Longest Subarray: CHEFSUB(Assignment)
link: https://www.codechef.com/CU1AP0009/problems/CHEFSUB
You are given an array A of N integers: A1, A2, …, 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 Ai, Ai+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 A1, A2, …, 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
Answer: Please login to see answer.
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.
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;
}