Given an array of size N consisting of only 0's and 1's ,which is sorted in such a manner that all the 1's are placed first and then they are followed by all the 0's. You have to find the count of all the 0's.
Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow.
The first line of each test case contains an integer N, where N is the size of the array A[ ].
The second line of each test case contains N space separated integers of all 1's follwed by all the 0's, denoting elements of the array A[ ].
Output:
Print out the number of 0's in the array.
Constraints:
1 <= T <= 100
1 <= N <= 30
0 <= A[i] <= 1
Example :
Input:
3
12
1 1 1 1 1 1 1 1 1 0 0 0
5
0 0 0 0 0
6
1 1 1 1 1 1
Output:
3
5
0
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def counter(arr,n):
count = 0
for i in arr:
if(i==0):
count +=1
return count
t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int,input().split()))
print(counter(arr,n))
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