I am solving this problem on hacker earth: https://www.hackerearth.com/practice/algorithms/graphs/depth-first-search/practice-problems/algorithm/just-c3-has-ended-1/description/ . It clearly seems like a BFS Problem. I solved a few test cases but stuck on others due to time complexity issues. This is what I have done so far:
from collections import Counter
import queue
def solve(students, pairsmap):
visited = Counter()
maximum = 1
pairs = 0
for i in range(students):
if(visited[i+1] == 0):
visited[i+1] = 1
pairs += 1
count = bfs(students, i+1, visited, 1, pairsmap, maximum)
maximum = count
return pairs, maximum
def bfs(students, curr, visited, count, pairsmap, currMax):
q = []
q.append(curr)
c = Counter()
while q:
element = q.pop()
for k, v in pairsmap.items():
if(k[0] == element and visited[k[1]] == 0):
q.append(k[1])
visited[k[1]] = 1
count += 1
elif (k[1] == element and visited[k[0]] == 0):
q.append(k[0])
visited[k[0]] = 1
count += 1
if currMax < count:
currMax = count
return currMax
students = input().split(" ")
n = int(students[0])
m = int(students[1])
pairsmap = Counter()
for i in range(m):
pairs = input().split()
# print(pairs)
u = int(pairs[0])
v = int(pairs[1])
pairsmap[(u, v)] = 1
# print(pairsmap)
res = solve(n, pairsmap)
print(res[0], res[1])
Can anyone point towards an optimized or better approach?
