Merge interval pattern: Check if two intervals overlap

Problem Statement:

You are given an array, you need to check if any two intervals overlap

Example:

Input: arr [] = [[1, 7], [2, 8], [3, 6]]

Output: Yes

Solution Explanation: Sorting approach

Sort the given intervals in increasing order of start time.
Then for each interval check if the start time is less than or equal to the end time of previous interval and return true.

Time Complexity: O(n logn)
Space Complexity: O(1)

Code Solution

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

void solution(vector<vector<int>>& arr)
{
	sort(arr.begin(), arr.end());

	for (int i = 1; i < arr.size(); i++) {
        if (arr[i][0] <= arr[i - 1][1]){
        	cout<<"True";
            return;
        }
    }
    
    cout<<"False";
    return;
}

int main()
{
    vector<vector<int>> arr = {{1, 7}, {2, 8}, {3, 6}};
    solution(arr);
}

Output

True