Count all prefixes of the given binary array which are divisible by x
Last Updated :
09 Jun, 2022
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[0] 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
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++
#include <bits/stdc++.h>
using namespace std;
int CntDivbyX(int arr[], int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++) {
number = number * 2 + arr[i];
if ((number % x == 0))
count += 1;
}
return count;
}
int main()
{
int arr[] = { 1, 0, 1, 0, 1, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 2;
cout << CntDivbyX(arr, n, x);
return 0;
}
|
Java
class GfG
{
static int CntDivbyX(int arr[], int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++)
{
number = number * 2 + arr[i];
if ((number % x == 0))
count += 1;
}
return count;
}
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));
}
}
|
Python3
def CntDivbyX(arr, n, x):
number = 0
count = 0
for i in range(n):
number = number * 2 + arr[i]
if ((number % x == 0)):
count += 1
return count
if __name__ == '__main__':
arr = [1, 0, 1, 0, 1, 1, 0]
n = len(arr)
x = 2
print(CntDivbyX(arr, n, x))
|
C#
using System;
class GfG
{
static int CntDivbyX(int[] arr, int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++)
{
number = number * 2 + arr[i];
if ((number % x == 0))
count += 1;
}
return count;
}
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));
}
}
|
PHP
<?php
function CntDivbyX($arr, $n, $x)
{
$number = 0;
$count = 0;
for ($i = 0; $i < $n; $i++)
{
$number = $number * 2 + $arr[$i];
if (($number % $x == 0))
$count += 1;
}
return $count;
}
$arr = array(1, 0, 1, 0, 1, 1, 0);
$n = sizeof($arr);
$x = 2;
echo CntDivbyX($arr, $n, $x);
|
Javascript
<script>
function CntDivbyX(arr, n, x)
{
let number = 0;
let count = 0;
for (let i = 0; i < n; i++) {
number = number * 2 + arr[i];
if ((number % x == 0))
count += 1;
}
return count;
}
let arr = [ 1, 0, 1, 0, 1, 1, 0 ];
let n = arr.length;
let x = 2;
document.write(CntDivbyX(arr, n, x));
</script>
|
Time Complexity: O(N) where N is the size of the array
Auxiliary Space: O(1)
Efficient Approach: As we see in the above approach we convert each binary prefix to a 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++
#include <bits/stdc++.h>
using namespace std;
int CntDivbyX(int arr[], int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++) {
number = (number * 2 + arr[i]) % x;
if (number == 0)
count += 1;
}
return count;
}
int main()
{
int arr[] = { 1, 0, 1, 0, 1, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 2;
cout << CntDivbyX(arr, n, x);
return 0;
}
|
Java
class GFG {
public static int CntDivbyX(int arr[], int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++) {
number = (number * 2 + arr[i]) % x;
if (number == 0)
count += 1;
}
return count;
}
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
def CntDivbyX(arr, n, x):
number = 0
count = 0
for i in range (0, n):
number = ( number * 2 + arr[i] ) % x
if number == 0:
count += 1
return count
arr = [1, 0, 1, 0, 1, 1, 0]
n = 7
x = 2
print( CntDivbyX(arr, n, x) )
|
C#
using System;
class GFG
{
public static int CntDivbyX(int []arr, int n, int x)
{
int number = 0;
int count = 0;
for (int i = 0; i < n; i++)
{
number = (number * 2 + arr[i]) % x;
if (number == 0)
count += 1;
}
return count;
}
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));
}
}
|
PHP
<?php
function CntDivbyX($arr, $n, $x)
{
$number = 0;
$count1 = 0;
for ($i = 0; $i < $n; $i++)
{
$number = ($number * 2 + $arr[$i]) % $x;
if ($number == 0)
$count1 += 1;
}
return $count1;
}
$arr = array(1, 0, 1, 0, 1, 1, 0);
$n = sizeof($arr);
$x = 2;
echo CntDivbyX($arr, $n, $x);
?>
|
Javascript
<script>
function CntDivbyX(arr, n, x)
{
let number = 0;
let count = 0;
for (let i = 0; i < n; i++) {
number = (number * 2 + arr[i]) % x;
if (number == 0)
count += 1;
}
return count;
}
let arr = [ 1, 0, 1, 0, 1, 1, 0 ];
let n = arr.length;
let x = 2;
document.write(CntDivbyX(arr, n, x));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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