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Subarray of length K having concatenation of its elements divisible by X

  • Last Updated : 01 Jun, 2021
Geek Week

Given an array arr[] consisting of N positive integers, the task is to find a subarray of length K such that concatenation of each element of the subarray is divisible by X. If no such subarray exists, then print “-1”. If more than one such subarray exists, print any one of them.

Examples:

Input: arr[] = {1, 2, 4, 5, 9, 6, 4, 3, 7, 8}, K = 4, X = 4
Output: 4 5 9 6
Explanation:
The elements of the subarray {4, 5, 9, 6} concatenates to form 4596, which is divisible by 4.

Input: arr[] = {2, 3, 5, 1, 3}, K = 3, X = 6
Output: -1

 

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays of length K and print that subarray whose concatenation of elements is divisible by X. If no such subarray exists, print “-1”. Otherwise, print any of these subarrays. 



Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using the Sliding Window Technique. Follow the steps below to solve the problem:

  1. Generate a number by concatenating the first K array elements. Store it in a variable, say num.
  2. Check if the number generated is divisible by X or not. If found to be true, then print the current subarray.
  3. Otherwise, traverse the array over the range [K, N] and for each element follow the steps below:
    • Add the digits of the element arr[i] to the variable num.
    • Remove the digits of the element arr[i – K] from the front of the num.
    • Now check if the current number formed is divisible by X or not. If found to be true, then print the current subarray in the range [i – K, i].
    • Otherwise, check for the next subarray.
  4. If no such subarray exists, then print “-1”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the starting
// index of the subarray whose
// concatenation is divisible by X
int findSubarray(vector<int> arr,
                 int K, int X)
{
    int i, num = 0;
 
    // Generate the concatenation
    // of first K length subarray
    for (i = 0; i < K; i++) {
        num = num * 10 + arr[i];
    }
 
    // If num is divisible by X
    if (num % X == 0) {
        return 0;
    }
 
    // Traverse the remaining array
    for (int j = i; j < arr.size(); j++) {
 
        // Append the digits of arr[i]
        num = (num % (int)pow(10, j - 1))
                  * 10
              + arr[j];
 
        // If num is divisible by X
        if (num % X == 0) {
            return j - i + 1;
        }
    }
 
    // No subarray exists
    return -1;
}
 
// Function to print the subarray in
// the range [answer, answer + K]
void print(vector<int> arr, int answer,
           int K)
{
    // No such subarray exists
    if (answer == -1) {
        cout << answer;
    }
 
    // Otherwise
    else {
 
        // Print the subarray in the
        // range [answer, answer + K]
        for (int i = answer;
             i < answer + K; i++) {
            cout << arr[i] << " ";
        }
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    vector<int> arr = { 1, 2, 4, 5, 9,
                        6, 4, 3, 7, 8 };
 
    int K = 4, X = 4;
 
    // Function Call
    int answer = findSubarray(arr, K, X);
 
    // Function Call to print subarray
    print(arr, answer, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to return the starting
// index of the subarray whose
// concatenation is divisible by X
static int findSubarray(ArrayList<Integer> arr, int K,
                                                int X)
{
    int i, num = 0;
 
    // Generate the concatenation
    // of first K length subarray
    for(i = 0; i < K; i++)
    {
        num = num * 10 + arr.get(i);
    }
 
    // If num is divisible by X
    if (num % X == 0)
    {
        return 0;
    }
 
    // Traverse the remaining array
    for(int j = i; j < arr.size(); j++)
    {
         
        // Append the digits of arr[i]
        num = (num % (int)Math.pow(10, j - 1)) *
                10 + arr.get(j);
 
        // If num is divisible by X
        if (num % X == 0)
        {
            return j - i + 1;
        }
    }
 
    // No subarray exists
    return -1;
}
 
// Function to print the subarray in
// the range [answer, answer + K]
static void print(ArrayList<Integer> arr, int answer,
                                          int K)
{
     
    // No such subarray exists
    if (answer == -1)
    {
        System.out.println(answer);
    }
 
    // Otherwise
    else
    {
         
        // Print the subarray in the
        // range [answer, answer + K]
        for(int i = answer; i < answer + K; i++)
        {
            System.out.print(arr.get(i) + " ");
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    ArrayList<Integer> arr = new ArrayList<Integer>(
        Arrays.asList(1, 2, 4, 5, 9, 6, 4, 3, 7, 8));
 
    int K = 4, X = 4;
 
    // Function call
    int answer = findSubarray(arr, K, X);
 
    // Function call to print subarray
    print(arr, answer, K);
}
}
 
// This code is contributed by akhilsaini

Python3




# Python3 program for the above approach
 
# Function to return the starting
# index of the subarray whose
# concatenation is divisible by X
def findSubarray(arr, K, X):
 
    num = 0
 
    # Generate the concatenation
    # of first K length subarray
    for i in range(0, K):
        num = num * 10 + arr[i]
 
    # If num is divisible by X
    if num % X == 0:
        return 0
       
    i = K
     
    # Traverse the remaining array
    for j in range(i, len(arr)):
         
        # Append the digits of arr[i]
        num = ((num % int(pow(10, j - 1))) *
                10 + arr[j])
 
        # If num is divisible by X
        if num % X == 0:
            return j - i + 1
 
    # No subarray exists
    return -1
 
# Function to print the subarray in
# the range [answer, answer + K]
def print_subarray(arr, answer, K):
     
    # No such subarray exists
    if answer == -1:
        print(answer)
 
    # Otherwise
    else:
         
        # Print the subarray in the
        # range [answer, answer + K]
        for i in range(answer, answer + K):
            print(arr[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
     
    # Given array arr[]
    arr = [ 1, 2, 4, 5, 9,
            6, 4, 3, 7, 8 ]
 
    K = 4
    X = 4
 
    # Function call
    answer = findSubarray(arr, K, X)
 
    # Function call to print subarray
    print_subarray(arr, answer, K)
 
# This code is contributed by akhilsaini

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the starting
// index of the subarray whose
// concatenation is divisible by X
static int findSubarray(List<int> arr, int K,
                                       int X)
{
    int i, num = 0;
 
    // Generate the concatenation
    // of first K length subarray
    for(i = 0; i < K; i++)
    {
        num = num * 10 + arr[i];
    }
 
    // If num is divisible by X
    if (num % X == 0)
    {
        return 0;
    }
 
    // Traverse the remaining array
    for(int j = i; j < arr.Count; j++)
    {
         
        // Append the digits of arr[i]
        num = (num % (int)Math.Pow(10, j - 1)) *
                10 + arr[j];
 
        // If num is divisible by X
        if (num % X == 0)
        {
            return j - i + 1;
        }
    }
     
    // No subarray exists
    return -1;
}
 
// Function to print the subarray in
// the range [answer, answer + K]
static void print(List<int> arr, int answer,
                                 int K)
{
     
    // No such subarray exists
    if (answer == -1)
    {
        Console.WriteLine(answer);
    }
 
    // Otherwise
    else
    {
         
        // Print the subarray in the
        // range [answer, answer + K]
        for(int i = answer; i < answer + K; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
}
 
// Driver Code
static public void Main()
{
     
    // Given array arr[]
    List<int> arr = new List<int>(){ 1, 2, 4, 5, 9,
                                     6, 4, 3, 7, 8 };
 
    int K = 4, X = 4;
 
    // Function call
    int answer = findSubarray(arr, K, X);
 
    // Function call to print subarray
    print(arr, answer, K);
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to return the starting
// index of the subarray whose
// concatenation is divisible by X
function findSubarray(arr, K, X)
{
    var i, num = 0;
 
    // Generate the concatenation
    // of first K length subarray
    for(i = 0; i < K; i++)
    {
        num = num * 10 + arr[i];
    }
 
    // If num is divisible by X
    if (num % X == 0)
    {
        return 0;
    }
 
    // Traverse the remaining array
    for(var j = i; j < arr.length; j++)
    {
         
        // Append the digits of arr[i]
        num = (num % parseInt(Math.pow(10, j - 1))) *
                10 + arr[j];
 
        // If num is divisible by X
        if (num % X == 0)
        {
            return j - i + 1;
        }
    }
 
    // No subarray exists
    return -1;
}
 
// Function to print the subarray in
// the range [answer, answer + K]
function print(arr, answer, K)
{
     
    // No such subarray exists
    if (answer == -1)
    {
        document.write(answer);
    }
     
    // Otherwise
    else
    {
         
        // Print the subarray in the
        // range [answer, answer + K]
        for(var i = answer;
                i < answer + K; i++)
        {
            document.write( arr[i] + " ");
        }
    }
}
 
// Driver Code
 
// Given array arr[]
var arr = [ 1, 2, 4, 5, 9,
            6, 4, 3, 7, 8 ];
var K = 4, X = 4;
 
// Function Call
var answer = findSubarray(arr, K, X);
 
// Function Call to print subarray
print(arr, answer, K);
 
// This code is contributed by itsok
 
</script>
Output: 
4 5 9 6

 

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

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