Longest subarray with elements divisible by k

Suppose you are given an array. You have to find the length of the longest subarray such that each and every element of it is divisible by k.
Examples:

Input : arr[] = { 1, 7, 2, 6, 8, 100, 3, 6, 16}, k=2
Output : 4

Input : arr[] = { 3, 11, 22, 32, 55, 100, 1, 5}, k=5
Output : 2 

Approach:

• Initialize two variables current_count and max_count with value 0;
• Iterate the array from left to right and check the divisibility of each element by k.
• If the element is divisible, then increment current_count, otherwise make current_count equal to 0
• Compare current_count with max_count at each element, if current_count is greater than max_count, assign the value of current_count to max_count
• Finally, output value of max_count

Below is the implementation of the above approach:

C++

// C++ program of above approach
#include <bits/stdc++.h>
using namespace std;

// function to find longest subarray
int longestsubarray(int arr[], int n, int k)
{
    int current_count = 0;
    // this will contain length of longest subarray found
    int max_count = 0;

    for (int i = 0; i < n; i++) {
        if (arr[i] % k == 0)
            current_count++;
        else
            current_count = 0;
        max_count = max(current_count, max_count);
    }
    return max_count;
}

// Driver code
int main()
{
    int arr[] = { 2, 5, 11, 32, 64, 88 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 8;
    cout << longestsubarray(arr, n, k);
    return 0;
}

Java

// Java program of above approach

import java.io.*;

class GFG {
    // function to find longest subarray
    static int longestsubarray(int arr[], int n, int k)
    {
        int current_count = 0;

        // this will contain length of
        // longest subarray found
        int max_count = 0;

        for (int i = 0; i < n; i++) {
            if (arr[i] % k == 0)
                current_count++;
            else
                current_count = 0;
            max_count = Math.max(current_count, max_count);
        }
        return max_count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 5, 11, 32, 64, 88 };
        int n = arr.length;
        int k = 8;
        System.out.println(longestsubarray(arr, n, k));
    }
}

// This code is contributed
// by ajit

Python3

# Python 3 program of above approach

# function to find longest subarray


def longestsubarray(arr, n, k):
    current_count = 0

    # this will contain length of
    # longest subarray found
    max_count = 0

    for i in range(0, n, 1):
        if (arr[i] % k == 0):
            current_count += 1
        else:
            current_count = 0
        max_count = max(current_count,
                        max_count)

    return max_count


# Driver code
if __name__ == '__main__':
    arr = [2, 5, 11, 32, 64, 88]
    n = len(arr)
    k = 8
    print(longestsubarray(arr, n, k))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program of above approach
using System;

class GFG
{
// function to find longest subarray
static int longestsubarray(int[] arr,
                           int n, int k)
{
    int current_count = 0;

    // this will contain length of
    // longest subarray found
    int max_count = 0;

    for (int i = 0; i < n; i++)
    {
        if (arr[i] % k == 0)
            current_count++;
        else
            current_count = 0;
        max_count = Math.Max(current_count,
                             max_count);
    }
    return max_count;
}

// Driver code
public static void Main()
{
    int[] arr = { 2, 5, 11, 32, 64, 88 };
    int n = arr.Length;
    int k = 8;
    Console.Write(longestsubarray(arr, n, k));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP program of above approach

// function to find longest subarray
function longestsubarray($arr, $n, $k)
{
    $current_count = 0;

    // This will contain length of
    // longest subarray found
    $max_count = 0;

    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] % $k == 0)
            $current_count++;
        else
            $current_count = 0;
        $max_count = max($current_count,
                         $max_count);
    }
    return $max_count;
}

// Driver code
$arr = array(2, 5, 11, 32, 64, 88 );
$n = sizeof($arr);
$k = 8;
echo longestsubarray($arr, $n, $k);

// This code is contributed
// by Sach_Code
?>

Javascript

<script>
    // Javascript program of above approach

    // function to find longest subarray
    function longestsubarray(arr, n, k)
    {
        let current_count = 0;
        // this will contain length of longest subarray found
        let max_count = 0;

        for (let i = 0; i < n; i++) {
            if (arr[i] % k == 0)
                current_count++;
            else
                current_count = 0;
            max_count = Math.max(current_count, max_count);
        }
        return max_count;
    }

    let arr = [ 2, 5, 11, 32, 64, 88 ];
    let n = arr.length;
    let k = 8;
    document.write(longestsubarray(arr, n, k));

</script>

Output

3

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