Second Max of Three Numbers CodeChef Solution

Problem – Second Max of Three Numbers CodeChef Solution

Problem Statement

Write a program that accepts sets of three numbers, and prints the second-maximum number among the three.

Input

• First line contains the number of triples, N.
• The next N lines which follow each have three space separated integers.

Output

For each of the N triples, output one new line which contains the second-maximum integer among the three.

Constraints

• 1 ≤ N ≤ 6
• 1 ≤ every integer ≤ 10000
• The three integers in a single triplet are all distinct. That is, no two of them are equal.

Sample 1:

Input: 3
1 2 3
10 15 5
100 999 500
Output: 2
10
500

Second Max of Three Numbers CodeChef Solution in C++17

#include <iostream>
using namespace std;

void solve(){
    int a,b,c;
    cin>>a>>b>>c;
    if(a>b && a<c){
        cout<<a<<"\n";
    }else if(a>c && a<b){
        cout<<a<<"\n";
    }else if(b>a && b<c){
        cout<<b<<"\n";
    }else if(b>c && b<a){
        cout<<b<<"\n";
    }else if(c>a && c<b){
        cout<<c<<"\n";
    }else if(c>b && c<a){
        cout<<c<<"\n";
    }
}

int main ()
{
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

Second Max of Three Numbers CodeChef Solution in Python3

n=int(input())
for i in range(n):
    x,y,z=map(int,input().split())
    a=max(x,y,z)
    b=min(x,y,z)
    if x<a and x>b:
        print(x)
    elif y<a and y>b:
        print(y)
    elif z<a and z>b:
        print(z)

Second Max of Three Numbers CodeChef Solution in Java

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

class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		for (int i = 1; i <= t; i++) {
		    int a =sc.nextInt();
		    int b = sc.nextInt();
		    int c = sc.nextInt();
		    
		    if((a > b && a < c) || (a > c && a < b))
		    System.out.println(a);
		    else if ((b > a && b < c) || (b >c && b < a))
		    System.out.println(b);
		    else System.out.println(c);  
		}
	}
}

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