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

Difficulty Level : Hard
Last Updated : 25 Aug, 2020

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}
Therfore 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 Occurence of each 
// element from 1 to 2*MAX 
void findCount(vector<int>& Arr1, 
               vector<int>& Arr2) 
{ 
    // Intialise 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 Occurence 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 Occurence 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; 
          
    # Prthe 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 Occurence 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  

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:

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.
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.
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.

So, $A(x) = x + x^2$

Similiarily

$B(x) = 2x + x^2$

Let

$P(x) = A(x) * B(x) = 2x^2 + 3x^3 + x^4$

For each term, power of x represents the resulting element and the coefficient represents its count.
If term is $k(x^i)$
Then Count[i] = k. Here Count is same as P(x).
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) 
{ 
    // Intialise 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); 
} 

Output:

2->2
3->3
4->1

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

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