The Smallest Pair Codechef Solution

Hello coders, today we are going to solve The Smallest Pair CodeChef Solution whose Problem Code is SMPAIR.

Problem

You are given a sequence a1, a2, …, aN. Find the smallest possible value of ai + aj, where 1 ≤ i < j ≤ N.

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. 

The first line of each description consists of a single integer N.

The second line of each description contains N space separated integers – a1, a2, …, aN respectively.

Output

For each test case, output a single line containing a single integer – the smallest possible sum for the corresponding test case.

Constraints

• T = 105, N = 2 : 13 points.
• T = 105, 2 ≤ N ≤ 10 : 16 points.
• T = 1000, 2 ≤ N ≤ 100 : 31 points.
• T = 10, 2 ≤ N ≤ 105 : 40 points.
• 1 ≤ ai ≤ 106

Example

Sample Input 

1
4
5 1 3 4

Sample Output

4

Explanation

Here we pick a2 and a3. Their sum equals to 1 + 3 = 4.

The Smallest Pair CodeChef Solution In Python

T = int(input())
for i in range(T):
    n = int(input())
    s = list(map(int, input().split()))
    s.sort()
    print(s[0] + s[1])

The Smallest Pair CodeChef Solution In Cpp

#include <iostream>
using namespace std;
int main(){
	int a;
	long b,c,d,low1,low2;
	cin >> a;
	while(a--){
		low1=low2=100000000;
		cin >> b;
		while(b--){
			cin >> c;
			if(c<low1){
				low2=low1;
				low1=c;
			}
			else if(c<low2)
				low2=c;
		}
		cout << low1+low2 << endl;
	}
}

The Smallest Pair CodeChef Solution In JAva

import java.io.FileNotFoundException;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) throws FileNotFoundException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        int a;
        int m1, m2;
        while(t-- > 0) {
            int n = sc.nextInt();
            m1 = Integer.MAX_VALUE;
            m2=Integer.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                a = sc.nextInt();
                if(a < m1) {
                    m2 = m1;
                    m1 = a;
                } else if(a >= m1 && a < m2) {
                    m2 = a;
                }
            }
            System.out.println(m1 + m2);
        }
        sc.close();
    }
}

Disclaimer: The above Problem (The Smallest Pair) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.