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1282. Group the People Given the Group Size They Belong To |
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
/**
* @param {number[]} groupSizes
* @return {number[][]}
*/
var groupThePeople = function(groupSizes) {
let map = new Map();
let ans = [];
for(let i=0;i<groupSizes.length;i++){
let cur = groupSizes[i];
if(!map.has(cur) || cur === ans[map.get(cur)].length){
map.set(cur, ans.push([])-1)
}
ans[map.get(cur)].push(i)
}
return ans
};
- Time Complexity:
O(n)
- Space Complexity:
O(n)