There are n
people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0
to n - 1
.
You are given an integer array groupSizes
, where groupSizes[i]
is the size of the group that person i
is in. For example, if groupSizes[1] = 3
, then person 1
must be in a group of size 3
.
Return a list of groups such that each person i
is in a group of size groupSizes[i]
.
Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.
Example 1:
Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation:
The first group is [5]. The size is 1, and groupSizes[5] = 1.
The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].
Example 2:
Input: groupSizes = [2,1,3,3,3,2]
Output: [[1],[0,5],[2,3,4]]
Constraints:
-
groupSizes.length == n
-
1 <= n <= 500
-
1 <= groupSizes[i] <= n
SOLUTION:
class Solution:
def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
numpeople = {}
for i, gsize in enumerate(groupSizes):
numpeople[gsize] = numpeople.get(gsize, []) + [i]
op = []
for size, people in numpeople.items():
n = len(people)
for i in range(0, n, size):
op.append(people[i:i+size])
return op
Top comments (0)