Problem
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
/**
* @param {number[]} groupSizes
* @return {number[][]}
*/
var groupThePeople = function(groupSizes) {
var map = Array(groupSizes.length + 1).fill(0).map(() => []);
var res = [];
for (var i = 0; i < groupSizes.length; i++) {
var size = groupSizes[i];
map[size].push(i);
if (map[size].length === size) {
res.push(map[size]);
map[size] = [];
}
}
return res;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).