Most Frequent Number in Intervals - Facebook Top Interview Questions

Problem Statement :

You are given a list of list of integers intervals where each element contains the inclusive interval [start, end].

Return the most frequently occurring number in the intervals. If there are ties, return the smallest number.


n ≤ 100,000 where n is the length of intervals

Example 1


intervals = [

    [1, 4],

    [3, 5],

    [6, 9],

    [7, 9]





The most frequent numbers are [3, 4, 7, 8, 9] and they all occur twice but 3 is the smallest one.

Solution :

                        Solution in C++ :

int solve(vector<vector<int>>& intervals) {
    map<int, int> counts;
    for (const auto& i : intervals) {
        counts[i[1] + 1]--;

    int level = 0;
    int max_level = 0;
    int ans = 0;

    for (const auto& p : counts) {
        level += p.second;
        if (level > max_level) {
            max_level = level;
            ans = p.first;

    return ans;

                        Solution in Java :

import java.util.*;

class Solution {
    private class Node {
        int val;
        int state;
        public Node(int v, int state) {
            this.val = v;
            this.state = state;
    private int res;
    private int ans = -1;
    private int count = 0;
    List<Node> list = new ArrayList();
    public int solve(int[][] intervals) {
        for (int[] interval : intervals) {
            list.add(new Node(interval[0], 0));
            list.add(new Node(interval[1], 1));
        Collections.sort(list, (n1, n2) -> {
            if (n1.val == n2.val)
                return n1.state - n2.state;
            return n1.val - n2.val;
        for (int i = 0; i < list.size(); i++) {
            Node node = list.get(i);
            int diff = (node.state == 0) ? 1 : -1;
            count += diff;
            if (count > ans) {
                ans = count;
                res = node.val;
        return res;

                        Solution in Python : 
class Solution:
    def solve(self, intervals):
        OPEN, CLOSE = 0, 1

        events = []
        for s, e in intervals:
            events.append([s, OPEN])
            events.append([e, CLOSE])

        active = max_active = 0
        for x, cmd in events:
            active += 1 if cmd == OPEN else -1
            max_active = max(max_active, active)

        for x, cmd in events:
            active += 1 if cmd == OPEN else -1
            if active == max_active:
                return x

        return 0

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