# Split Array into min number of subsets with difference between each pair greater than 1

• Difficulty Level : Expert
• Last Updated : 25 Jan, 2022

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 using namespace std; // Function to Split the array into// minimum number of subsets with// difference strictly > 1void 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 partition            count = 2;            break;        }    }     // Print the count of partitions    cout << count << endl;} // Driver Codeint 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 approachimport java.util.*; class GFG{     // Function to split the array into// minimum number of subsets with// difference strictly > 1static 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 partition            count = 2;            break;        }    }     // Print the count of partitions    System.out.print(count);} // Driver Codepublic 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 > 1def 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 partition            count = 2            break     # Print the count of partitions    print(count) # Driver Codeif __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 approachusing System;class GFG{      // Function to split the array into// minimum number of subsets with// difference strictly > 1static 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 partition            count = 2;            break;        }    }      // Print the count of partitions    Console.Write(count);}  // Driver Codepublic 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

## Javascript



Output:

1

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

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