Airline Restrictions | CodeChef Solution

Hello coders, today we are going to solve Airline Restrictions CodeChef Solution whose Problem Code is AIRLINE.

Task

Chef has 3 bags that she wants to take on a flight. They weigh A, B, and C kgs respectively. She wants to check-in exactly two of these bags and carry the remaining one bag with her.

The airline restrictions says that the total sum of the weights of the bags that are checked-in cannot exceed D kgs and the weight of the bag which is carried cannot exceed E kgs. Find if Chef can take all the three bags on the flight.

Input Format

• The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
• Each testcase contains a single line of input, five space separated integers A, B, C, D, E.

Output Format

For each testcase, output in a single line answer “YES” if Chef can take all the three bags with her or “NO” if she cannot.

You may print each character of the string in uppercase or lowercase (for example, the strings “yEs”, “yes”, “Yes” and “YES” will all be treated as identical).

Constraints

• 1 ≤ T ≤ 36000
• 1 ≤ A, B, C ≤ 10
• 15 ≤ D ≤ 20
• 5 ≤ E ≤ 10

Subtasks

Subtask #1 (100 points): original constraints

Sample Input 1

3
1 1 1 15 5
8 7 6 15 5
8 5 7 15 6

Sample Output 1

YES
NO
YES

Explanation

Test case 1: Chef can check-in the first and second bag (since 1 + 1 = 2 ≤ 15) and carry the third bag with her (since 1 ≤ 5).

Test case 2: None of the three bags can be carried in hand without violating the airport restrictions.

Test case 3: Chef can check-in the first and the third bag (since 8 + 7 ≤ 15) and carry the second bag with her (since 5 ≤ 6).

Solution – Airline Restrictions

C++

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t; cin>>t;
	while(t--){
	    int a,b,c,d,e; 
	    cin >> a >> b >> c >> d >> e;
	     if(a<=e || b<=e || c<=e){
		        if(a+b <= d && c<=e)cout << "YES\n";
		        else if(b+c<=d && a<=e)cout << "YES\n";
		        else if(a+c<=d && b<=e)cout << "YES\n";
		        else cout << "NO\n";
		    }
		    else cout << "NO\n";
		
	}
	return 0;
}

Python

# Airline Restrictions | CodeChef Solution 
T = int(input())
for i in range(T):
    a, b, c, d, e = map(int, input().split())
    if (((a + b) <= d and c <= e) or ((b + c) <= d and a <= e) or ((a + c) <= d and b <= e)):
        print("Yes")
    else:
        print("No")

Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-- > 0){
		    int a = sc.nextInt();
		    int b = sc.nextInt();
		    int c = sc.nextInt();
		    int d = sc.nextInt();
		    int e = sc.nextInt();
		    
		    if(a<=e || b<=e || c<=e){
		        if(a+b <= d && c<=e)System.out.println("YES");
		        else if(b+c<=d && a<=e)System.out.println("YES");
		        else if(a+c<=d && b<=e)System.out.println("YES");
		        else System.out.println("NO");
		    }
		    else System.out.println("NO");
		}
		
	}
}

Disclaimer: The above Problem (Airline Restrictions) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.