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Count distinct elements after adding each element of First Array with Second Array

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Given two arrays arr1[] and arr2[]. We can generate another array arr3[] by adding each element of the array arr1[] to each element arr2[]. The task is to find the count of distinct element in the array arr3[].

Examples:  

Input: Arr1[] = {1, 2}, Arr2[] = {3, 4}, MAX = 4 
Output: 
4 -> 1 
5 -> 2 
6 -> 1 
Explanation: 
Here the third array will be Arr3[] = {1+3, 1+4, 2+3, 2+4} = {4, 5, 5, 6}

Input: Arr1[] = {1, 2}, Arr2[] = {1, 2, 1}, MAX = 2 
Output: 
2 -> 2 
3 -> 3 
4 -> 1 
Explanation: 
Here the third array is Arr3[] = {1+1, 1+2, 1+1, 2+1, 2+2, 2+1} = {2, 3, 2, 3, 4, 3} 
Therefore Count of elements from 1 to 2*2 (4) are {0, 2, 3, 1}

Naive Approach: The naive approach is to find the sum of all possible pairs from the given two arrays and insert that sum into the array arr3[]. Print the frequency of all the elements of the array arr3[].

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find Occurrence of each
// element from 1 to 2*MAX
void findCount(vector<int>& Arr1,
               vector<int>& Arr2)
{
    // Initialise MAX
    int MAX = max(*max_element(Arr1.begin(),
                               Arr1.end()),
                  *max_element(Arr2.begin(),
                               Arr2.end()));
 
    // Count vector to store count of
    // each element from 1 to 2*MAX
    vector<int> Count(2 * MAX + 1, 0);
 
    // Size of Arr1 and Arr2
    int n = Arr1.size(), m = Arr2.size();
 
    // Find the elements of arr3[] and
    // increase count of element by 1
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < m; j++) {
 
            int element = Arr1[i] + Arr2[j];
 
            Count[element]++;
        }
    }
 
    // Print the result
    for (int i = 1; i <= 2 * MAX; i++) {
 
        if (Count[i] > 0) {
            cout << i << "->"
                 << Count[i] << endl;
        }
    }
}
 
// Driver Code
int main()
{
    // Given arrays arr1[] and arr2[]
    vector<int> arr1 = { 1, 2 };
    vector<int> arr2 = { 1, 2, 1 };
 
    // Function Call
    findCount(arr1, arr2);
}

                    

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find Occurrence of each
// element from 1 to 2*MAX
static void findCount(int[] Arr1, int[]Arr2)
{
     
    // Initialise MAX
    int MAX = Math.max(Arrays.stream(Arr1).max().getAsInt(),
                       Arrays.stream(Arr2).max().getAsInt());
 
    // Count vector to store count of
    // each element from 1 to 2*MAX
    int[] Count = new int[2 * MAX + 1];
 
    // Size of Arr1 and Arr2
    int n = Arr1.length, m = Arr2.length;
 
    // Find the elements of arr3[] and
    // increase count of element by 1
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            int element = Arr1[i] + Arr2[j];
 
            Count[element]++;
        }
    }
 
    // Print the result
    for(int i = 1; i <= 2 * MAX; i++)
    {
        if (Count[i] > 0)
        {
            System.out.print(i + "->" +
                      Count[i] + "\n");
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given arrays arr1[] and arr2[]
    int[] arr1 = { 1, 2 };
    int[] arr2 = { 1, 2, 1 };
 
    // Function call
    findCount(arr1, arr2);
}
}
 
// This code is contributed by sapnasingh4991

                    

Python3

# Python3 program for the above approach
 
# Function to find Occurrence of each
# element from 1 to 2*MAX
def findCount(Arr1, Arr2):
 
    # Initialise MAX
    MAX = max(max(Arr1), max(Arr2));
 
    # Count vector to store count of
    # each element from 1 to 2*MAX
    #Count = new int[2 * MAX + 1];
    Count = [0 for i in range(2 * MAX + 1)]
     
    # Size of Arr1 and Arr2
    n = len(Arr1);
    m = len(Arr2);
 
    # Find the elements of arr3 and
    # increase count of element by 1
    for i in range(n):
        for j in range(m):
            element = Arr1[i] + Arr2[j];
 
            Count[element]+=1;
         
    # Print the result
    for i in range(1,2*MAX+1):
        if (Count[i] > 0):
            print(i , "->" , Count[i]);
         
# Driver Code
if __name__ == '__main__':
 
    # Given arrays arr1 and arr2
    arr1 = [1, 2 ];
    arr2 = [ 1, 2, 1 ];
 
    # Function call
    findCount(arr1, arr2);
 
# This code is contributed by Rohit_ranjan

                    

C#

// C# program for the above approach
using System;
using System.Linq;
 
class GFG{
 
// Function to find Occurrence of each
// element from 1 to 2*MAX
static void findCount(int[] Arr1, int[]Arr2)
{
     
    // Initialise MAX
    int MAX = Math.Max(Arr1.Max(), Arr2.Max());
 
    // Count vector to store count of
    // each element from 1 to 2*MAX
    int[] Count = new int[2 * MAX + 1];
 
    // Size of Arr1 and Arr2
    int n = Arr1.Length, m = Arr2.Length;
 
    // Find the elements of arr3[] and
    // increase count of element by 1
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            int element = Arr1[i] + Arr2[j];
            Count[element]++;
        }
    }
 
    // Print the result
    for(int i = 1; i <= 2 * MAX; i++)
    {
        if (Count[i] > 0)
        {
            Console.Write(i + "->" +
                   Count[i] + "\n");
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given arrays arr1[] and arr2[]
    int[] arr1 = { 1, 2 };
    int[] arr2 = { 1, 2, 1 };
 
    // Function call
    findCount(arr1, arr2);
}
}
 
// This code is contributed by Princi Singh

                    

Javascript

<script>
// JavaScript program for the above approach
 
// Function to find Occurrence of each
// element from 1 to 2*MAX
function findCount(Arr1, Arr2)
{
       
    // Initialise MAX
    let MAX = Math.max(Math.max(...Arr1),
                       Math.max(...Arr2));
   
    // Count vector to store count of
    // each element from 1 to 2*MAX
    let Count = Array.from({length: 2 * MAX + 1}, (_, i) => 0);
   
    // Size of Arr1 and Arr2
    let n = Arr1.length, m = Arr2.length;
   
    // Find the elements of arr3[] and
    // increase count of element by 1
    for(let i = 0; i < n; i++)
    {
        for(let j = 0; j < m; j++)
        {
            let element = Arr1[i] + Arr2[j];
   
            Count[element]++;
        }
    }
   
    // Print the result
    for(let i = 1; i <= 2 * MAX; i++)
    {
        if (Count[i] > 0)
        {
            document.write(i + "->" +
                      Count[i] + "<br/>");
        }
    }
}
       
// Driver Code
     
    // Given arrays arr1[] and arr2[]
    let arr1 = [ 1, 2 ];
    let arr2 = [ 1, 2, 1 ];
   
    // Function call
    findCount(arr1, arr2);
               
</script>

                    

Output: 
2->2
3->3
4->1

 

Time Complexity: O(N2
Space Complexity: O(N)

Efficient Solution: The given task can be efficiently done with the help of FFT(Fast Fourier Transform). Below are the steps:  

  1. Consider the examples Arr1[] = {1, 2} and Arr2[] = {1, 2, 1}. Let Count be the frequency array i.e., Count[i] represents the frequency of i in resultant array.
  2. When Arr1[i] is added to Arr2[j], we increment Count[s] where s = Arr1[i]+Arr2[j]. This is similar to multiplying polynomials as there power get added.
  3. Let A(x) be the polynomial represented by Arr1[]. Elements of Arr1 represents power of x and their count in Arr1 are coefficients terms with that power in polynomial.
  4. For each term, power of x represents the resulting element and the coefficient represents its count.
  5. If term isk(x^i)
  6. Then Count[i] = k. Here Count is same as P(x).
  7. To calculate the value of P(x), we can simply multiply A(x) and B(x).

The Naive method of polynomial multiplication takes O(N2). To make the multiplication faster we can use FFT(Fast Fourier Transform).

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
 
// Value of PI need in FFT
const double PI = acos(-1);
 
// Function to implement the FFT
void fft(vector<cd>& a, bool invert)
{
    int n = a.size();
    if (n == 1)
        return;
 
    vector<cd> a0(n / 2), a1(n / 2);
    for (int i = 0; 2 * i < n; i++) {
        a0[i] = a[2 * i];
        a1[i] = a[2 * i + 1];
    }
 
    // Recursively find fft
    fft(a0, invert);
    fft(a1, invert);
 
    double ang = 2 * PI / n * (invert ? -1 : 1);
 
    cd w(1), wn(cos(ang), sin(ang));
 
    for (int i = 0; 2 * i < n; i++) {
        a[i] = a0[i] + w * a1[i];
        a[i + n / 2] = a0[i] - w * a1[i];
        if (invert) {
            a[i] /= 2;
            a[i + n / 2] /= 2;
        }
        w *= wn;
    }
}
 
// Function to multiply two polynomials
// A(x) and B(x) using FFT
vector<int> multiply(vector<int> const& a,
                     vector<int> const& b)
{
    vector<cd> fa(a.begin(), a.end()),
        fb(b.begin(), b.end());
 
    int n = 1;
 
    while (n < a.size() + b.size()) {
        n <<= 1;
    }
 
    // Resize fa and fb
    fa.resize(n);
    fb.resize(n);
 
    // Assign initially false
    fft(fa, false);
    fft(fb, false);
 
    for (int i = 0; i < n; i++)
        fa[i] *= fb[i];
 
    fft(fa, true);
 
    // To store the result
    vector<int> result(n);
 
    for (int i = 0; i < n; i++)
        result[i] = round(fa[i].real());
 
    // Return result
    return result;
}
 
// Function to find the Count of each
// element from 1 to 2*MAX
void findCount(vector<int>& Arr1,
               vector<int>& Arr2)
{
    // Initialise MAX
    int MAX = max(*max_element(Arr1.begin(),
                               Arr1.end()),
                  *max_element(Arr2.begin(),
                               Arr2.end()));
 
    int n = Arr1.size();
    int m = Arr2.size();
 
    // vector for Polynomial A(x) from Arr1
    vector<int> A(MAX + 1);
 
    for (int i = 0; i < n; i++) {
        A[Arr1[i]]++;
    }
 
    // Vector for Polynomial B(x) from Arr2
    vector<int> B(MAX + 1);
 
    for (int i = 0; i < m; i++) {
        B[Arr2[i]]++;
    }
 
    // Vector to store the result of
    // multiplication of A(x) and B(x)
    vector<int> P;
 
    // Multiplying Arr1 and Arr2 and
    // storing in P is same as Count
    P = multiply(A, B);
 
    // Print the result
    for (int i = 1; i <= 2 * MAX; i++) {
        if (P[i] > 0) {
            cout << i << "->"
                 << P[i] << endl;
        }
    }
 
    cout << '\n';
}
 
// Driver Code
int main()
{
    // Given arrays arr1[] and arr2[]
    vector<int> arr1 = { 1, 2 };
    vector<int> arr2 = { 1, 2, 1 };
 
    // Function Call
    findCount(arr1, arr2);
}

                    

Java

import java.util.*;
 
// Complex class definition
class Complex {
 
  // Data members
  public double real, imag;
 
  // Constructor
  public Complex(double real, double imag)
  {
    this.real = real;
    this.imag = imag;
  }
 
  // Method to add two Complex objects
  public Complex add(Complex other)
  {
    return new Complex(real + other.real,
                       imag + other.imag);
  }
 
  // Method to subtract two Complex objects
  public Complex subtract(Complex other)
  {
    return new Complex(real - other.real,
                       imag - other.imag);
  }
 
  // Method to multiply two Complex objects
  public Complex multiply(Complex other)
  {
    double re = real * other.real - imag * other.imag;
    double im = real * other.imag + imag * other.real;
    return new Complex(re, im);
  }
 
  // Method to divide two Complex objects
  public Complex divide(Complex other)
  {
    double denom = other.real * other.real
      + other.imag * other.imag;
    double re = (real * other.real + imag * other.imag)
      / denom;
    double im = (imag * other.real - real * other.imag)
      / denom;
    return new Complex(re, im);
  }
}
 
class GFG {
 
  // Function to implement the FFT
  public static void fft(ArrayList<Complex> a,
                         boolean invert)
  {
    int n = a.size();
    if (n == 1) {
      return;
    }
 
    ArrayList<Complex> a0 = new ArrayList<>();
    ArrayList<Complex> a1 = new ArrayList<>();
    for (int i = 0; 2 * i < n; i++) {
      a0.add(a.get(2 * i));
      a1.add(a.get(2 * i + 1));
    }
 
    // Recursively find fft
    fft(a0, invert);
    fft(a1, invert);
 
    double ang = 2 * Math.PI / n * (invert ? -1 : 1);
 
    Complex w = new Complex(1, 0);
    Complex wn
      = new Complex(Math.cos(ang), Math.sin(ang));
 
    for (int i = 0; 2 * i < n; i++) {
      a.set(i, a0.get(i).add(w.multiply(a1.get(i))));
      a.set(i + n / 2, a0.get(i).subtract(
        w.multiply(a1.get(i))));
      if (invert) {
        a.set(i,
              a.get(i).divide(new Complex(2, 0)));
        a.set(i + n / 2, a.get(i + n / 2).divide(
          new Complex(2, 0)));
      }
      w = w.multiply(wn);
    }
  }
 
  // Function to multiply two polynomials
  // A(x) and B(x) using FFT
  public static ArrayList<Integer> multiply(int[] a,
                                            int[] b)
  {
    ArrayList<Complex> fa = new ArrayList<>();
    ArrayList<Complex> fb = new ArrayList<>();
    for (int i = 0; i < a.length; i++) {
      fa.add(new Complex(a[i], 0));
    }
    for (int i = 0; i < b.length; i++) {
      fb.add(new Complex(b[i], 0));
    }
 
    int n = 1;
 
    while (n < a.length + b.length) {
      n <<= 1;
    }
 
    // Resize fa and fb
    while (fa.size() < n) {
      fa.add(new Complex(0, 0));
    }
    while (fb.size() < n) {
      fb.add(new Complex(0, 0));
    }
 
    // Assign initially false
    fft(fa, false);
    fft(fb, false);
 
    for (int i = 0; i < n; i++) {
      fa.set(i, fa.get(i).multiply(fb.get(i)));
    }
 
    fft(fa, true);
 
    // To store the result
    ArrayList<Integer> result = new ArrayList<>();
    for (int i = 0; i < n; i++) {
      result.add((int)Math.round(fa.get(i).real));
    }
 
    // Return result
    return result;
  }
 
  // Function to find the Count of each
  // element from 1 to 2*MAX
  public static void findCount(ArrayList<Integer> arr1,
                               ArrayList<Integer> arr2)
  {
    // Initialise MAX
    int MAX = Math.max(Collections.max(arr1),
                       Collections.max(arr2));
 
    int n = arr1.size();
    int m = arr2.size();
 
    // vector for Polynomial A(x) from Arr1
    int[] A = new int[MAX + 1];
 
    for (int i = 0; i < n; i++) {
      A[arr1.get(i)]++;
    }
 
    // Vector for Polynomial B(x) from Arr2
    int[] B = new int[MAX + 1];
 
    for (int i = 0; i < m; i++) {
      B[arr2.get(i)]++;
    }
 
    // Vector to store the result of
    // multiplication of A(x) and B(x)
    ArrayList<Integer> P = multiply(A, B);
 
    // Print the result
    for (int i = 1; i <= 2 * MAX; i++) {
      if (P.get(i) > 0) {
        System.out.println(i + "->" + P.get(i));
      }
    }
 
    System.out.println();
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Given arrays arr1[] and arr2[]
    ArrayList<Integer> arr1 = new ArrayList<Integer>();
    arr1.add(1);
    arr1.add(2);
 
    ArrayList<Integer> arr2 = new ArrayList<Integer>();
    arr2.add(1);
    arr2.add(2);
    arr2.add(1);
 
    // Function Call
    findCount(arr1, arr2);
  }
}

                    

Python3

import math
 
# Python3 program to implement the above approach
 
# Value of PI need in FFT
PI = math.acos(-1)
 
# Function to implement the FFT
def fft(a, invert):
    n = len(a)
    if (n == 1):
        return
     
    a0 = [0] * (n // 2)
    a1 = [0] * (n // 2)
     
    for i in range(n // 2):
        a0[i] = a[2 * i]
        a1[i] = a[2 * i + 1]
     
    # Recursively find fft
    fft(a0, invert)
    fft(a1, invert)
     
    ang = 2 * PI / n * (-1 if invert else 1)
     
    w = complex(1)
    wn = complex(math.cos(ang), math.sin(ang))
     
    for i in range(n // 2):
        a[i] = a0[i] + w * a1[i]
        a[i + n // 2] = a0[i] - w * a1[i]
        if (invert):
            a[i] /= 2
            a[i + n // 2] /= 2
        w *= wn
 
# Function to multiply two polynomials
# A(x) and B(x) using FFT
def multiply(a, b):
    fa = [complex(i, 0) for i in a]
    fb = [complex(i, 0) for i in b]
 
    n = 1
 
    while (n < len(a) + len(b)):
        n <<= 1
 
    # Resize fa and fb
    fa += [complex(0)] * (n - len(fa))
    fb += [complex(0)] * (n - len(fb))
 
    # Assign initially false
    fft(fa, False)
    fft(fb, False)
 
    for i in range(n):
        fa[i] *= fb[i]
 
    fft(fa, True)
 
    # To store the result
    result = [0] * n
 
    for i in range(n):
        result[i] = round(fa[i].real)
 
    # Return result
    return result
 
# Function to find the Count of each
# element from 1 to 2*MAX
def findCount(Arr1, Arr2):
    # Initialise MAX
    MAX = max(max(Arr1), max(Arr2))
 
    n = len(Arr1)
    m = len(Arr2)
 
    # vector for Polynomial A(x) from Arr1
    A = [0] * (MAX + 1)
 
    for i in range(n):
        A[Arr1[i]] += 1
 
    # Vector for Polynomial B(x) from Arr2
    B = [0] * (MAX + 1)
 
    for i in range(m):
        B[Arr2[i]] += 1
 
    # Vector to store the result of
    # multiplication of A(x) and B(x)
    P = []
 
    # Multiplying Arr1 and Arr2 and
    # storing in P is same as Count
    P = multiply(A, B)
 
    # Print the result
    for i in range(1, 2 * MAX + 1):
        if (P[i] > 0):
            print(str(i) + "->" + str(P[i]))
 
    print()
 
# Driver Code
if __name__ == "__main__":
    # Given arrays arr1[] and arr2[]
    arr1 =[1, 2]
    arr2 = [1, 2, 1]
 
    # Function Call
    findCount(arr1, arr2)

                    

C#

using System;
using System.Linq;
using System.Collections.Generic;
 
// Complex class definition
class Complex {
    // Data members
    public double real, imag;
 
    // Constructor
    public Complex(double real, double imag)
    {
        this.real = real;
        this.imag = imag;
    }
 
    // Method to Add two Complex objects
    public Complex Add(Complex other)
    {
        return new Complex(real + other.real,
                           imag + other.imag);
    }
 
    // Method to subtract two Complex objects
    public Complex Subtract(Complex other)
    {
        return new Complex(real - other.real,
                           imag - other.imag);
    }
 
    // Method to multiply two Complex objects
    public Complex Multiply(Complex other)
    {
        double re = real * other.real - imag * other.imag;
        double im = real * other.imag + imag * other.real;
        return new Complex(re, im);
    }
 
    // Method to divide two Complex objects
    public Complex Divide(Complex other)
    {
        double denom = other.real * other.real
                       + other.imag * other.imag;
        double re = (real * other.real + imag * other.imag)
                    / denom;
        double im = (imag * other.real - real * other.imag)
                    / denom;
        return new Complex(re, im);
    }
}
 
class GFG {
    // Function to implement the FFT
    public static void fft(List<Complex> a, bool invert)
    {
        int n = a.Count;
        if (n == 1)
            return;
 
        List<Complex> a0 = new List<Complex>();
        List<Complex> a1 = new List<Complex>();
        for (int i = 0; 2 * i < n; i++) {
            a0.Add(a[2 * i]);
            a1.Add(a[2 * i + 1]);
        }
 
        // Recursively find fft
        fft(a0, invert);
        fft(a1, invert);
 
        double ang = 2 * Math.PI / n * (invert ? -1 : 1);
 
        Complex w = new Complex(1, 0);
        Complex wn
            = new Complex(Math.Cos(ang), Math.Sin(ang));
 
        for (int i = 0; 2 * i < n; i++) {
            a[i] = a0[i].Add(w.Multiply(a1[i]));
            a[i + n / 2]
                = a0[i].Subtract(w.Multiply(a1[i]));
            if (invert) {
                a[i] = a[i].Divide(new Complex(2, 0));
                a[i + n / 2] = a[i + n / 2].Divide(
                    new Complex(2, 0));
            }
            w = w.Multiply(wn);
        }
    }
 
    // Function to multiply two polynomials A(x) and B(x)
    // using FFT
    public static List<int> Multiply(int[] a, int[] b)
    {
        List<Complex> fa = new List<Complex>();
        List<Complex> fb = new List<Complex>();
        for (int i = 0; i < a.Length; i++)
            fa.Add(new Complex(a[i], 0));
        for (int i = 0; i < b.Length; i++)
            fb.Add(new Complex(b[i], 0));
 
        int n = 1;
        while (n < a.Length + b.Length)
            n <<= 1;
 
        // Resize fa and fb
        while (fa.Count < n)
            fa.Add(new Complex(0, 0));
        while (fb.Count < n)
            fb.Add(new Complex(0, 0));
 
        // Assign initially false
        fft(fa, false);
        fft(fb, false);
 
        for (int i = 0; i < n; i++) {
            fa[i] = fa[i].Multiply(fb[i]);
        }
 
        fft(fa, true);
 
        // To store the result
        List<int> result = new List<int>();
        for (int i = 0; i < n; i++) {
            result.Add((int)Math.Floor(fa[i].real));
        }
 
        // Return result
        return result;
    }
 
    // Function to find the Count of each
    // element from 1 to 2*MAX
    public static void FindCount(List<int> arr1,
                                 List<int> arr2)
    {
        // Initialise MAX
        int MAX = Math.Max(arr1.Max(), arr2.Max());
 
        int n = arr1.Count;
        int m = arr2.Count;
 
        // vector for Polynomial A(x) from Arr1
        int[] A = new int[MAX + 1];
 
        for (int i = 0; i < n; i++) {
            A[arr1[i]]++;
        }
 
        // Vector for Polynomial B(x) from Arr2
        int[] B = new int[MAX + 1];
 
        for (int i = 0; i < m; i++) {
            B[arr2[i]]++;
        }
 
        // Vector to store the result of
        // multiplication of A(x) and B(x)
        List<int> P = Multiply(A, B);
 
        // Print the result
        for (int i = 1; i <= 2 * MAX; i++) {
            if (P[i] > 0) {
                Console.WriteLine(i + "->" + P[i]);
            }
        }
 
        Console.WriteLine(" ");
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        // Given arrays arr1[] and arr2[]
        List<int> arr1 = new List<int>();
        arr1.Add(1);
        arr1.Add(2);
 
        List<int> arr2 = new List<int>();
        arr2.Add(1);
        arr2.Add(2);
        arr2.Add(1);
 
        // Function Call
        FindCount(arr1, arr2);
    }
}

                    

Javascript

// Complex class definition
class Complex {
 
    // Data members
    constructor(real, imag) {
        this.real = real;
        this.imag = imag;
    }
 
    // Method to add two Complex objects
    add(other) {
        return new Complex(this.real + other.real,
            this.imag + other.imag);
    }
 
    // Method to subtract two Complex objects
    subtract(other) {
        return new Complex(this.real - other.real,
            this.imag - other.imag);
    }
 
    // Method to multiply two Complex objects
    multiply(other) {
        let re = this.real * other.real - this.imag * other.imag;
        let im = this.real * other.imag + this.imag * other.real;
        return new Complex(re, im);
    }
 
    // Method to divide two Complex objects
    divide(other) {
        let denom = other.real * other.real +
            other.imag * other.imag;
        let re = (this.real * other.real + this.imag * other.imag) /
            denom;
        let im = (this.imag * other.real - this.real * other.imag) /
            denom;
        return new Complex(re, im);
    }
}
 
 
// Function to implement the FFT
function fft(a, invert) {
    let n = a.length;
    if (n == 1) {
        return;
    }
 
 
    let a0 = [];
    let a1 = [];
    for (let i = 0; 2 * i < n; i++) {
        a0.push(a[2 * i]);
        a1.push(a[2 * i + 1]);
    }
 
    // Recursively find fft
    fft(a0, invert);
    fft(a1, invert);
 
    let ang = 2 * Math.PI / n * (invert ? -1 : 1);
 
    let w = new Complex(1, 0);
    let wn = new Complex(Math.cos(ang), Math.sin(ang));
 
    for (let i = 0; 2 * i < n; i++) {
        a[i] = a0[i].add(w.multiply(a1[i]));
        a[i + n / 2] = a0[i].subtract(
            w.multiply(a1[i]));
        if (invert) {
            a[i] = a[i].divide(new Complex(2, 0));
            a[i + n / 2] = a[i + n / 2].divide(
                new Complex(2, 0));
        }
        w = w.multiply(wn);
    }
 
}
 
// Function to multiply two polynomials
// A(x) and B(x) using FFT
function multiply(a, b) {
    let fa = [];
    let fb = [];
    for (let i = 0; i < a.length; i++) {
        fa.push(new Complex(a[i], 0));
    }
    for (let i = 0; i < b.length; i++) {
        fb.push(new Complex(b[i], 0));
    }
 
 
    let n = 1;
 
    while (n < a.length + b.length) {
        n <<= 1;
    }
 
    // Resize fa and fb
    while (fa.length < n) {
        fa.push(new Complex(0, 0));
    }
    while (fb.length < n) {
        fb.push(new Complex(0, 0));
    }
 
    // Assign initially false
    fft(fa, false);
    fft(fb, false);
 
    for (let i = 0; i < n; i++) {
        fa[i] = fa[i].multiply(fb[i]);
    }
 
    fft(fa, true);
 
    // To store the result
    let result = []
    for (var i = 0; i < n; i++) {
        result.push(Math.floor(fa[i].real));
    }
 
    // Return result
    return result;
}
 
 
// Function to find the Count of each
// element from 1 to 2*MAX
function findCount(arr1, arr2) {
    // Initialise MAX
    let MAX = Math.max(Math.max(...arr1),
        Math.max(...arr2));
 
    let n = arr1.length;
    let m = arr2.length;
 
    // vector for Polynomial A(x) from Arr1
    let A = new Array(MAX + 1).fill(0);
 
    for (var i = 0; i < n; i++) {
        A[arr1[i]]++;
    }
 
    // Vector for Polynomial B(x) from Arr2
    let B = new Array(MAX + 1).fill(0);
 
    for (var i = 0; i < m; i++) {
        B[arr2[i]]++;
    }
 
    // Vector to store the result of
    // multiplication of A(x) and B(x)
    let P = multiply(A, B);
 
    // Print the result
    for (var i = 1; i <= 2 * MAX; i++) {
        if (P[i] > 0) {
            console.log(i + "->" + P[i]);
        }
    }
 
    console.log(" ")
}
 
// Driver Code
 
// Given arrays arr1[] and arr2[]
let arr1 = []
arr1.push(1);
arr1.push(2);
 
let arr2 = []
arr2.push(1);
arr2.push(2);
arr2.push(1);
 
// Function Call
findCount(arr1, arr2);

                    

Output: 
2->2
3->3
4->1

 

Time Complexity: O(N*log N) 
Auxiliary Space: O(N)
 



Last Updated : 16 Mar, 2023
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