Check if any subarray of length M repeats at least K times consecutively or not

Given an array arr[] consisting of N integers and two positive integers M and K, the task is to check if there exists any subarray of length M that repeats consecutively at least K times. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {2, 1, 2, 1, 1, 1, 3}, M = 2, K = 2
Output: Yes
Explanation: The subarray {2, 1} of length 2 repeats at least K(= 2) times consecutively.

Input: arr[] = {7, 1, 3, 1, 1, 1, 1, 3}, M = 1, K = 3
Output: Yes

Naive Approach: The simplest approach is to generate all possible subarrays of length M and check for each subarray, whether on concatenating it exactly K times is present as a subarray in the given array or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to check if there exists
// any subarray of length M repeating
// at least K times consecutively
bool check(int arr[], int M, int K,
           int ind)
{
    // Iterate from i equal 0 to M
    for (int i = 0; i < M; i++) {

        // Iterate from j equals 1 to K
        for (int j = 1; j < K; j++) {

            // If elements at pos + i and
            // pos + i + j * M are not equal
            if (arr[ind + i]
                != arr[ind + i + j * M]) {

                return false;
            }
        }
    }
    return true;
}

// Function to check if a subarray repeats
// at least K times consecutively or not
bool SubarrayRepeatsKorMore(
    int arr[], int N, int M, int K)
{
    // Iterate from ind equal 0 to M
    for (int ind = 0;
         ind <= N - M * K; ind++) {

        // Check if subarray arr[i, i + M]
        // repeats atleast K times or not
        if (check(arr, M, K, ind)) {
            return true;
        }
    }

    // Otherwise, return false
    return false;
}

// Driver Code
int main()
{
    int arr[] = { 2, 1, 2, 1, 1, 1, 3 };
    int M = 2, K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);

    if (SubarrayRepeatsKorMore(
            arr, N, M, K)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to check if there exists
// any subarray of length M repeating
// at least K times consecutively
static boolean check(int arr[], int M,
                     int K, int ind)
{

    // Iterate from i equal 0 to M
    for(int i = 0; i < M; i++)
    {

        // Iterate from j equals 1 to K
        for(int j = 1; j < K; j++)
        {

            // If elements at pos + i and
            // pos + i + j * M are not equal
            if (arr[ind + i] != arr[ind + i + j * M])
            {
                return false;
            }
        }
    }
    return true;
}

// Function to check if a subarray repeats
// at least K times consecutively or not
static boolean SubarrayRepeatsKorMore(int arr[], int N,
                                      int M, int K)
{

    // Iterate from ind equal 0 to M
    for(int ind = 0; ind <= N - M * K; ind++)
    {

        // Check if subarray arr[i, i + M]
        // repeats atleast K times or not
        if (check(arr, M, K, ind))
        {
            return true;
        }
    }

    // Otherwise, return false
    return false;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 2, 1, 1, 1, 3 };
    int M = 2, K = 2;
    int N = arr.length;

    if (SubarrayRepeatsKorMore(arr, N, M, K))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to check if there exists
# any subarray of length M repeating
# at least K times consecutively
def check(arr, M, K, ind):

    # Iterate from i equal 0 to M
    for i in range(M):

        # Iterate from j equals 1 to K
        for j in range(1, K, 1):

            # If elements at pos + i and
            # pos + i + j * M are not equal
            if (arr[ind + i] != arr[ind + i + j * M]):
                return False

    return True

# Function to check if a subarray repeats
# at least K times consecutively or not
def SubarrayRepeatsKorMore(arr, N, M, K):

    # Iterate from ind equal 0 to M
    for ind in range(N - M * K + 1):

        # Check if subarray arr[i, i + M]
        # repeats atleast K times or not
        if (check(arr, M, K, ind)):
            return True

    # Otherwise, return false
    return False

# Driver Code
if __name__ == '__main__':

    arr =  [2, 1, 2, 1, 1, 1, 3]
    M = 2
    K = 2
    N = len(arr)

    if (SubarrayRepeatsKorMore(arr, N, M, K)):
        print("Yes")
    else:
        print("No")

# This code is contributed by bgangwar59

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if there exists
// any subarray of length M repeating
// at least K times consecutively
static bool check(int[] arr, int M, int K,
                  int ind)
{

    // Iterate from i equal 0 to M
    for(int i = 0; i < M; i++)
    {

        // Iterate from j equals 1 to K
        for(int j = 1; j < K; j++)
        {

            // If elements at pos + i and
            // pos + i + j * M are not equal
            if (arr[ind + i] != arr[ind + i + j * M])
            {
                return false;
            }
        }
    }
    return true;
}

// Function to check if a subarray repeats
// at least K times consecutively or not
static bool SubarrayRepeatsKorMore(int[] arr, int N,
                                   int M, int K)
{

    // Iterate from ind equal 0 to M
    for(int ind = 0; ind <= N - M * K; ind++)
    {

        // Check if subarray arr[i, i + M]
        // repeats atleast K times or not
        if (check(arr, M, K, ind))
        {
            return true;
        }
    }

    // Otherwise, return false
    return false;
}

// Driver code
static void Main()
{
    int[] arr = { 2, 1, 2, 1, 1, 1, 3 };
    int M = 2, K = 2;
    int N = arr.Length;

    if (SubarrayRepeatsKorMore(
            arr, N, M, K))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program for the above approach

// Function to check if there exists
// any subarray of length M repeating
// at least K times consecutively
function check(arr, M, K, ind)
{

    // Iterate from i equal 0 to M
    for(let i = 0; i < M; i++)
    {

        // Iterate from j equals 1 to K
        for(let j = 1; j < K; j++)
        {

            // If elements at pos + i and
            // pos + i + j * M are not equal
            if (arr[ind + i] !=
                arr[ind + i + j * M])
            {
                return false;
            }
        }
    }
    return true;
}

// Function to check if a subarray repeats
// at least K times consecutively or not
function SubarrayRepeatsKorMore(arr, N, M, K)
{

    // Iterate from ind equal 0 to M
    for(let ind = 0;
            ind <= N - M * K; ind++)
    {

        // Check if subarray arr[i, i + M]
        // repeats atleast K times or not
        if (check(arr, M, K, ind))
        {
            return true;
        }
    }

    // Otherwise, return false
    return false;
}

// Driver Code
let arr = [ 2, 1, 2, 1, 1, 1, 3 ];
let M = 2, K = 2;
let N = arr.length;

if (SubarrayRepeatsKorMore(arr, N, M, K))
{
    document.write("Yes");
}
else
{
    document.write("No");
}

// This code is contributed by subhammahato348

</script>

Output:

Yes

Time Complexity: O(N*M*K)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using Two Pointers Technique. Follow the steps below to solve the problem:

• Initialize a variable, say count as 0.
• Traverse the given array arr[] over the range of indices [0, N – M] using a variable, say i, and perform the following steps:
  • If the value of arr[i] is equal to arr[i + M], then increment count by 1, as there is a match in the subarray.
  • Otherwise, update count to 0 as there is a break in the contiguous subarrays.
  • If the value of count is M * (K – 1), then it means that there are K consecutively equal subarrays of length M. Therefore, print “Yes” and break out of the loop.
• After completing the above steps, if the count never becomes M * (K – 1), then print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to check if any subarray
// of length M repeats at least
// K times consecutively or not
bool checkExists(int arr[], int N,
                 int M, int K)
{
    // Stores the required count
    // of repeated subarrays
    int count = 0;

    for (int i = 0; i < N - M; i++) {

        // Check if the next continuous
        // subarray has equal elements
        if (arr[i] == arr[i + M])
            count++;
        else
            count = 0;

        // Check if K continuous subarray
        // of length M are found or not
        if (count == M * (K - 1))
            return true;
    }

    // If no subarrays are found
    return false;
}

// Driver Code
int main()
{
    int arr[] = { 2, 1, 2, 1, 1, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 2, K = 2;

    if (checkExists(arr, N, M, K)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to check if any subarray
// of length M repeats at least
// K times consecutively or not
static boolean checkExists(int arr[], int N,
                           int M, int K)
{

    // Stores the required count
    // of repeated subarrays
    int count = 0;

    for(int i = 0; i < N - M; i++)
    {

        // Check if the next continuous
        // subarray has equal elements
        if (arr[i] == arr[i + M])
            count++;
        else
            count = 0;

        // Check if K continuous subarray
        // of length M are found or not
        if (count == M * (K - 1))
            return true;
    }

    // If no subarrays are found
    return false;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 2, 1, 1, 1, 3 };
    int M = 2, K = 2;
    int N = arr.length;

    if (checkExists(arr, N, M, K))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to check if any subarray
# of length M repeats at least
# K times consecutively or not
def checkExists(arr, N, M, K):

    # Stores the required count
    # of repeated subarrays
    count = 0

    for i in range(N - M):

        # Check if the next continuous
        # subarray has equal elements
        if (arr[i] == arr[i + M]):
            count += 1
        else:
            count = 0

        # Check if K continuous subarray
        # of length M are found or not
        if (count == M * (K - 1)):
            return True

    # If no subarrays are found
    return False

# Driver Code
if __name__ == '__main__':

    arr = [ 2, 1, 2, 1, 1, 1, 3 ]
    N = len(arr)
    M = 2
    K = 2

    if (checkExists(arr, N, M, K)):
        print("Yes")
    else:
        print("No")

# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if any subarray
// of length M repeats at least
// K times consecutively or not
public static bool checkExists(int []arr, int N,
                               int M, int K)
{

    // Stores the required count
    // of repeated subarrays
    int count = 0;

    for(int i = 0; i < N - M; i++)
    {

        // Check if the next continuous
        // subarray has equal elements
        if (arr[i] == arr[i + M])
            count++;
        else
            count = 0;

        // Check if K continuous subarray
        // of length M are found or not
        if (count == M * (K - 1))
            return true;
    }

    // If no subarrays are found
    return false;
}

// Driver Code
public static void Main()
{
    int []arr = { 2, 1, 2, 1, 1, 1, 3 };
    int N = arr.Length;
    int M = 2, K = 2;

    if (checkExists(arr, N, M, K))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by mohit kumar 29

Javascript

<script>

// Javascript program for the above approach

// Function to check if any subarray
// of length M repeats at least
// K times consecutively or not
function checkExists(arr, N, M, K)
{
    // Stores the required count
    // of repeated subarrays
    let count = 0;

    for (let i = 0; i < N - M; i++) {

        // Check if the next continuous
        // subarray has equal elements
        if (arr[i] == arr[i + M])
            count++;
        else
            count = 0;

        // Check if K continuous subarray
        // of length M are found or not
        if (count == M * (K - 1))
            return true;
    }

    // If no subarrays are found
    return false;
}

// Driver Code
    let arr = [ 2, 1, 2, 1, 1, 1, 3 ];
    let N = arr.length;
    let M = 2, K = 2;

    if (checkExists(arr, N, M, K)) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }

</script>

Output:

Yes

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

Related Topic: Subarrays, Subsequences, and Subsets in Array