Find if it is possible to reach the end through given transitions

Given, n points on X-axis and the list of allowed transition between the points. Find if it is possible to reach the end from starting point through these transitions only. 

Note: If there is a transition between points x1 and x2, then you can move from point x to any intermediate points between x1 and x2 or directly to x2. 

Examples:

Input :  n = 5 ,  
Transitions allowed: 0 -> 2
                     2 -> 4
                     3 -> 5
Output : YES
Explanation : We can move from 0 to 5 using the 
allowed transitions. 0->2->3->5

Input : n = 7 ,  
Transitions allowed: 0 -> 4
                     2 -> 5
                     6 -> 7
Output : NO
Explanation : We can't move from 0 to 7 as there is 
no transition between 5 and 6.  

The idea to solve this problem is to first sort this list according to first element of the pairs. Then start traversing from the second pair of the list and check if the first element of this pair is in between second element of previous pair and second element of current pair or not. This condition is used to check if there is a path between two consecutive pairs. At the end check if the point we have reached is the destination point and the point from which we have started is start point. If so, print YES otherwise print NO. 

// C++ implementation of above idea
#include<bits/stdc++.h>
using namespace std;

// function to check if it is possible to 
// reach the end through given points
bool checkPathPairs(int n, vector<pair<int, int> > vec)
{   
    // sort the list of pairs 
    // according to first element
    sort(vec.begin(),vec.end());
    
    int start = vec[0].first;
    
    int end=vec[0].second;
    
    // start traversing from 2nd pair    
    for (int i=1; i<n; i++)
    {   
        // check if first element of current pair
        // is in between second element of previous
        // and current pair
        if (vec[i].first > end)        
            break;        
    
        end=max(end,vec[i].second);
    }
        
    return (n <= end && start==0);
}

// Driver code
int main()
{
    vector<pair<int, int> > vec;    
    vec.push_back(make_pair(0,4));
    vec.push_back(make_pair(2,5));
    vec.push_back(make_pair(6,7));
    
    if (checkPathPairs(7,vec))
        cout << "YES";
    else
        cout << "NO";
    
    return 0;
}

# Python3 implementation of above idea

# function to check if it is possible to
# reach the end through given points
def checkPathPairs(n: int, vec: list) -> bool:

    # sort the list of pairs
    # according to first element
    vec.sort(key = lambda a: a[0])

    start = vec[0][0]
    end = vec[0][1]

    # start traversing from 2nd pair
    for i in range(1, n):

        # check if first element of 
        # current pair is in between 
        # second element of previous 
        # and current pair
        if vec[i][1] > end:
            break

        end = max(end, vec[i][1])

    return (n <= end and start == 0)

# Driver Code
if __name__ == "__main__":
    vec = []
    vec.append((0, 4))
    vec.append((2, 5))
    vec.append((6, 7))

    if checkPathPairs(7, vec):
        print("YES")
    else:
        print("NO")

# This code is contributed by
# sanjeev2552

// C# implementation of above idea
using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
    // function to check if it is possible to 
    // reach the end through given points
    public static bool CheckPathPairs(int n, List<(int, int)> vec)
    {
        // sort the list of pairs 
        // according to first element
        vec = vec.OrderBy(p => p.Item1).ToList();

        int start = vec[0].Item1;
        int end = vec[0].Item2;

        // start traversing from 2nd pair    
        for (int i = 1; i < n; i++)
        {
            // check if first element of current pair
            // is in between second element of previous
            // and current pair
            if (vec[i].Item1 > end)
            {
                break;
            }

            end = Math.Max(end, vec[i].Item2);
        }

        return (n <= end && start == 0);
    }

    // Driver code
    public static void Main(string[] args)
    {
        List<(int, int)> vec = new List<(int, int)>();
        vec.Add((0, 4));
        vec.Add((2, 5));
        vec.Add((6, 7));

        if (CheckPathPairs(7, vec))
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }
    }
}

<script>

// JavaScript program to implement above approach

// function to check if it is possible to
// reach the end through given points
function checkPathPairs(n,vec){

    // sort the list of pairs
    // according to first element
    vec.sort((a,b)=>a-b)

    let start = vec[0][0]
    let end = vec[0][1]

    // start traversing from 2nd pair
    for(let i = 1; i < n; i++)
    {

        // check if first element of
        // current pair is in between
        // second element of previous
        // and current pair
        if(vec[i][1] > end)
            break

        end = Math.max(end, vec[i][1])
    }

    return (n <= end && start == 0)
}

// driver program
        
let vec = []
vec.push([0, 4])
vec.push([2, 5])
vec.push([6, 7])

if(checkPathPairs(7, vec))
    document.write("YES")
else
    document.write("NO")

// This code is contributed by shinjanpatra

</script>

import java.util.Arrays;

class Main {
    // function to check if it is possible to reach the end through given points
    public static boolean checkPathPairs(int n, int[][] vec) {
        // sort the list of pairs according to first element
        Arrays.sort(vec, (a, b) -> Integer.compare(a[0], b[0]));

        int start = vec[0][0];
        int end = vec[0][1];

        // start traversing from 2nd pair
        for (int i = 1; i < n; i++) {
            // check if first element of current pair is in between
            // second element of previous and current pair
            if (vec[i][1] > end)
                break;

            end = Math.max(end, vec[i][1]);
        }

        return (n <= end && start == 0);
    }

    public static void main(String[] args) {
        int[][] vec = new int[][] { { 0, 4 }, { 2, 5 }, { 6, 7 } };
        int n = 7;

        if (checkPathPairs(n, vec))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

NO

Time Complexity: O(n log n)
Auxiliary Space: O(1)