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")
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