# 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 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 cnt = new Dictionary();        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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