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.


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
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
    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)        
    return (n <= end && start==0);
// Driver code
int main()
    vector<pair<int, int> > vec;    
    if (checkPathPairs(7,vec))
        cout << "YES";
        cout << "NO";
    return 0;




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