Cuente elementos distintos después de agregar cada elemento de First Array con Second Array

Dadas dos arrays arr1[] y arr2[] . Podemos generar otra array arr3[] agregando cada elemento de la array arr1[] a cada elemento arr2[] . La tarea es encontrar el recuento de elementos distintos en la array arr3[] .

Ejemplos:  

Entrada: Arr1[] = {1, 2}, Arr2[] = {3, 4}, MAX = 4 
Salida: 
4 -> 1 
5 -> 2 
6 -> 1 
Explicación: 
Aquí la tercera array será Arr3[] = {1+3, 1+4, 2+3, 2+4} = {4, 5, 5, 6}

Entrada: Arr1[] = {1, 2}, Arr2[] = {1, 2, 1}, MAX = 2 
Salida: 
2 -> 2 
3 -> 3 
4 -> 1 
Explicación: 
Aquí la tercera array es Arr3[ ] = {1+1, 1+2, 1+1, 2+1, 2+2, 2+1} = {2, 3, 2, 3, 4, 3} 
Por lo tanto Cuenta de elementos de 1 a 2* 2 (4) son {0, 2, 3, 1}

Enfoque ingenuo: el enfoque ingenuo es encontrar la suma de todos los pares posibles de las dos arrays dadas e insertar esa suma en la array arr3[] . Imprime la frecuencia de todos los elementos del arreglo arr3[] .

A continuación se muestra la implementación del enfoque anterior: 

// 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 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 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# 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

<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>

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

Complejidad de tiempo: O(N ² ) 
Complejidad de espacio: O(N)

Solución eficiente: la tarea dada se puede realizar de manera eficiente con la ayuda de FFT (Fast Fourier Transform) . A continuación se muestran los pasos:  

1. Considere los ejemplos Arr1[] = {1, 2} y Arr2[] = {1, 2, 1} . Sea Count la array de frecuencias, es decir, Count[i] representa la frecuencia de i en la array resultante.
2. Cuando Arr1[i] se agrega a Arr2[j], incrementamos Count[s] donde s = Arr1[i]+Arr2[j]. Esto es similar a multiplicar polinomios a medida que se agregan potencias.
3. Sea A(x) el polinomio representado por Arr1[]. Los elementos de Arr1 representan la potencia de x y su cuenta en Arr1 son términos de coeficientes con esa potencia en polinomio.
4. Para cada término, la potencia de x representa el elemento resultante y el coeficiente representa su cuenta.
5. Si el término es
6. Entonces Cuenta[i] = k. Aquí Count es lo mismo que P(x).
7. Para calcular el valor de P(x), simplemente podemos multiplicar A(x) y B(x).

El método Naive de multiplicación de polinomios toma O(N ² ). Para hacer la multiplicación más rápida podemos usar FFT (Fast Fourier Transform) .

A continuación se muestra la implementación del enfoque anterior: 

// 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);
}

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

Complejidad de tiempo: O(N*log N) 
Espacio auxiliar: O(N)