LeetCode #1282 — MEDIUM

Group the People Given the Group Size They Belong To

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[3,3,3,3,3,1,3]

Example 2

[2,1,3,3,3,2]

Core Insight

What unlocks the optimal approach

Put people's IDs with same groupSize into buckets, then split each bucket into groups.
Greedy fill until you need a new group.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1282: Group the People Given the Group Size They Belong To
class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        int n = groupSizes.length;
        List<Integer>[] g = new List[n + 1];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 0; i < n; ++i) {
            g[groupSizes[i]].add(i);
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < g.length; ++i) {
            List<Integer> v = g[i];
            for (int j = 0; j < v.size(); j += i) {
                ans.add(v.subList(j, j + i));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1282: Group the People Given the Group Size They Belong To
func groupThePeople(groupSizes []int) [][]int {
	n := len(groupSizes)
	g := make([][]int, n+1)
	for i, v := range groupSizes {
		g[v] = append(g[v], i)
	}
	ans := [][]int{}
	for i, v := range g {
		for j := 0; j < len(v); j += i {
			ans = append(ans, v[j:j+i])
		}
	}
	return ans
}

# Accepted solution for LeetCode #1282: Group the People Given the Group Size They Belong To
class Solution:
    def groupThePeople(self, groupSizes: List[int]) -> List[List[int]]:
        g = defaultdict(list)
        for i, v in enumerate(groupSizes):
            g[v].append(i)
        return [v[j : j + i] for i, v in g.items() for j in range(0, len(v), i)]

// Accepted solution for LeetCode #1282: Group the People Given the Group Size They Belong To
impl Solution {
    pub fn group_the_people(group_sizes: Vec<i32>) -> Vec<Vec<i32>> {
        let n: usize = group_sizes.len();
        let mut g: Vec<Vec<usize>> = vec![Vec::new(); n + 1];

        for (i, &size) in group_sizes.iter().enumerate() {
            g[size as usize].push(i);
        }

        let mut ans: Vec<Vec<i32>> = Vec::new();
        for (i, v) in g.into_iter().enumerate() {
            for j in (0..v.len()).step_by(i.max(1)) {
                ans.push(
                    v[j..(j + i).min(v.len())]
                        .iter()
                        .map(|&x| x as i32)
                        .collect(),
                );
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #1282: Group the People Given the Group Size They Belong To
function groupThePeople(groupSizes: number[]): number[][] {
    const n: number = groupSizes.length;
    const g: number[][] = Array.from({ length: n + 1 }, () => []);

    for (let i = 0; i < groupSizes.length; i++) {
        const size: number = groupSizes[i];
        g[size].push(i);
    }
    const ans: number[][] = [];
    for (let i = 1; i <= n; i++) {
        const group: number[] = [];
        for (let j = 0; j < g[i].length; j += i) {
            group.push(...g[i].slice(j, j + i));
            ans.push([...group]);
            group.length = 0;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.