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

• Difficulty Level : Easy
• Last Updated : 11 Jul, 2022

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 = 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 ways. So number of ways would be 5 * ways.

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 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 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 :

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 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 = 1LL;``    ``ways = 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 > 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.``            ``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.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);``    ``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 ``=` `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.``            ``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`

Time Complexity: O(nlogn)

Auxiliary Space: O(MAX + n)

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