Count Contained Intervals - Amazon Top Interview Questions

Problem Statement :

You are given a two-dimensional list of integers intervals. Each value is unique and contains an inclusive interval [start, end]. Return the number of intervals are contained by another interval. An interval that's contained by multiple other intervals should only be counted once.

An interval [s0, e0] contains another interval [s1, e1] if s0 ≤ s1 and e0 ≥ e1.


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

Example 1


intervals = [
    [1, 5],
    [2, 3],
    [3, 6],
    [4, 4]




[2, 3] and [4, 4] are contained by [1, 5]. [4, 4] is also contained by [3, 6] but we don't count it again.

Example 2


intervals = [
    [1, 2],
    [2, 3],
    [1, 5]




Both [1, 2] and [2, 3] are contained by [1, 5]

Solution :


                        Solution in C++ :

int solve(vector<vector<int>>& nums) {
    sort(nums.begin(), nums.end(), [](vector<int>& a, vector<int>& b) {
        if (a[0] == b[0]) return a[1] > b[1];
        return a[0] < b[0];
    int count = 0, pe = INT_MIN;
    for (auto& n : nums) {
        if (n[1] <= pe) {
            count += 1;
        } else {
            pe = n[1];
    return count;

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

        for i, (start, end) in enumerate(intervals):
            events.append((start, OPEN, end, i))
            events.append((end, CLOSE, end, i))

        events.sort(key=lambda x: (x[0], -x[2]))

        ans = 0
        balance = 0
        max_end = float("-inf")

        for time, event_type, end, idx in events:
            if event_type == OPEN and end <= max_end:
                ans += 1

            if event_type == OPEN:
                max_end = max(max_end, end)

        return ans

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