Geeks#1-Missing number in array

Problem:

Given an array C of size N-1 and given that there are numbers from 1 to N with one element missing, the missing number is to be found.

Input:
The first line of input contains an integer T denoting the number of test cases. For each test case first line contains N(size of array). The subsequent line contains N-1 array elements.

Output:
Print the missing number in array.

Constraints:
1 ≤ T ≤ 200
1 ≤ N ≤ 107
1 ≤ C[i] ≤ 107

Example:
Input:
2
5
1 2 3 5
10
1 2 3 4 5 6 7 8 10

Output:
4
9

Explanation:
Testcase 1: Given array : 1 2 3 5. Missing element is 4.

Solution:

import java.util.Arrays;

import java.util.Scanner;

public class MissinNumbers {

public static void main(String[] args) {

Scanner in =new Scanner(System.in);

long i=in.nextLong();

boolean flag=true;

while(i-->0) {

int j = in.nextInt();

long sum=0;

int arr[] =new int[j];

for(int l=0;l<j-1;l++) {

sum+=in.nextInt();

}

long nan = j*(j+1)/2;

System.out.println(nan -sum);

}

}

}