# Factor of 3 Codevita 9 Solutions

## Factor of 3 Codevita 9 Solutions

Given an array arr, of size N, find whether it is possible to rearrange the elements of array such that sum of no two adjacent elements is divisible by 3.

Constraints

1 <= T <= 10
2 <= N <= 10^5
1 <= arr[i] <= 10^5

Input

First line contains integer T denoting the number of testcases. Each test cases consists of 2 lines as follows- First line contains integer N denoting the size of the array. Second line contains N space separated integers.

Output

For each test case print either “Yes” or “No” (without quotes) on new line.

Time Limit

1

Examples

Example 1
Input
1
4
1 2 3 3
Output
Yes

Explanation

Some of the rearrangements can be {2,1,3,3}, {3,3,1,2}, {2,3,3,1}, {1,3,3,2},… We can see that there exist at least 1 combination {3,2,3,1} where
sum of 2 adjacent number is not divisible by 3. Other combinations
can be {1,3,2,3}, {2,3,1,3}. Hence the output is Yes.

Example 2
Input
1
4
3 6 1 9
Output
No

Explanation

All possible combination of {3,6,1,9} are
{1,3,6,9}, {1,3,9,6}, {1,6,9,3}, {1,6,3,9}, {1,9,3,6}, {1,9,6,3}, {6,1,3,9}, {6,1, 9,3}, {6,3,1,9}, {6,3,9,1}, {6,9,1,3}, {6,9,3,1}, {3,1,6,9}, {3,1,9,6}, {3,9,1,6}, {3,9,6,1}, {3,6,1,9}, {3,6,9,1}, {9,1,3,6}, {9,1,6,3}, {9,3,1,6}, {9,3,6,1}, {9,6,1,3}, {9,6,3,1}.

Since none of these combinations satisfy the condition, the output is No

### SOLUTION

Program: Factor of 3 Codevita 9 Solutions in Python

``````for i in range (int(input("Enter Test Case"))):
N = int(input())
list1 = list(map(int,input().split()))
array = []
for i in range(N):
array.append(list1[i]%3)
x = array.count(0)
y = array.count(1)
z = array.count(2)
if x == 0 and y != 0 and z !=0:
print("NO")
elif x == 0 and y == 0 and z !=0:
print("YES")
elif x == 0 and y != 0 and z ==0:
print("YES")
elif x<=(z+y):
print("YES")
else:
print("No")``````