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++
// 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 pointsbool 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 codeint 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
# Python3 implementation of above idea# function to check if it is possible to# reach the end through given pointsdef 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 Codeif __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#
// C# implementation of above ideausing 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"); } }} |
Javascript
<script>// JavaScript program to implement above approach// function to check if it is possible to// reach the end through given pointsfunction 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> |
Java
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"); }} |
Output:
NO
Time Complexity: O(n log n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
