Algorithm
Problem Name: 851. Loud and Rich
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.
Code Examples
#1 Code Example with Python Programming
Code -
Python Programming
class Solution:
def loudAndRich(self, richer, quiet):
edges, memo, res = collections.defaultdict(list), {}, [i for i in range(len(quiet))]
for r, p in richer: edges[p].append(r)
def explore(i):
if i in memo: return memo[i]
cur_min = i
for v in edges[i]:
cur = explore(v)
if quiet[cur] < quiet[cur_min]: cur_min = cur
res[i] = memo[i] = cur_min
return cur_min
for i in range(len(quiet)): explore(i)
return res
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