Akash and Equal Mean Solution | CodeChef Problem Solution 2022

Akash and Equal Mean Solution | CodeChef Problem Solution 2022 

Akash goes to class everyday. His classes are held on the 9th$9^{th}$ floor so he uses a lift. One day N$N$ people get on the lift. However, since the lift can carry at most N−2$N-2$ people, 2$2$ of them need to get off.

Given an array A$A$ of the weights of the N$N$ people, Akash wonders how many ways of removing 2$2$ people exist such that the mean of weights of the people in the lift remains constant. In other words, the mean of weights of the N−2$N-2$ people remaining in the lift must be the same as the mean of weights of all the N$N$ people who were initially on the lift. Help Akash find how many such ways exist.

Input Format

• The first line contains a single integer T$T$ - the number of test cases. Then the test cases follow.
• The first line of each test case contains an integer N$N$ - the number of people who get on the lift initially.
• The second line of each test case contains N$N$ space-separated integers A1,A2,…,AN$A_1,A_2,\dots ,A_N$ denoting the weights of the N$N$ people on the lift.

Output Format

For each test case, print a single integer denoting the number of ways of removing 2$2$ people such that the mean reamins the same.

Constraints

• 1≤T≤1000$1\le T\le 1000$
• 3≤N≤105$3\le N\le {10}^5$
• 1≤Ai≤109$1\le A_i\le {10}^9$
• Sum of N$N$ over all test cases does not exceed 2⋅105$2\cdot {10}^5$

Sample Input 1 

3
4
1 3 5 7
4
1 7 7 7
6
1 1 1 1 1 1

Sample Output 1 

2
0
15

Explanation

Test case 1$1$: The mean of weights of all N$N$ students is 1+3+5+74=164=4$\frac{1+3+5+7}{4}=\frac{16}{4}=4$.

• Consider removing the people with weights (1,7)$(1,7)$. The mean of weights of the N−2$N-2$ people remaining in the lift is 3+52=82=4$\frac{3+5}{2}=\frac{8}{2}=4$, which is same as the mean of all N$N$ people.
• Consider removing the people with weights (3,5)$(3,5)$. The mean of weights of the N−2$N-2$ people remaining in the lift is 1+72=82=4$\frac{1+7}{2}=\frac{8}{2}=4$, which is same as the mean of all N$N$ people.

There is no other way to remove 2$2$ people such that the mean reamins the same. Hence the answer is 2$2$.

Test case 2$2$: The mean of weights of all N$N$ students is 1+7+7+74=5.5$\frac{1+7+7+7}{4}=5.5$. It can be seen that there is no way to remove 2$2$ people such that the mean reamins the same. Hence the answer is 0$0.$

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