Longest subarray with elements divisible by k
Last Updated :
16 Oct, 2022
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++
#include <bits/stdc++.h>
using namespace std;
int longestsubarray( int arr[], int n, int k)
{
int current_count = 0;
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;
}
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
import java.io.*;
class GFG {
static int longestsubarray( int arr[], int n, int k)
{
int current_count = 0 ;
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;
}
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));
}
}
|
Python3
def longestsubarray(arr, n, k):
current_count = 0
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
if __name__ = = '__main__' :
arr = [ 2 , 5 , 11 , 32 , 64 , 88 ]
n = len (arr)
k = 8
print (longestsubarray(arr, n, k))
|
C#
using System;
class GFG
{
static int longestsubarray( int [] arr,
int n, int k)
{
int current_count = 0;
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;
}
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));
}
}
|
PHP
<?php
function longestsubarray( $arr , $n , $k )
{
$current_count = 0;
$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 ;
}
$arr = array (2, 5, 11, 32, 64, 88 );
$n = sizeof( $arr );
$k = 8;
echo longestsubarray( $arr , $n , $k );
?>
|
Javascript
<script>
function longestsubarray(arr, n, k)
{
let current_count = 0;
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>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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