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

• Difficulty Level : Medium
• Last Updated : 05 Sep, 2022

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 :
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 :
No any valid order is possible hence answer is 0.

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 fulfill 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 differencesint 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 Codeint main(){    int N = 5;     int arr[N] = { 2, 4, 4, 0, 2 };     cout<

## Java

 // Java implementation of the above approachimport java.util.*; class GFG{ // Function to find the number of permutations// possible of the original array to satisfy// the given absolute differencesstatic 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 Codepublic 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 differencesdef 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 CodeN = 5 arr = [2, 4, 4, 0, 2 ] print(totalways(arr, N)) # This code is contributed by Mohit Kumar

## C#

 // C# implementation of the above approachusing System;using System.Collections.Generic; class GFG{ // Function to find the number of permutations// possible of the original array to satisfy// the given absolute differencesstatic 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 Codepublic 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

## Javascript



Output:

4

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N) for using a hashmap to store the frequency of the given elements.

My Personal Notes arrow_drop_up