Ways to form n/2 pairs such that difference of pairs is minimum

Given an array arr of N integers, the task is to split the elements of the array into N/2 pairs (each pair having 2 elements) such that the absolute difference between two elements of any pair is as minimum as possible. Note: N will always be even. Examples:

Input: arr[] = {1, 7, 3, 8} Output: 1 There is only one way to form pairs with minimum difference, (7, 8) and (1, 3). Input: arr[] = {1, 1, 1, 1, 2, 2, 2, 2} Output: 9 Here all elements with value 2 will pair amongst themselves (3 ways possible) and all elements with value 1 will make pairs between them (3 ways possible) Therefore, number of ways = 3 * 3 = 9 Input: arr[] = {2, 3, 2, 2} Output: 3

Approach: This problem involves the fundamental principle of counting and some basic understanding of permutations and combinations. Before going any further, let’s count the number of ways for the array = [3, 3] Answer = 1, since only 1 combination is possible. Now, lets modify the array to [3, 3, 3, 3]. The ways for this as discussed in the above examples are:

(1, 2), (3, 4) (1, 3), (2, 4) (1, 4), (2, 3) Answer = 3 ways.

Further modifying the array to [3, 3, 3, 3, 3, 3]

(1, 2), (3, 4), (5, 6) (1, 2), (3, 5), (4, 6) (1, 2), (3, 6), (4, 5) (1, 3), (2, 4), (5, 6) (1, 3), (2, 5), (4, 6) (1, 3), (2, 6), (4, 5) (1, 4), (2, 3), (5, 6) (1, 4), (2, 5), (3, 6) (1, 4), (2, 6), (3, 5) (1, 5), (2, 3), (4, 6) (1, 5), (2, 4), (3, 6) (1, 5), (2, 6), (3, 4) (1, 6), (2, 3), (4, 5) (1, 6), (2, 4), (3, 5) (1, 6), (2, 5), (3, 4) Answer = 15 ways.

Here we obtain a generalized result by simple observation. If there are K elements in the array that have the same value and K is even, then the number of ways to form pairs amongst them:

For size 2, count = 1 (1) For size 4, count = 3 (1 * 3) For size 6, count = 15 (1 * 3 * 5) And so on. Hence, number of ways to form pairs for size K where K is even = 1 * 3 * 5 * …. * (K-1)

We can precompute this result as follows. Let ways[] be the array such that ways[i] stores the number of ways for size ‘i’.

ways[2] = 1; for(i = 4; i < 1e5 + 1; i += 2) ways[i] = ways[i – 2] * (i – 1); For example, we consider array [3, 3, 3, 3, 3] To compute the number of ways, we fix the first element with any of the remaining 5. So we form one pair. Now 4 elements are left that can be paired in ways[4] ways. So number of ways would be 5 * ways[4].

Now, it may not be necessary that counts may always be even in number. Therefore, if need to solve this for a general array, we need to do two things.

1. Sort the array in ascending order.
2. Analyze the count of each group having the same value.

Let our array = [2, 3, 3, 3, 3, 4, 4, 4, 4, 4]. This array is sorted.

1. Consider the element with value 4. Since there are 5 elements with value 4, they can pair amongst themselves in ways[4] ways.
2. The left out element will have to pair with an element with value 3. This happens in 4 ways since there are 4 elements with value 3.
3. One element of value 3 will be reserved to pair with an element of 4, so 3 elements with value 3 remain.
4. Two of the remaining elements with value 3 will pair amongst themselves in ways[2] ways.
5. One element of value 3 will be left, which will pair with the element with value 2 in 1 way.
6. The lone element will be selected from the remaining 3 elements in 3 ways.
7. Therefore, the total number of ways to pair the elements in the given sorted array is the product of the ways calculated in steps 1-6.

• Therefore from point 1, number of ways will be :

Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define S second
#define ll long long

using namespace std;

// Using mod because the number
// of ways might be very large
const int mod = 1000000007;

const int MAX = 100000;

// ways is serving the same
// purpose as discussed
ll ways[MAX + 1];

void preCompute()
{
    // pairing up zero people
    // requires one way.
    ways[0] = 1LL;
    ways[2] = 1LL;
    for (int i = 4; i <= MAX; i += 2) {
        ways[i] = (1LL * (i - 1) * ways[i - 2]) % mod;
    }
}

void countWays(int* arr, int n)
{

    // map count stores count of s.
    map<int, int> count;
    for (int i = 0; i < n; i++)
        count[arr[i]]++;

    vector<pair<int, int> > count_vector;
    map<int, int>::iterator it;
    for (it = count.begin(); it != count.end(); it++) {
        count_vector.pb(mp(it->first, it->second));
    }

    // vector count_vector stores a
    // pair < value, count of value>

    // sort according to value
    sort(count_vector.begin(), count_vector.end());

    ll ans = 1;

    // Iterating backwards.
    for (int i = count_vector.size() - 1; i > 0; i--) {

        int current_count = count_vector[i].S;
        int prev_count = count_vector[i - 1].S;

        // Checking if current count is odd.
        if (current_count & 1) {

            // if current count = 5, multiply ans by ways[4].
            ans = (ans * ways[current_count - 1]) % mod;

            // left out person will be selected
            // in current_count ways
            ans = (ans * current_count) % mod;

            // left out person will pair with previous
            //  person in previous_count ways
            ans = (ans * prev_count) % mod;

            /* if previous count is odd,
             * then multiply answer by ways[prev_count-1].
             * since one has already been reserved,
             * remaining will be even.
             * reduce prev_count = 0, since we don't need it now.*/
            if (prev_count & 1) {
                ans = (ans * ways[prev_count - 1]) % mod;
                count_vector[i - 1].S = 0;
            }
            else {

                /* if prev count is even, one will be reserved,
                 * therefore decrement by 1.
                 * In the next iteration, prev_count will become odd
                 * and it will be handled in the same way.*/
                count_vector[i - 1].S--;
            }
        }
        else {

            /* if current count is even,
             * then simply multiply ways[current_count]
             * to answer.*/
            ans = (ans * ways[current_count]) % mod;
        }
    }

    /* multiply answer by ways[first__count] since
       that is left out, after iterating the array.*/
    ans = (ans * ways[count_vector[0].S]) % mod;
    cout << ans << "\n";
}

// Driver code
int main()
{
    preCompute();
    int arr[] = { 2, 3, 3, 3, 3, 4, 4, 4, 4, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countWays(arr, n);
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG
{

  // Using mod because the number
  // of ways might be very large
  static int mod = 1000000007;

  static int MAX = 100000;

  // ways is serving the same
  // purpose as discussed
  static long[] ways = new long[MAX + 1];

  static void preCompute()
  {
    // pairing up zero people
    // requires one way.
    ways[0] = 1;
    ways[2] = 1;
    for (int i = 4; i <= MAX; i += 2) {
      ways[i] = ((i - 1) * ways[i - 2]) % mod;
    }
  }

  static void countWays(int[] arr, int n)
  {

    // map count stores count of s.
    TreeMap<Integer, Integer> count = new TreeMap<Integer, Integer>();
    for (int i = 0; i < n; i++)
    {
      if (!count.containsKey(arr[i]))
        count.put(arr[i], 0);
      count.put(arr[i], count.get(arr[i]) + 1);
    }

    ArrayList<ArrayList<Integer>> count_vector = new ArrayList<ArrayList<Integer>>();

    for (Map.Entry<Integer, Integer> entry : count.entrySet())
    {
        ArrayList<Integer> l = new ArrayList<Integer>();
        l.add(entry.getKey());
        l.add(entry.getValue());
      count_vector.add(l);
    }


    // vector count_vector stores a
    // pair < value, count of value>

    // sort according to value
    Collections.sort(count_vector, new Comparator<ArrayList<Integer>> () {
    @Override
    public int compare(ArrayList<Integer> a, ArrayList<Integer> b) {
        if (a.get(0) == b.get(0))
            return a.get(1).compareTo(b.get(1));
        return a.get(0).compareTo(b.get(0));
    }
});


    long ans = 1;

    // Iterating backwards.
    for (int i = count_vector.size() - 1; i > 0; i--) {

      int current_count = count_vector.get(i).get(1);
      int prev_count = count_vector.get(i - 1).get(1);

      // Checking if current count is odd.
      if ((current_count & 1) != 0) {

        // if current count = 5, multiply ans by ways[4].
        ans = (ans * ways[current_count - 1]) % mod;

        // left out person will be selected
        // in current_count ways
        ans = (ans * current_count) % mod;

        // left out person will pair with previous
        //  person in previous_count ways
        ans = (ans * prev_count) % mod;

        /* if previous count is odd,
                 * then multiply answer by ways[prev_count-1].
                 * since one has already been reserved,
                 * remaining will be even.
                 * reduce prev_count = 0, since we don't need it now.*/
        if ((prev_count & 1) != 0) {
          ans = (ans * ways[prev_count - 1]) % mod;
          count_vector.get(i - 1).set(1, 0);
        }
        else {

          /* if prev count is even, one will be reserved,
                     * therefore decrement by 1.
                     * In the next iteration, prev_count will become odd
                     * and it will be handled in the same way.*/
          count_vector.get(i - 1).set(1, count_vector.get(i - 1).get(1) - 1); ;
        }
      }
      else {

        /* if current count is even,
                 * then simply multiply ways[current_count]
                 * to answer.*/
        ans = (ans * ways[current_count]) % mod;
      }
    }

    /* multiply answer by ways[first__count] since
           that is left out, after iterating the array.*/
    ans = (ans * ways[count_vector.get(0).get(1)]) % mod;
    System.out.println(ans);
  }

  // Driver code
  public static void main(String[] args)
  {
    preCompute();
    int[] arr = { 2, 3, 3, 3, 3, 4, 4, 4, 4, 4 };
    int n = arr.length;
    countWays(arr, n);
  }
}


// This code is contributed by phasing17

Python3

# Python3 implementation of the
# above approach
from collections import defaultdict

# Using mod because the number
# of ways might be very large
mod = 1000000007
MAX = 100000

# ways is serving the same
# purpose as discussed
ways = [None] * (MAX + 1)

def preCompute():

    # pairing up zero people
    # requires one way.
    ways[0] = 1
    ways[2] = 1
    for i in range(4, MAX + 1, 2):
        ways[i] = ((1 * (i - 1) *
                    ways[i - 2]) % mod)

def countWays(arr, n):

    # map count stores count of s.
    count = defaultdict(lambda:0)
    for i in range(0, n):
        count[arr[i]] += 1

    count_vector = []
    for key in count:
        count_vector.append([key, count[key]])

    # vector count_vector stores a
    # pair < value, count of value>

    # sort according to value
    count_vector.sort()
    ans = 1

    # Iterating backwards.
    for i in range(len(count_vector) - 1, -1, -1):

        current_count = count_vector[i][1]
        prev_count = count_vector[i - 1][1]

        # Checking if current count is odd.
        if current_count & 1:

            # if current count = 5, multiply
            # ans by ways[4].
            ans = (ans * ways[current_count - 1]) % mod

            # left out person will be selected
            # in current_count ways
            ans = (ans * current_count) % mod

            # left out person will pair with previous
            # person in previous_count ways
            ans = (ans * prev_count) % mod

            # if previous count is odd,
            # then multiply answer by ways[prev_count-1].
            # since one has already been reserved,
            # remaining will be even.
            # reduce prev_count = 0, since we
            # don't need it now.
            if prev_count & 1:
                ans = (ans * ways[prev_count - 1]) % mod
                count_vector[i - 1][1] = 0

            else:

                # if prev count is even, one will be
                # reserved, therefore decrement by 1.
                # In the next iteration, prev_count
                # will become odd and it will be
                # handled in the same way.
                count_vector[i - 1][1] -= 1

        else:

            # if current count is even, then simply
            # multiply ways[current_count] to answer.
            ans = (ans * ways[current_count]) % mod

    # multiply answer by ways[first__count] since
    # that is left out, after iterating the array.
    ans = (ans * ways[count_vector[0][1]]) % mod
    print(ans)

# Driver code
if __name__ == "__main__":

    preCompute()
    arr = [2, 3, 3, 3, 3, 4, 4, 4, 4, 4]
    n = len(arr)
    countWays(arr, n)

# This code is contributed by Rituraj Jain

C#

// C# implementation of the above approach
using System;
using System.Linq;
using System.Collections.Generic;

class GFG
{

  // Using mod because the number
  // of ways might be very large
  static int mod = 1000000007;

  static int MAX = 100000;

  // ways is serving the same
  // purpose as discussed
  static long[] ways = new long[MAX + 1];

  static void preCompute()
  {
    // pairing up zero people
    // requires one way.
    ways[0] = 1;
    ways[2] = 1;
    for (int i = 4; i <= MAX; i += 2) {
      ways[i] = ((i - 1) * ways[i - 2]) % mod;
    }
  }

  static void countWays(int[] arr, int n)
  {

    // map count stores count of s.
    Dictionary<int, int> count = new Dictionary<int, int>();
    for (int i = 0; i < n; i++)
    {
      if (!count.ContainsKey(arr[i]))
        count[arr[i]] = 0;
      count[arr[i]]++;
    }

    List<int []> count_vector = new List<int[]>();

    List<int> keys = new List<int>(count.Keys);
    keys.Sort();

    foreach (var key in keys)
      count_vector.Add(new int[]{key, count[key]});


    // vector count_vector stores a
    // pair < value, count of value>

    // sort according to value
    count_vector = count_vector.OrderBy(r => r[0])
      .ThenBy(r => r[1])
      .ToList();


    long ans = 1;

    // Iterating backwards.
    for (int i = count_vector.Count - 1; i > 0; i--) {

      int current_count = count_vector[i][1];
      int prev_count = count_vector[i - 1][1];

      // Checking if current count is odd.
      if ((current_count & 1) != 0) {

        // if current count = 5, multiply ans by ways[4].
        ans = (ans * ways[current_count - 1]) % mod;

        // left out person will be selected
        // in current_count ways
        ans = (ans * current_count) % mod;

        // left out person will pair with previous
        //  person in previous_count ways
        ans = (ans * prev_count) % mod;

        /* if previous count is odd,
                 * then multiply answer by ways[prev_count-1].
                 * since one has already been reserved,
                 * remaining will be even.
                 * reduce prev_count = 0, since we don't need it now.*/
        if ((prev_count & 1) != 0) {
          ans = (ans * ways[prev_count - 1]) % mod;
          count_vector[i - 1][1] = 0;
        }
        else {

          /* if prev count is even, one will be reserved,
                     * therefore decrement by 1.
                     * In the next iteration, prev_count will become odd
                     * and it will be handled in the same way.*/
          count_vector[i - 1][1]--;
        }
      }
      else {

        /* if current count is even,
                 * then simply multiply ways[current_count]
                 * to answer.*/
        ans = (ans * ways[current_count]) % mod;
      }
    }

    /* multiply answer by ways[first__count] since
           that is left out, after iterating the array.*/
    ans = (ans * ways[count_vector[0][1]]) % mod;
    Console.WriteLine(ans);
  }

  // Driver code
  public static void Main(string[] args)
  {
    preCompute();
    int[] arr = { 2, 3, 3, 3, 3, 4, 4, 4, 4, 4 };
    int n = arr.Length;
    countWays(arr, n);
  }
}

// This code is contributed by phasing17.

Javascript

// JavaScript implementation of the
// above approach

// Using mod because the number
// of ways might be very large
let mod = 1000000007
let MAX = 100000

// ways is serving the same
// purpose as discussed
let ways = new Array(MAX + 1)

function preCompute()
{
    // pairing up zero people
    // requires one way.
    ways[0] = 1
    ways[2] = 1
    for (var i = 4; i < MAX + 1; i += 2)
        ways[i] = ((1 * (i - 1) * ways[i - 2]) % mod)
}

function countWays(arr, n)
{
    // map count stores count of s.
    let count = {}
    for (var i = 0; i < n; i++)
    {
        if (!count.hasOwnProperty(arr[i]))
            count[arr[i]] = 0;
        count[arr[i]] += 1
    }

    let count_vector = []
    for (var [key, value]  of  Object.entries(count))
        count_vector.push([key, value])

    // vector count_vector stores a
    // pair < value, count of value>

    // sort according to value
    count_vector.sort()
    let ans = 1

    // Iterating backwards.
    for (var i = (count_vector).length - 1; i > 0; i--)
    {
        let current_count = count_vector[i][1]
        let prev_count = count_vector[i - 1][1]

        // Checking if current count is odd.
        if (current_count & 1)
        {
            // if current count = 5, multiply
            // ans by ways[4].
            ans = (ans * ways[current_count - 1]) % mod

            // left out person will be selected
            // in current_count ways
            ans = (ans * current_count) % mod

            // left out person will pair with previous
            // person in previous_count ways
            ans = (ans * prev_count) % mod

            // if previous count is odd,
            // then multiply answer by ways[prev_count-1].
            // since one has already been reserved,
            // remaining will be even.
            // reduce prev_count = 0, since we
            // don't need it now.
            if (prev_count & 1)
            {
                ans = (ans * ways[prev_count - 1]) % mod
                count_vector[i - 1][1] = 0
            }

            else
            {
                // if prev count is even, one will be
                // reserved, therefore decrement by 1.
                // In the next iteration, prev_count
                // will become odd and it will be
                // handled in the same way.
                count_vector[i - 1][1] -= 1
            }
        }
        else
        {
            // if current count is even, then simply
            // multiply ways[current_count] to answer.
            ans = (ans * ways[current_count]) % mod
        }
    }

    // multiply answer by ways[first__count] since
    // that is left out, after iterating the array.
    ans = (ans * ways[count_vector[0][1]]) % mod
    console.log(ans)
}

// Driver code

preCompute()
let arr = [2, 3, 3, 3, 3, 4, 4, 4, 4, 4]
let n = arr.length
countWays(arr, n)


// This code is contributed by phasing17

Output:

180

Time Complexity:
The preCompute function has a time complexity of O(MAX), where MAX is the maximum value of the input array. The countWays function has a time complexity of O(n log n), where n is the size of the input array, due to the sorting operation. Therefore, the overall time complexity of the program is O(MAX + n log n).

Auxiliary Space:
The program uses an array called ways of size MAX + 1, so the auxiliary space complexity is O(MAX). Additionally, the program uses a map and a vector to store the count of elements and their frequencies, respectively. The space required for the map and vector is proportional to the number of distinct elements in the input array, which is at most MAX. Therefore, the overall auxiliary space complexity is O(MAX).

Space Complexity:
The space complexity of the program is the sum of the time complexity and the auxiliary space complexity. Therefore, the space complexity of the program is O(MAX + n log n).