# Number of possible permutations when absolute difference between number of elements to the right and left are given

Given an array of N elements where each element i, the absolute difference between total elements to the right and left of it are given. Find the number of possible ordering of the actual array elements.

Examples:

Input : N = 5, arr[] = {2, 4, 4, 0, 2}
Output : 4
There are four possible orders, as follows:
2, 1, 4, 5, 3
2, 5, 4, 1, 3
3, 1, 4, 5, 2
3, 5, 4, 1, 2

Input : N = 7, arr[] = {6, 4, 0, 2, 4, 0, 2}
Output : 0
No any valid order is possible hence answer is 0.

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

Approach: Divide the problem into two parts. When N is odd and when N is even.

• Case 1 : When N is odd.
Consider N = 7, there are 7 empty spaces and the absolute difference between the elements to the left and right must be like [6 4 2 0 2 4 6]. Observe that the element which is at the middle must have absolute difference 0, while other elements are from 2 to N-1 and each of their counts should be 2. If it doesn’t fulfil it then there is no valid order else for each element i from 2 to N-1 we have 2 ways to fill the spaces, hence total ways will be the product of all the ways.
• Case 2 : When N is even.
Consider N = 6, There are 6 spaces and it will be like [5 3 1 1 3 5], where a[i] gives the absolute difference between the number of elements to the left and right. For each a[i] we have 2 ways, hence answer will be the product of all the ways.

Below is the implementation of the approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of permutations  ` `// possible of the original array to satisfy  ` `// the given absolute differences ` `int` `totalways(``int``* arr, ``int` `n) ` `{ ` `    ``// To store the count of each ` `    ``// a[i] in a map ` `    ``unordered_map<``int``, ``int``> cnt; ` `    ``for` `(``int` `i = 0; i < n; ++i) { ` `        ``cnt[arr[i]]++; ` `    ``} ` ` `  `    ``// if n is odd ` `    ``if` `(n % 2 == 1) { ` `        ``int` `start = 0, endd = n - 1; ` ` `  `        ``// check the count of each whether ` `        ``// it satisfy the given criteria or not ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2) { ` `            ``if` `(i == 0) { ` ` `  `                ``// there is only 1 way ` `                ``// for middle element. ` `                ``if` `(cnt[i] != 1) { ` `                    ``return` `0; ` `                ``} ` `            ``} ` `            ``else` `{ ` ` `  `                ``// for others there are 2 ways. ` `                ``if` `(cnt[i] != 2) { ` `                    ``return` `0; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// now find total ways ` `        ``int` `ways = 1; ` `        ``start = 2, endd = n - 1; ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2) { ` `            ``ways = ways * 2; ` `        ``} ` `        ``return` `ways; ` `    ``} ` ` `  `    ``// When n is even. ` `    ``else` `if` `(n % 2 == 0) { ` ` `  `        ``// there will be no middle element so ` `        ``// for each a[i] there will be 2 ways ` `        ``int` `start = 1, endd = n - 1; ` `        ``for` `(``int` `i = 1; i <= endd; i = i + 2) { ` `            ``if` `(cnt[i] != 2) ` `                ``return` `0; ` `        ``} ` `        ``int` `ways = 1; ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2) { ` `            ``ways = ways * 2; ` `        ``} ` `        ``return` `ways; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `N = 5; ` ` `  `    ``int` `arr[N] = { 2, 4, 4, 0, 2 }; ` ` `  `    ``cout<

## Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the number of permutations  ` `// possible of the original array to satisfy  ` `// the given absolute differences ` `static` `int` `totalways(``int``[] arr, ``int` `n) ` `{ ` `    ``// To store the count of each ` `    ``// a[i] in a map ` `    ``HashMapcnt = ``new` `HashMap(); ` ` `  `    ``for` `(``int` `i = ``0` `; i < n; i++) ` `    ``{ ` `        ``if``(cnt.containsKey(arr[i])) ` `        ``{ ` `            ``cnt.put(arr[i], cnt.get(arr[i])+``1``); ` `        ``} ` `        ``else` `        ``{ ` `            ``cnt.put(arr[i], ``1``); ` `        ``} ` `    ``} ` `     `  `    ``// if n is odd ` `    ``if` `(n % ``2` `== ``1``) ` `    ``{ ` `        ``int` `start = ``0``, endd = n - ``1``; ` ` `  `        ``// check the count of each whether ` `        ``// it satisfy the given criteria or not ` `        ``for` `(``int` `i = start; i <= endd; i = i + ``2``)  ` `        ``{ ` `            ``if` `(i == ``0``)  ` `            ``{ ` ` `  `                ``// there is only 1 way ` `                ``// for middle element. ` `                ``if` `(cnt.get(i) != ``1``) ` `                ``{ ` `                    ``return` `0``; ` `                ``} ` `            ``} ` `            ``else`  `            ``{ ` ` `  `                ``// for others there are 2 ways. ` `                ``if` `(cnt.get(i) != ``2``)  ` `                ``{ ` `                    ``return` `0``; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// now find total ways ` `        ``int` `ways = ``1``; ` `        ``start = ``2``; endd = n - ``1``; ` `        ``for` `(``int` `i = start; i <= endd; i = i + ``2``)  ` `        ``{ ` `            ``ways = ways * ``2``; ` `        ``} ` `        ``return` `ways; ` `    ``} ` ` `  `    ``// When n is even. ` `    ``else` `if` `(n % ``2` `== ``0``)  ` `    ``{ ` ` `  `        ``// there will be no middle element so ` `        ``// for each a[i] there will be 2 ways ` `        ``int` `start = ``1``, endd = n - ``1``; ` `        ``for` `(``int` `i = ``1``; i <= endd; i = i + ``2``)  ` `        ``{ ` `            ``if` `(cnt.get(i) != ``2``) ` `                ``return` `0``; ` `        ``} ` `        ``int` `ways = ``1``; ` `        ``for` `(``int` `i = start; i <= endd; i = i + ``2``)  ` `        ``{ ` `            ``ways = ways * ``2``; ` `        ``} ` `        ``return` `ways; ` `    ``} ` `    ``return` `Integer.MIN_VALUE; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `N = ``5``; ` ` `  `    ``int` `[]arr = { ``2``, ``4``, ``4``, ``0``, ``2` `}; ` ` `  `    ``System.out.println(totalways(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function to find the number of permutations ` `# possible of the original array to satisfy ` `# the given absolute differences ` `def` `totalways(arr, n): ` `     `  `    ``# To store the count of each ` `    ``# a[i] in a map ` `    ``cnt ``=` `dict``() ` `    ``for` `i ``in` `range``(n): ` `        ``cnt[arr[i]] ``=` `cnt.get(arr[i], ``0``) ``+` `1` ` `  `    ``# if n is odd ` `    ``if` `(n ``%` `2` `=``=` `1``): ` `        ``start, endd ``=` `0``, n ``-` `1` ` `  `        ``# check the count of each whether ` `        ``# it satisfy the given criteria or not ` `        ``for` `i ``in` `range``(start, endd ``+` `1``, ``2``): ` `            ``if` `(i ``=``=` `0``): ` ` `  `                ``# there is only 1 way ` `                ``# for middle element. ` `                ``if` `(cnt[i] !``=` `1``): ` `                    ``return` `0` `            ``else``: ` ` `  `                ``# for others there are 2 ways. ` `                ``if` `(cnt[i] !``=` `2``): ` `                    ``return` `0` ` `  `        ``# now find total ways ` `        ``ways ``=` `1` `        ``start ``=` `2` `        ``endd ``=` `n ``-` `1` `        ``for` `i ``in` `range``(start, endd ``+` `1``, ``2``): ` `            ``ways ``=` `ways ``*` `2` `        ``return` `ways ` ` `  `    ``# When n is even. ` `    ``elif` `(n ``%` `2` `=``=` `0``): ` ` `  `        ``# there will be no middle element so ` `        ``# for each a[i] there will be 2 ways ` `        ``start ``=` `1` `        ``endd ``=` `n ``-` `1` `        ``for` `i ``in` `range``(``1``, endd ``+` `1``, ``2``): ` `            ``if` `(cnt[i] !``=` `2``): ` `                ``return` `0` `        ``ways ``=` `1` `        ``for` `i ``in` `range``(start, endd ``+` `1``, ``2``): ` `            ``ways ``=` `ways ``*` `2` `        ``return` `ways ` ` `  `# Driver Code ` `N ``=` `5` ` `  `arr ``=` `[``2``, ``4``, ``4``, ``0``, ``2` `] ` ` `  `print``(totalways(arr, N)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the number of permutations  ` `// possible of the original array to satisfy  ` `// the given absolute differences ` `static` `int` `totalways(``int``[] arr, ``int` `n) ` `{ ` `    ``// To store the count of each ` `    ``// a[i] in a map ` `    ``Dictionary<``int``, ` `               ``int``> cnt = ``new` `Dictionary<``int``, ` `                                         ``int``>(); ` ` `  `    ``for` `(``int` `i = 0 ; i < n; i++) ` `    ``{ ` `        ``if``(cnt.ContainsKey(arr[i])) ` `        ``{ ` `            ``cnt[arr[i]] = cnt[arr[i]] + 1; ` `        ``} ` `        ``else` `        ``{ ` `            ``cnt.Add(arr[i], 1); ` `        ``} ` `    ``} ` `     `  `    ``// if n is odd ` `    ``if` `(n % 2 == 1) ` `    ``{ ` `        ``int` `start = 0, endd = n - 1; ` ` `  `        ``// check the count of each whether ` `        ``// it satisfy the given criteria or not ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2)  ` `        ``{ ` `            ``if` `(i == 0)  ` `            ``{ ` ` `  `                ``// there is only 1 way ` `                ``// for middle element. ` `                ``if` `(cnt[i] != 1) ` `                ``{ ` `                    ``return` `0; ` `                ``} ` `            ``} ` `            ``else` `            ``{ ` ` `  `                ``// for others there are 2 ways. ` `                ``if` `(cnt[i] != 2)  ` `                ``{ ` `                    ``return` `0; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``// now find total ways ` `        ``int` `ways = 1; ` `        ``start = 2; endd = n - 1; ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2)  ` `        ``{ ` `            ``ways = ways * 2; ` `        ``} ` `        ``return` `ways; ` `    ``} ` ` `  `    ``// When n is even. ` `    ``else` `if` `(n % 2 == 0)  ` `    ``{ ` ` `  `        ``// there will be no middle element so ` `        ``// for each a[i] there will be 2 ways ` `        ``int` `start = 1, endd = n - 1; ` `        ``for` `(``int` `i = 1; i <= endd; i = i + 2)  ` `        ``{ ` `            ``if` `(cnt[i] != 2) ` `                ``return` `0; ` `        ``} ` `         `  `        ``int` `ways = 1; ` `        ``for` `(``int` `i = start; i <= endd; i = i + 2)  ` `        ``{ ` `            ``ways = ways * 2; ` `        ``} ` `        ``return` `ways; ` `    ``} ` `    ``return` `int``.MinValue; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `N = 5; ` ` `  `    ``int` `[]arr = { 2, 4, 4, 0, 2 }; ` ` `  `    ``Console.WriteLine(totalways(arr, N)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```4
```

Time Complexity : O(N)

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