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# Split array into minimum number of subsets such that elements of all pairs are present in different subsets at least once

Given an array arr[] consisting of N distinct integers, the task is to find the minimum number of times the array needs to be split into two subsets such that elements of each pair are present into two different subsets at least once.

Examples:

Input: arr[] = { 3, 4, 2, 1, 5 }
Output:
Explanation:
Possible pairs are { (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5) }
Splitting the array into { 1, 2 } and { 3, 4, 5 }
Elements of each of the pairs { (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5) } are present into two different subsets.
Splitting the array into { 1, 3 } and { 2, 4, 5 }
Elements of each of the pairs { (1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5) } are present into two different subsets.
Splitting the array into { 1, 3, 4 } and { 2, 5 }
Elements of each of the pairs { (1, 2), (1, 5), (2, 3), (3, 5), (2, 4), (4, 5) } are present into two different subsets.
Since elements of each pair of the array is present in two different subsets at least once, the required output is 3.

Input: arr[] = { 2, 1, 3 }
Output:

Approach: The idea is to always split the array into two subsets of size floor(N / 2) and ceil(N / 2). Before each partition just swap the value of arr[i] with arr[N / 2 + i]. Follow the steps given below to solve the problem:

Below is the C++ implementation of the above approach:

## C++

 // C++ program to to implement// the above approach#include using namespace std; // Function to find minimum count of ways to split// the array into two subset such that elements of// each pair occurs in two different subsetint MinimumNoOfWays(int arr[], int n){     // Stores minimum count of ways to split array    // into two subset such that elements of    // each pair occurs in two different subset    int mini_no_of_ways;     // If N is odd    if (n % 2 == 0) {        mini_no_of_ways = n / 2;    }    else {        mini_no_of_ways = n / 2 + 1;    }    return mini_no_of_ways;} // Driver Codeint main(){    int arr[] = { 3, 4, 2, 1, 5 };    int N = sizeof(arr) / sizeof(arr[0]);    cout << MinimumNoOfWays(arr, N);    return 0;}

## Java

 // Java program to implement// the above approachimport java.util.*;   class GFG{   // Function to find minimum count of ways to split// the array into two subset such that elements of// each pair occurs in two different subsetstatic int MinimumNoOfWays(int arr[], int n){      // Stores minimum count of ways to split array    // into two subset such that elements of    // each pair occurs in two different subset    int mini_no_of_ways;      // If N is odd    if (n % 2 == 0) {        mini_no_of_ways = n / 2;    }    else {        mini_no_of_ways = n / 2 + 1;    }    return mini_no_of_ways;}   // Driver codepublic static void main(String[] args){    int arr[] = { 3, 4, 2, 1, 5 };    int N = arr.length;    System.out.print(MinimumNoOfWays(arr, N));}} // This code is contributed by sanjoy_62

## Python3

 # Python program to to implement# the above approach # Function to find minimum count of ways to split# the array into two subset such that elements of# each pair occurs in two different subsetdef MinimumNoOfWays(arr, n):       # Stores minimum count of ways to split array    # into two subset such that elements of    # each pair occurs in two different subset    min_no_of_ways = 0         # if n is even    if (n % 2 == 0):        mini_no_of_ways = n // 2             # n is odd    else:        mini_no_of_ways = n // 2 + 1             return mini_no_of_ways # driver codeif __name__ == '__main__':    arr = [3, 4, 1, 2, 5]    n = len(arr)    print(MinimumNoOfWays(arr, n)) # This code is contributed by MuskanKalra1

## C#

 // C# program to implement// the above approachusing System;class GFG{   // Function to find minimum count of ways to split// the array into two subset such that elements of// each pair occurs in two different subsetstatic int MinimumNoOfWays(int []arr, int n){      // Stores minimum count of ways to split array    // into two subset such that elements of    // each pair occurs in two different subset    int mini_no_of_ways;      // If N is odd    if (n % 2 == 0)    {        mini_no_of_ways = n / 2;    }    else    {        mini_no_of_ways = n / 2 + 1;    }    return mini_no_of_ways;}     // Driver code  public static void Main(string[] args)  {      int[] arr = { 3, 4, 2, 1, 5 };      int N = arr.Length;      Console.WriteLine(MinimumNoOfWays(arr, N));  }} // This code is contributed by AnkThon

## Javascript



Output:

3

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