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Split Array into min number of subsets with difference between each pair greater than 1
  • Difficulty Level : Expert
  • Last Updated : 03 Aug, 2020

Given an array arr[] of size N, the task is to split the array into a minimum number of subset such that each pair of elements in each subset have the difference strictly greater than 1.
Note: All elements in the array are distinct.

Examples:

Input: arr = {5, 10, 6, 50}
Output: 2
Explanation: 
Possible partitions are: {5, 10, 50}, { 6 }

Input: arr = { 2, 4, 6 }
Output: 1
Explanation: 
Possible partitions are: {2, 4, 6}

 

Approach: The idea is to observe that if there is no such pair i, j such that |arr[i] – arr[j]| = 1, then it is possible to put all the elements in the same partition, otherwise divide them into two partitions. So the required minimum number of partitions is always 1 or 2.

  1. Sort the given array.
  2. Compare the adjacent elements. If at any point their difference to be equal to 1, then print “2” as the required number of subset partition will always be 2 as we can put one of the elements from the above pair into another subset.
  3. If we traversed all the array didn’t found any adjacent pair with a difference less than 2 then print “1” without splitting the array into subsets we can have all possible pairs difference at least 2.

Below is the implementation for the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to Split the array into
// minimum number of subsets with
// difference strictly > 1
void split(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
    int count = 1;
 
    // Traverse through the sorted array
    for (int i = 1; i < n; i++) {
 
        // Check the pairs of elements
        // with difference 1
        if (arr[i] - arr[i - 1] == 1) {
 
            // If we find even a single
            // pair with difference equal
            // to 1, then 2 partitions
            // else only 1 partiton
            count = 2;
            break;
        }
    }
 
    // Print the count of partitions
    cout << count << endl;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 6 };
 
    // Size of the array
    int n = sizeof(arr) / sizeof(int);
 
    // Function Call
    split(arr, n);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to split the array into
// minimum number of subsets with
// difference strictly > 1
static void split(int arr[], int n)
{
     
    // Sort the array
    Arrays.sort(arr);
    int count = 1;
 
    // Traverse through the sorted array
    for(int i = 1; i < n; i++)
    {
         
        // Check the pairs of elements
        // with difference 1
        if (arr[i] - arr[i - 1] == 1)
        {
             
            // If we find even a single
            // pair with difference equal
            // to 1, then 2 partitions
            // else only 1 partiton
            count = 2;
            break;
        }
    }
 
    // Print the count of partitions
    System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 2, 4, 6 };
 
    // Size of the array
    int n = arr.length;
 
    // Function call
    split(arr, n);
}
}
 
// This code is contributed by jrishabh99

Python3




# Python3 implementation of
# the above approach
 
# Function to Split the array into
# minimum number of subsets with
# difference strictly > 1
def split(arr, n):
 
    # Sort the array
    arr.sort()
    count = 1
 
    # Traverse through the sorted array
    for i in range(1, n):
 
        # Check the pairs of elements
        # with difference 1
        if(arr[i] - arr[i - 1] == 1):
 
            # If we find even a single
            # pair with difference equal
            # to 1, then 2 partitions
            # else only 1 partiton
            count = 2
            break
 
    # Print the count of partitions
    print(count)
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    arr = [ 2, 4, 6 ]
 
    # Size of the array
    n = len(arr)
 
    # Function call
    split(arr, n)
 
# This code is contributed by Shivam Singh

C#




// C# program for the above approach
using System;
class GFG{
      
// Function to split the array into
// minimum number of subsets with
// difference strictly > 1
static void split(int []arr, int n)
{
    // Sort the array
    Array.Sort(arr);
    int count = 1;
  
    // Traverse through the sorted array
    for(int i = 1; i < n; i++)
    {
          
        // Check the pairs of elements
        // with difference 1
        if (arr[i] - arr[i - 1] == 1)
        {
            // If we find even a single
            // pair with difference equal
            // to 1, then 2 partitions
            // else only 1 partiton
            count = 2;
            break;
        }
    }
  
    // Print the count of partitions
    Console.Write(count);
}
  
// Driver Code
public static void Main(string[] args)
{
      
    // Given array
    int[] arr = new int[]{ 2, 4, 6 };
  
    // Size of the array
    int n = arr.Length;
  
    // Function call
    split(arr, n);
}
}
  
// This code is contributed by Ritik Bansal
Output: 
1


 

Time Complexity: O(N log N), where N is the length of the array.
Auxiliary Space: O(1)

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