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# Binary Prefix Divisible By 5 in Python

Suppose we have an array A of 0s and 1s, consider N[i] is the i-th subarray from index A[0] to A[i] interpreted as a binary number. We have to find a list of boolean answers, where answer[i] is true if and only if N[i] is divisible by 5.

So, if the input is like [0,1,1,1,1,1], then the output will be [true,false,false,false,true,false]

To solve this, we will follow these steps −

- length := size of A
- ans:= make an array of size length, and fill with false
- number:= a binary value by concatenating each element from A
- for i in range 0 to length, do
- if number mod 5 is same as 0, then
- ans[length-i-1] := True

- number:= number / 2

- if number mod 5 is same as 0, then
- return ans

Let us see the following implementation to get better understanding −

## Example

class Solution: def prefixesDivBy5(self, A): length=len(A) ans=[False]*length number=int("".join(map(str,A)),2) for i in range(length): if number%5==0: ans[length-i-1]=True number=number>>1 return ans ob = Solution() print(ob.prefixesDivBy5([0,1,1,1,1,1]))

## Input

[0,1,1,1,1,1]

## Output

[True, False, False, False, True, False]

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