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committedSep 11, 2023
feat: solve No.1282
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# 1282. Group the People Given the Group Size They Belong To
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- Difficulty: Medium.
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- Related Topics: Array, Hash Table.
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- Similar Questions: Maximum Number of Groups With Increasing Length.
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## Problem
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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`.
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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`.
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Return **a list of groups such that each person `i` is in a group of size `groupSizes[i]`**.
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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.
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Example 1:
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```
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Input: groupSizes = [3,3,3,3,3,1,3]
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Output: [[5],[0,1,2],[3,4,6]]
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Explanation:
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The first group is [5]. The size is 1, and groupSizes[5] = 1.
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The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
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The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
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Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].
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```
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Example 2:
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```
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Input: groupSizes = [2,1,3,3,3,2]
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Output: [[1],[0,5],[2,3,4]]
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```
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**Constraints:**
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- `groupSizes.length == n`
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- `1 <= n <= 500`
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- `1 <= groupSizes[i] <= n`
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## Solution
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```javascript
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/**
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* @param {number[]} groupSizes
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* @return {number[][]}
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*/
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var groupThePeople = function(groupSizes) {
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var map = Array(groupSizes.length + 1).fill(0).map(() => []);
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var res = [];
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for (var i = 0; i < groupSizes.length; i++) {
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var size = groupSizes[i];
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map[size].push(i);
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if (map[size].length === size) {
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res.push(map[size]);
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map[size] = [];
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}
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}
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return res;
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(n).

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