LeetCode #2406 — MEDIUM

Divide Intervals Into Minimum Number of Groups

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

The Problem

Problem Statement

You are given a 2D integer array intervals where intervals[i] = [left_i, right_i] represents the inclusive interval [left_i, right_i].

You have to divide the intervals into one or more groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other.

Return the minimum number of groups you need to make.

Two intervals intersect if there is at least one common number between them. For example, the intervals [1, 5] and [5, 8] intersect.

Example 1:

Input: intervals = [[5,10],[6,8],[1,5],[2,3],[1,10]]
Output: 3
Explanation: We can divide the intervals into the following groups:
- Group 1: [1, 5], [6, 8].
- Group 2: [2, 3], [5, 10].
- Group 3: [1, 10].
It can be proven that it is not possible to divide the intervals into fewer than 3 groups.

Example 2:

Input: intervals = [[1,3],[5,6],[8,10],[11,13]]
Output: 1
Explanation: None of the intervals overlap, so we can put all of them in one group.

Constraints:

1 <= intervals.length <= 10⁵
intervals[i].length == 2
1 <= left_i <= right_i <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array intervals where intervals[i] = [lefti, righti] represents the inclusive interval [lefti, righti]. You have to divide the intervals into one or more groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other. Return the minimum number of groups you need to make. Two intervals intersect if there is at least one common number between them. For example, the intervals [1, 5] and [5, 8] intersect.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Greedy

Example 1

[[5,10],[6,8],[1,5],[2,3],[1,10]]

Example 2

[[1,3],[5,6],[8,10],[11,13]]

Core Insight

What unlocks the optimal approach

Can you find a different way to describe the question?
The minimum number of groups we need is equivalent to the maximum number of intervals that overlap at some point. How can you find that?

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 #2406: Divide Intervals Into Minimum Number of Groups
class Solution {
    public int minGroups(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        PriorityQueue<Integer> q = new PriorityQueue<>();
        for (var e : intervals) {
            if (!q.isEmpty() && q.peek() < e[0]) {
                q.poll();
            }
            q.offer(e[1]);
        }
        return q.size();
    }
}

// Accepted solution for LeetCode #2406: Divide Intervals Into Minimum Number of Groups
func minGroups(intervals [][]int) int {
	sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
	q := hp{}
	for _, e := range intervals {
		if q.Len() > 0 && q.IntSlice[0] < e[0] {
			heap.Pop(&q)
		}
		heap.Push(&q, e[1])
	}
	return q.Len()
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

# Accepted solution for LeetCode #2406: Divide Intervals Into Minimum Number of Groups
class Solution:
    def minGroups(self, intervals: List[List[int]]) -> int:
        q = []
        for left, right in sorted(intervals):
            if q and q[0] < left:
                heappop(q)
            heappush(q, right)
        return len(q)

// Accepted solution for LeetCode #2406: Divide Intervals Into Minimum Number of Groups
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2406: Divide Intervals Into Minimum Number of Groups
// class Solution {
//     public int minGroups(int[][] intervals) {
//         Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
//         PriorityQueue<Integer> q = new PriorityQueue<>();
//         for (var e : intervals) {
//             if (!q.isEmpty() && q.peek() < e[0]) {
//                 q.poll();
//             }
//             q.offer(e[1]);
//         }
//         return q.size();
//     }
// }

// Accepted solution for LeetCode #2406: Divide Intervals Into Minimum Number of Groups
function minGroups(intervals: number[][]): number {
    intervals.sort((a, b) => a[0] - b[0]);
    const q = new PriorityQueue({ compare: (a, b) => a - b });
    for (const [left, right] of intervals) {
        if (!q.isEmpty() && q.front() < left) {
            q.dequeue();
        }
        q.enqueue(right);
    }
    return q.size();
}

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 × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

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.