Longest Subarray: CHEFSUB(Assignment)

link: https://www.codechef.com/CU1AP0009/problems/CHEFSUB

You are given an array A of N integers: A1A2, …, AN. You need to find a longest contiguous subarray in the array such that each integer in this subarray is an even integer, and output its length.

A contiguous subarray is of the form AiAi+1, …, Aj, for some i and j.

Input

  • The first line of the input contains an integer T denoting the number of test cases. The description of each test case follows.
  • The first line of each test case contains a single integer N denoting the number of elements in the given array.
  • The second line contains N space-separated integers A1A2, …, AN denoting the array A.

Output

For each test case, output a single line containing the answer.

Constraints

  • 1 ≤ T ≤ 10
  • 1 ≤ N ≤ 105
  • 0 ≤ A1, A2, … , AN ≤ 105

Information to score partial points

  • For 16% of the score, it is guaranteed that N ≤ 100.
  • For further 16% of the score, it is guaranteed that N ≤ 1000.
  • For the rest of the 68% of the score, no extra guarantees. That is, N ≤ 105.

Example

Input:
3
4
1 2 2 4
3
2 4 6
5
2 3 2 2 5

Output:
3
3
2

Explanation

Testcase 1: The longest contiguous subarray that has all its elements even will be the subarray consisting of the 3 elements [2, 2, 4]. Hence, the answer will be 3.

Testcase 2: The longest contiguous subarray that has all its elements even will be the subarray consisting of the 3 elements [2, 4, 6]. Hence, the answer will be 3.

Testcase 3: The longest contiguous subarray that has all its elements even will be the subarray consisting of the 2 elements [2, 2]. Hence, the answer will be 2.

See also  Which Division: WHICHDIV

Solution:

#include <bits/stdc++.h>
using namespace std;

bool is_even(int a)
{
    if (a % 2 == 0)
    {
        return true;
    }
    else
    {
        return false;
    }
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        int a[n];
        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
        }

        // sort(a, a + n);
        int count = 0;
        int ans = -1 ;

        for (int i = 0; i < n; i++)
        {
            /* code */
            if (is_even(a[i]) )
            {
                count++;
                ans = max(count ,ans);
            }
            else{
                count = 0;
            }
        }

        cout << ans << endl;
    }

    return 0;
}