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Check if an array can be split into K consecutive non-overlapping subarrays of length M consisting of single distinct element
  • Last Updated : 02 Mar, 2021

Given two integers M and K and an array arr[] consisting of N positive integers, the task is to check if the array can be split into K consecutive non-overlapping subarrays of length M such that each subarray consists of a single distinct element. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {6, 1, 3, 3, 3, 3}, M = 1, K = 3
Output: Yes
Explanation:
The K consecutive non-overlapping subarrays are {6}, {1}, {3, 3, 3, 3}.

Input: arr[] = {3, 5, 3, 5, 3, 1}, M = 2, K = 3
Output: No

 

Approach: The given problem can be solved by using a simple array traversal and check whether the element at the current index i and the element at the index (i + M) are the same or not. Follow the steps below to solve the problem:



  • Initialize two variables, count and t with 1 and 0 respectively to store the total count of pattern matched and the current length of pattern matched respectively.
  • Traverse the given array over the range [0, N – M – 1] using the variable i and do the following:
    • If the value of arr[i] and arr[i + M] is the same then, increment t by 1 and if t is the same as m then update t to 0 and increment the count. If the value of count is K then print “Yes” and break out of the loop.
    • Else, if t is M then increment the count by 1.
  • After the above steps if the value of count is not the same as K then print “No” as there doesn’t exist any such pattern.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if array can be split
// into K consecutive and non-overlapping
// subarrays of length M consisting of a
// single distinct element
string checkPattern(int arr[], int m,
                    int k, int n)
{
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++) {
 
        // Check if arr[i] is the
        // same as arr[i + m]
        if (arr[i] == arr[i + m]) {
 
            // Increment current length
            // t of pattern matched by 1
            t++;
 
            // Check if t is equal to m,
            // increment count of total
            // repeated pattern
            if (t == m) {
 
                t = 0;
                count++;
 
                // Return true if length of
                // total repeated pattern is k
                if (count == k) {
                    return "Yes";
                }
            }
        }
 
        else {
 
            // Update length of the
            // current pattern
            t = 0;
 
            // Update count to 1
            count = 1;
        }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << checkPattern(arr, M, K, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if array can be split
// into K consecutive and non-overlapping
// subarrays of length M consisting of a
// single distinct element
static String checkPattern(int arr[], int m,
                    int k, int n)
{
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++)
    {
 
        // Check if arr[i] is the
        // same as arr[i + m]
        if (arr[i] == arr[i + m])
        {
 
            // Increment current length
            // t of pattern matched by 1
            t++;
 
            // Check if t is equal to m,
            // increment count of total
            // repeated pattern
            if (t == m)
            {
                t = 0;
                count++;
 
                // Return true if length of
                // total repeated pattern is k
                if (count == k)
                {
                    return "Yes";
                }
            }
        }
        else
        {
 
            // Update length of the
            // current pattern
            t = 0;
 
            // Update count to 1
            count = 1;
        }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = arr.length;
    System.out.print(checkPattern(arr, M, K, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program for the above approach
 
# Function to check if array can be split
# into K consecutive and non-overlapping
# subarrays of length M consisting of a
# single distinct element
def checkPattern(arr, m, k, n):
    count = 1
    t = 0
 
    # Traverse over the range [0, N - M - 1]
    for i in range(n - m):
       
        # Check if arr[i] is the
        # same as arr[i + m]
        if (arr[i] == arr[i + m]):
           
            # Increment current length
            # t of pattern matched by 1
            t += 1
 
            # Check if t is equal to m,
            # increment count of total
            # repeated pattern
            if (t == m):
                t = 0
                count += 1
 
                # Return true if length of
                # total repeated pattern is k
                if (count == k):
                    return "Yes"
 
        else:
            # Update length of the
            # current pattern
            t = 0
 
            # Update count to 1
            count = 1
 
    # Finally return false if
    # no pattern found
    return "No"
 
# Driver Code
if __name__ == '__main__':
    arr  =  [6, 1, 3, 3, 3, 3]
    M = 1
    K = 3
    N = len(arr)
 
    print(checkPattern(arr, M, K, N))
     
    # This code is contributed by bgangwar59.

C#




// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to check if array can be split
  // into K consecutive and non-overlapping
  // subarrays of length M consisting of a
  // single distinct element
  static String checkPattern(int []arr, int m,
                             int k, int n)
  {
    int count = 1, t = 0;
 
    // Traverse over the range [0, N - M - 1]
    for (int i = 0; i < n - m; i++)
    {
 
      // Check if arr[i] is the
      // same as arr[i + m]
      if (arr[i] == arr[i + m])
      {
 
        // Increment current length
        // t of pattern matched by 1
        t++;
 
        // Check if t is equal to m,
        // increment count of total
        // repeated pattern
        if (t == m)
        {
          t = 0;
          count++;
 
          // Return true if length of
          // total repeated pattern is k
          if (count == k)
          {
            return "Yes";
          }
        }
      }
      else
      {
 
        // Update length of the
        // current pattern
        t = 0;
 
        // Update count to 1
        count = 1;
      }
    }
 
    // Finally return false if
    // no pattern found
    return "No";
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 6, 1, 3, 3, 3, 3 };
    int M = 1, K = 3;
    int N = arr.Length;
    Console.Write(checkPattern(arr, M, K, N));
  }
}
 
// This code is contributed by Rajput-Ji
Output: 
Yes

 

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

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