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.

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



Below is the implementation of the approach:

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// C++ implementation of the above approach
#include <bits/stdc++.h>
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<<totalways(arr, N);
  
    return 0;
}
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// 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
    HashMap<Integer, 
            Integer>cnt = new HashMap<Integer, 
                                      Integer>();
  
    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
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# 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
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// 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
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Output:
4

Time Complexity : O(N)

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