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 : 4Input : 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 subarrayint 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 codeint 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 approachimport 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 subarraydef 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 codeif __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 approachusing System;class GFG{// function to find longest subarraystatic 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 codepublic 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)
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