runtime of 9.14092 sec.
How can I improve the code quality and further optimize runtime duration?
In case of equal values, I need to sort the nodes according to their indices in ascending order. Currently I'm doing it by looping over the sorted list and then comparing values again side by side maintaining two pointers.
Can you please suggest some elegant way of doing this.