# Count all prefixes of the given binary array which are divisible by x

Given a binary array arr[] and an integer x, the task is to count all the prefixes of the given array which are divisible by x.
Note: The ith prefix from arr to arr[i] is interpreted as a binary number (from most-significant-bit to least-significant-bit.)

Examples:

Input: arr[] = {0, 1, 0, 1, 1}, x = 5
Output: 2
0 = 0
01 = 1
010 = 2
0101 = 5
01011 = 11
0 and 0101 are the only prefixes divisible by 5.

Input: arr[] = {1, 0, 1, 0, 1, 1, 0}, x = 2
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Iterate from 0 to i to convert each binary prefix to decimal and check whether the number is divisible by x or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of total ` `// binary prefix which are divisible by x ` `int` `CntDivbyX(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` `  `    ``// Initialize with zero ` `    ``int` `number = 0; ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Convert all prefixes to decimal ` `        ``number = number * 2 + arr[i]; ` ` `  `        ``// If number is divisible by x ` `        ``// then increase count ` `        ``if` `((number % x == 0)) ` `            ``count += 1; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 0, 1, 0, 1, 1, 0 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `x = 2; ` `    ``cout << CntDivbyX(arr, n, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GfG ` `{ ` ` `  `    ``// Function to return the count of total  ` `    ``// binary prefix which are divisible by x  ` `    ``static` `int` `CntDivbyX(``int` `arr[], ``int` `n, ``int` `x)  ` `    ``{  ` `     `  `        ``// Initialize with zero  ` `        ``int` `number = ``0``;  ` `        ``int` `count = ``0``;  ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `     `  `            ``// Convert all prefixes to decimal  ` `            ``number = number * ``2` `+ arr[i];  ` `     `  `            ``// If number is divisible by x  ` `            ``// then increase count  ` `            ``if` `((number % x == ``0``))  ` `                ``count += ``1``;  ` `        ``}  ` `     `  `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `         `  `        ``int` `arr[] = { ``1``, ``0``, ``1``, ``0``, ``1``, ``1``, ``0` `};  ` `        ``int` `n = arr.length;  ` `        ``int` `x = ``2``;  ` `        ``System.out.println(CntDivbyX(arr, n, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by Rituraj Jain `

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the count of total ` `# binary prefix which are divisible by x ` `def` `CntDivbyX(arr, n, x): ` `     `  `    ``# Initialize with zero ` `    ``number ``=` `0` `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Convert all prefixes to decimal ` `        ``number ``=` `number ``*` `2` `+` `arr[i] ` ` `  `        ``# If number is divisible by x ` `        ``# then increase count ` `        ``if` `((number ``%` `x ``=``=` `0``)): ` `            ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``1``, ``0``, ``1``, ``0``, ``1``, ``1``, ``0``] ` `    ``n ``=` `len``(arr) ` `    ``x ``=` `2` `    ``print``(CntDivbyX(arr, n, x)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GfG ` `{ ` ` `  `    ``// Function to return the count of total  ` `    ``// binary prefix which are divisible by x  ` `    ``static` `int` `CntDivbyX(``int``[] arr, ``int` `n, ``int` `x)  ` `    ``{  ` `     `  `        ``// Initialize with zero  ` `        ``int` `number = 0;  ` `        ``int` `count = 0;  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `     `  `            ``// Convert all prefixes to decimal  ` `            ``number = number * 2 + arr[i];  ` `     `  `            ``// If number is divisible by x  ` `            ``// then increase count  ` `            ``if` `((number % x == 0))  ` `                ``count += 1;  ` `        ``}  ` `     `  `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `         `  `        ``int``[] arr = { 1, 0, 1, 0, 1, 1, 0 };  ` `        ``int` `n = arr.Length;  ` `        ``int` `x = 2;  ` `        ``Console.WriteLine(CntDivbyX(arr, n, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## PHP

 `

Output:

```3
```

Efficient Approach: As we see in the above approach we convert each binary prefix to decimal number like 0, 01, 010, 0101…. but as the value of n(size of array) increases then the resultant number will be very large and no. will be out of range of data type so we can make use of the modular properties.
Instead of doing number = number * 2 + arr[ i ] , we can do better as number = (number * 2 + arr[ i ] ) % x
Explanation: We start with number = 0 and repeatedly do number = number * 2 + arr[ i ] then in each iteration we’ll get a new term of the above sequence.

A = {1, 0, 1, 0, 1, 1, 0}
“1” = 0*2 + 1 = 1
“10” = 1*2 + 0 = 2
“101” = 2*2 + 1 = 5
“1010” = 5*2 + 0 = 10
“10101” = 10*2 + 1 = 21
“101011” = 21*2 + 1 = 43
“1010110” = 43*2 + 0 =86

Since we are repeatedly taking the remainders of the number at each step, at each step we have, newNum = oldNum * 2 + arr[i] .By the rules of modular arithmetic (a * b + c) % m is same as ((a * b) % m + c % m) % m. So, it doesn’t matter whether oldNum is the remainder or the original number, the answer would be correct.
Note: Similar article discussed here.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of total ` `// binary prefix which are divisible by x ` `int` `CntDivbyX(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` `  `    ``// Initialize with zero ` `    ``int` `number = 0; ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Instead of converting all prefixes ` `        ``// to decimal, take reminder with x ` `        ``number = (number * 2 + arr[i]) % x; ` ` `  `        ``// If number is divisible by x ` `        ``// then reminder = 0 ` `        ``if` `(number == 0) ` `            ``count += 1; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 1, 0, 1, 0, 1, 1, 0 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `x = 2; ` `    ``cout << CntDivbyX(arr, n, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the count of total ` `    ``// binary prefix which are divisible by x ` `    ``public` `static` `int` `CntDivbyX(``int` `arr[], ``int` `n, ``int` `x) ` `    ``{ ` ` `  `        ``// Initialize with zero ` `        ``int` `number = ``0``; ` `        ``int` `count = ``0``; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``// Instead of converting all prefixes ` `            ``// to decimal, take reminder with x ` `            ``number = (number * ``2` `+ arr[i]) % x; ` ` `  `            ``// If number is divisible by x ` `            ``// then reminder = 0 ` `            ``if` `(number == ``0``) ` `                ``count += ``1``; ` `        ``} ` ` `  `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``0``, ``1``, ``0``, ``1``, ``1``, ``0` `}; ` `        ``int` `n = ``7``; ` `        ``int` `x = ``2``; ` `        ``System.out.print(CntDivbyX(arr, n, x)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of total ` `# binary prefix which are divisible by x ` `def` `CntDivbyX(arr, n, x): ` ` `  `    ``number ``=` `0` ` `  `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range` `(``0``, n): ` `         `  `        ``# Instead of converting all prefixes  ` `        ``# to decimal, take reminder with x ` `        ``number ``=` `( number ``*` `2` `+` `arr[i] ) ``%` `x ` `     `  `        ``# If number is divisible by x  ` `        ``# then reminder = 0 ` `        ``if` `number ``=``=` `0``: ` `            ``count ``+``=` `1` `     `  `    ``return` `count ` ` `  `# Driver code ` `arr ``=` `[``1``, ``0``, ``1``, ``0``, ``1``, ``1``, ``0``] ` `n ``=` `7` `x ``=` `2` `print``( CntDivbyX(arr, n, x) ) `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the count of total ` `    ``// binary prefix which are divisible by x ` `    ``public` `static` `int` `CntDivbyX(``int` `[]arr, ``int` `n, ``int` `x) ` `    ``{ ` ` `  `        ``// Initialize with zero ` `        ``int` `number = 0; ` `        ``int` `count = 0; ` ` `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` ` `  `            ``// Instead of converting all prefixes ` `            ``// to decimal, take reminder with x ` `            ``number = (number * 2 + arr[i]) % x; ` ` `  `            ``// If number is divisible by x ` `            ``// then reminder = 0 ` `            ``if` `(number == 0) ` `                ``count += 1; ` `        ``} ` ` `  `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = { 1, 0, 1, 0, 1, 1, 0 }; ` `        ``int` `n = 7; ` `        ``int` `x = 2; ` `         `  `        ``Console.Write(CntDivbyX(arr, n, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` `

Output:

```3
```

Time Complexity: O(N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.