Count of possible arrays from prefix-sum and suffix-sum arrays

Given 2*N integers which are elements of a prefix and suffix array(in shuffled order) of an array of size N, the task is to find the no of possible array’s of the size N which can be made from these elements

Examples:

Input: arr[] = {5, 2, 3, 5}
Output: 2
Explanation:
1st array can be : {2, 3}
Its prefix array:{2, 5}
Its suffix array:{5, 3}
2nd array can be : {3, 2}
Its prefix array : {3, 5}
Its suffix array : {5, 2}

Input: arr[] = {-1, -1, -1, 0, 1, 0, 1, 0, 1, 0, 0, 0}
Output: 80

Approach:



  • One insight which can be drawn is that if the sum of all elements of the given array is divided by n+1, then the last and the first element of a prefix and suffix array is obtained respectively.
  • This conclusion can be drawn by observing the elements of the prefix and suffix array. The sum of 1st element of prefix array and 2nd element of suffix array is equal to the sum of 2nd element of prefix array and 3rd element of suffix array(if there is a third element in the suffix array) and so on.

  • In the image, the first array is the given array, the second is the prefix array and the third is suffix array.
  • The sum of these pairs is equal to the sum of all elements of the array to be found.
  • If it is assumed that the sum of the pairs is s1 and the sum of all prefix and suffix elements is s then:
    s1 * (n-1) + 2 * s1 = s
    s1 = s / (n+1)

    where s1 is the last element of prefix array and 1st element of suffix array.
  • Now, all other pairs whose sum will be equal to s1 need to be found which can be done using hash maps.
  • If these pairs are shuffled linearly along with the array then we can get the answer as
    (n-1)! / (k1! * k2! … kr!)
    where k1, k2 … kr are the number of similar pairs
  • Each pair can also be interchanged among itself in the prefix and the suffix array(If the elements of pair are not equal) so the answer becomes
    (n-1)! * (2^p) / (k1!*k2!…kr!)
    where p is the no of distinct pairs in the array whose sum is equal to s1.

Below is the implementation of the above approach.

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find power of
// a number.
int power(int a, int b)
{
    int result = 1;
    while (b > 0) {
        if (b % 2 == 1) {
            result = result * a;
        }
        a = a * a;
        b = b / 2;
    }
    return result;
}
  
// Function to find
// factorial of a number.
int factorial(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact = fact * i;
    }
    return fact;
}
  
// Function to print no of arrays
void findNoOfArrays(int* a, int n)
{
    // c variable counts the no of pairs
    int sum = 0, s1, c = 0;
  
    // Map to store the frequency
    // of each element
    map<int, int> mp;
  
    for (int i = 0; i < 2 * n; i++) {
        mp[a[i]]++;
  
        // Sum of all elements of the array
        sum = sum + a[i];
    }
  
    // Variable to check if it is
    // possible to make any array
    bool isArrayPossible = true;
    int ans = factorial(n - 1);
  
    // First element of suffix array
    // and the last element of prefix array
    s1 = sum / (n + 1);
  
    // Check if the element exists in the map
    if (mp[s1] >= 2) {
        mp[s1] = mp[s1] - 2;
    }
    else {
        isArrayPossible = false;
    }
    if (isArrayPossible) {
        for (auto i : mp) {
  
            // If elements of any pair are equal
            // and their frequency is not divisible by 2
            // update the isArrayPossible variable
            // to false and break through the loop
  
            if (i.first == s1 - i.first) {
                if (mp[i.first] % 2 != 0) {
                    isArrayPossible = false;
                    break;
                }
            }
  
            // If elements of any pair are not equal
            // and their frequency is not same
            // update the isArrayPossible variable
            // to false and break through the loop
  
            if (i.first != s1 - i.first) {
                if (mp[i.first]
                    != mp[s1 - i.first]) {
                    isArrayPossible = false;
                    break;
                }
            }
            // Check if frequency is greater than zero
            if (i.second > 0) {
                if (i.first != s1 - i.first) {
                    // update the count of pairs
  
                    c = c + i.second;
  
                    // Multiply the answer by
                    // 2^(frequency of pairs) since
                    // the elements of the pair are
                    // not the same in this condition
  
                    ans = ans * power(2, i.second);
  
                    // Divide the answer by the factorial
                    // of no of similar pairs
  
                    ans = ans / factorial(i.second);
  
                    // Make frequency of both these elements 0
  
                    mp[i.first] = 0;
                    mp[s1 - i.first] = 0;
                }
                if (i.first == s1 - i.first) {
                    // Update the count of pairs
  
                    c = c + i.second / 2;
  
                    // Divide the answer by the factorial
                    // of no. of similar pairs
  
                    ans = ans / factorial(i.second / 2);
  
                    // Make frequency of this element 0
                    mp[i.first] = 0;
                }
            }
        }
    }
  
    // Check if it is possible to make the
    // array and there are n-1 pairs
    // whose sum will be equal to s1
    if (c < n - 1 || isArrayPossible == false) {
        cout << "0" << endl;
    }
    else {
        cout << ans << endl;
    }
}
  
// Driver code
int main()
{
    int arr1[] = { 5, 2, 3, 5 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
  
    // Function calling
    findNoOfArrays(arr1, n1 / 2);
  
    int arr2[] = { -1, -1, -1, 0, 1, 0,
                   1, 0, 1, 0, 0, 0 };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    findNoOfArrays(arr2, n2 / 2);
    return 0;
}

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Java

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// Java implementation of the above approach
import java.util.*;
  
class GFG{
      
// Function to find power of
// a number.
static int power(int a, int b)
{
    int result = 1;
    while (b > 0) {
        if (b % 2 == 1) {
            result = result * a;
        }
        a = a * a;
        b = b / 2;
    }
    return result;
}
  
// Function to find
// factorial of a number.
static int factorial(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact = fact * i;
    }
    return fact;
}
  
// Function to print no of arrays
static void findNoOfArrays(int[] a, int n)
{
    // c variable counts the no of pairs
    int sum = 0, s1, c = 0;
  
    // Map to store the frequency
    // of each element
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();         
  
    for (int i = 0; i < 2 * n; i++) {
        if(mp.get(a[i])==null)
          mp.put(a[i], 1);
        else
          mp.put(a[i], mp.get(a[i]) + 1);
  
        // Sum of all elements of the array
        sum = sum + a[i];
    }
  
    // Variable to check if it is
    // possible to make any array
    boolean isArrayPossible = true;
    int ans = factorial(n - 1);
  
    // First element of suffix array
    // and the last element of prefix array
    s1 = sum / (n + 1);
  
    // Check if the element exists in the map
    if (mp.get(s1) >= 2) {
        mp.replace(s1, mp.get(s1) - 2); 
    }
    else {
        isArrayPossible = false;
    }
    if (isArrayPossible) {
        for (Map.Entry<Integer,Integer> m:mp.entrySet()) {
  
            // If elements of any pair are equal
            // and their frequency is not divisible by 2
            // update the isArrayPossible variable
            // to false and break through the loop
  
            if (m.getKey() == s1-m.getKey()) {
                if (mp.get(m.getKey()) % 2 != 0) {
                    isArrayPossible = false;
                    break;
                }
            }
  
            // If elements of any pair are not equal
            // and their frequency is not same
            // update the isArrayPossible variable
            // to false and break through the loop
  
            if (m.getKey() != s1 - m.getKey()) {
                if (mp.get(m.getKey())
                    != mp.get(s1 - m.getKey())) {
                    isArrayPossible = false;
                    break;
                }
            }
            // Check if frequency is greater than zero
            if (m.getValue() > 0) {
                if (m.getKey() != s1 - m.getKey()) {
                    // update the count of pairs
  
                    c = c + m.getValue();
  
                    // Multiply the answer by
                    // 2^(frequency of pairs) since
                    // the elements of the pair are
                    // not the same in this condition
                    ans = ans * power(2, m.getValue());
  
                    // Divide the answer by the factorial
                    // of no of similar pairs
                    ans = ans / factorial(m.getValue());
  
                    // Make frequency of both these elements 0
                    mp.replace(m.getKey(),0);
                    mp.replace(s1 - m.getKey(),0);
                }
                if (m.getKey() == s1 - m.getKey()) {
                    // Update the count of pairs
  
                    c = c + m.getValue() / 2;
  
                    // Divide the answer by the factorial
                    // of no. of similar pairs
                    ans = ans / factorial(m.getValue() / 2);
  
                    // Make frequency of this element 0
                    mp.replace(m.getKey(),0);
                }
            }
        }
    }
  
    // Check if it is possible to make the
    // array and there are n-1 pairs
    // whose sum will be equal to s1
    if (c < n - 1 && isArrayPossible == false) {
        System.out.println("0");
    }
    else {
        System.out.println(ans);
    }
}
  
// Driver code
public static void main(String args[])
{
    int[] arr1 = { 5, 2, 3, 5 };
    int n1 = arr1.length;
  
    // Function calling
    findNoOfArrays(arr1, n1 / 2);
  
    int []arr2 = { -1, -1, -1, 0, 1, 0,
                1, 0, 1, 0, 0, 0 };
    int n2 = arr2.length;
    findNoOfArrays(arr2, n2 / 2);
}
}
  
// This code is contributed by Surendra_Gangwar

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Python3

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# Python3 implementation of the above approach 
  
# Function to find power of 
# a number. 
def power(a, b) :
  
    result = 1
    while (b > 0) :
        if (b % 2 == 1) : 
            result = result * a; 
        a = a * a; 
        b = b // 2
      
    return result; 
  
# Function to find 
# factorial of a number. 
def factorial(n) : 
  
    fact = 1
    for i in range(1, n + 1) :
        fact = fact * i; 
      
    return fact; 
  
# Function to print no of arrays 
def findNoOfArrays(a, n) :
  
    # c variable counts the no of pairs 
    sum = 0; c = 0
  
    # Map to store the frequency 
    # of each element 
    mp = dict.fromkeys(a, 0); 
  
    for i in range(2 * n) :
        mp[a[i]] += 1
  
        # Sum of all elements of the array 
        sum = sum + a[i];
  
    # Variable to check if it is 
    # possible to make any array 
    isArrayPossible = True
    ans = factorial(n - 1); 
  
    # First element of suffix array 
    # and the last element of prefix array 
    s1 = sum // (n + 1); 
  
    # Check if the element exists in the map 
    if (mp[s1] >= 2) :
        mp[s1] = mp[s1] - 2
          
    else :
        isArrayPossible = False
      
    if (isArrayPossible) : 
        for first,second in mp.items() : 
              
            # If elements of any pair are equal 
            # and their frequency is not divisible by 2 
            # update the isArrayPossible variable 
            # to false and break through the loop 
            if (first == s1 - first) :
                if (mp[first] % 2 != 0) :
                    isArrayPossible = False;
                    break
  
            # If elements of any pair are not equal 
            # and their frequency is not same 
            # update the isArrayPossible variable 
            # to false and break through the loop 
            if (first != s1 - first) :
                if s1 - first in mp :
                    if (mp[first] != mp[s1 - first]) :
                        isArrayPossible = False;
                        break
              
            # Check if frequency is greater than zero 
            if (second > 0) :
                if (first != s1 - first) :
  
                    # update the count of pairs 
                    c = c + second; 
  
                    # Multiply the answer by 
                    # 2^(frequency of pairs) since 
                    # the elements of the pair are 
                    # not the same in this condition 
                    ans = ans * power(2, second); 
  
                    # Divide the answer by the factorial 
                    # of no of similar pairs 
                    ans = ans / factorial(second); 
  
                    # Make frequency of both these elements 0 
                    mp[first] = 0
                    mp[s1 - first] = 0
                  
                if (first == s1 - first) : 
  
                    # Update the count of pairs 
                    c = c + second // 2
  
                    # Divide the answer by the factorial 
                    # of no. of similar pairs 
                    ans = ans // factorial(second // 2); 
  
                    # Make frequency of this element 0 
                    mp[first] = 0
  
    # Check if it is possible to make the 
    # array and there are n-1 pairs 
    # whose sum will be equal to s1 
    if (c < n - 1 or isArrayPossible == False) :
        print("0"); 
    else
        print(ans); 
  
# Driver code 
if __name__ == "__main__"
  
    arr1 = [ 5, 2, 3, 5 ]; 
    n1 = len(arr1); 
  
    # Function calling 
    findNoOfArrays(arr1, n1 // 2); 
  
    arr2 = [ -1, -1, -1, 0, 1, 0
                1, 0, 1, 0, 0, 0 ]; 
    n2 = len(arr2); 
    findNoOfArrays(arr2, n2 // 2); 
      
# This code is contributed by AnkitRai01

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Output:

2
80

Time complexity: O(N Log(N))

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