There is a group of n
people labeled from 0
to n - 1
where each person has a different amount of money and a different level of quietness.
You are given an array richer
where richer[i] = [ai, bi]
indicates that ai
has more money than bi
and an integer array quiet
where quiet[i]
is the quietness of the ith
person. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x
is richer than y
and y
is richer than x
at the same time).
Return an integer array answer
where answer[x] = y
if y
is the least quiet person (that is, the person y
with the smallest value of quiet[y]
) among all people who definitely have equal to or more money than the person x
.
Example 1:
Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation:
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but it is not clear if they have more money than person 0.
answer[7] = 7.
Among all people that definitely have equal to or more money than person 7 (which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x]) is person 7.
The other answers can be filled out with similar reasoning.
Example 2:
Input: richer = [], quiet = [0]
Output: [0]
Constraints:
-
n == quiet.length
-
1 <= n <= 500
-
0 <= quiet[i] < n
- All the values of
quiet
are unique. -
0 <= richer.length <= n * (n - 1) / 2
-
0 <= ai, bi < n
-
ai != bi
- All the pairs of
richer
are unique. - The observations in
richer
are all logically consistent.
SOLUTION:
class Solution:
def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
n = len(quiet)
graph = {}
for a, b in richer:
graph[b] = graph.get(b, []) + [a]
answer = [0] * n
for beg in range(n):
paths = [beg]
visited = {beg}
while len(paths) > 0:
curr = paths.pop()
for j in graph.get(curr, []):
if j not in visited:
visited.add(j)
paths.append(j)
answer[beg] = min(visited, key = lambda p: quiet[p])
return answer
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