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

• Difficulty Level : Medium
• Last Updated : 04 Jan, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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)

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)

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.

• Considering element with value 4. Since there are 5 elements, 4 elements will pair amongst themselves in ways[4] ways. The left out element can be chosen in 5 ways. The element left will have to pair with an element with value 3. This happens in 4 ways since there are 4 elements with value 3. Therefore, one element of value 3 will be reserved to pair with an element of 4. So 3 elements with value 3 remain. 2 will pair amongst themselves in ways[2] ways and one 1 will pair with the element with value 2 in 1 way. Again, the lone element will be selected in 3 ways.
• Therefore from point 1, number of ways will be :

ways[4] * 5 * 4 * ways[2] * 3 * 1 = 180

Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach#include   #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 largeconst int mod = 1000000007;  const int MAX = 100000;  // ways is serving the same// purpose as discussedll 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 count;    for (int i = 0; i < n; i++)        count[arr[i]]++;      vector > count_vector;    map::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 codeint 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;}

## Python3

 # Python3 implementation of the # above approach from collections import defaultdict  # Using mod because the number # of ways might be very large mod = 1000000007MAX = 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
Output:
180

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