Cuente los pares de una array que tenga el producto de su suma y diferencia igual a 0

Dada una array arr[] de tamaño N , la tarea es contar los posibles pares de elementos de la array (arr[i], arr[j]) tales que (arr[i] + arr[j]) * (arr[i] – arr[j]) es 0 .

Ejemplos:

Entrada: arr[] = {2, -2, 1, 1}
Salida: 2
Explicación:
(arr[0] + arr[1]) * (arr[0] – arr[1]) = 0
(arr[3 ] + arr[4]) * (arr[3] – arr[4]) = 0

Entrada: arr[] = {5, 9, -9, -9}
Salida: 3

Planteamiento: Se puede observar que la ecuación (arr[i] + arr[j]) * (arr[i] – arr[j]) = 0 se puede reducir a arr[i] ² = arr[j] ² . Por lo tanto, la tarea se reduce a contar pares que tienen un valor absoluto igual. Siga los pasos a continuación para resolver el problema:  

• Inicialice una array hash[] para almacenar la frecuencia del valor absoluto de cada elemento de la array.
• Calcule el recuento de pares agregando (hash[x] * (hash[x] – 1))/ 2 para cada array de valores absolutos distintos.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAXN 100005
 
// Function to count required
// number of pairs
int countPairs(int arr[], int N)
{
    // Stores count of pairs
    int desiredPairs = 0;
 
    // Initialize hash with 0
    int hash[MAXN] = { 0 };
 
    // Count frequency of each element
    for (int i = 0; i < N; i++) {
        hash[abs(arr[i])]++;
    }
 
    // Calculate desired number of pairs
    for (int i = 0; i < MAXN; i++) {
        desiredPairs
            += ((hash[i]) * (hash[i] - 1)) / 2;
    }
 
    // Print desired pairs
    cout << desiredPairs;
}
 
// Driver Code
int main()
{
    // Given arr[]
    int arr[] = { 2, -2, 1, 1 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countPairs(arr, N);
 
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG{
   
static int MAXN = 100005;
 
// Function to count required
// number of pairs
static void countPairs(int arr[], int N)
{
     
    // Stores count of pairs
    int desiredPairs = 0;
 
    // Initialize hash with 0
    int hash[] = new int[MAXN];
    Arrays.fill(hash, 0);
 
    // Count frequency of each element
    for(int i = 0; i < N; i++)
    {
        hash[Math.abs(arr[i])]++;
    }
 
    // Calculate desired number of pairs
    for(int i = 0; i < MAXN; i++)
    {
        desiredPairs += ((hash[i]) *
                         (hash[i] - 1)) / 2;
    }
 
    // Print desired pairs
    System.out.print(desiredPairs);
}  
   
// Driver Code
public static void main (String[] args)
{
     
    // Given arr[]
    int arr[] = { 2, -2, 1, 1 };
 
    // Size of the array
    int N = arr.length;
 
    // Function call
    countPairs(arr, N);
}
}
 
// This code is contributed by code_hunt

# Python3 program for
# the above approach
MAXN = 100005
 
# Function to count required
# number of pairs
def countPairs(arr, N):
 
    # Stores count of pairs
    desiredPairs = 0
 
    # Initialize hash with 0
    hash = [0] * MAXN
 
    # Count frequency of
    # each element
    for i in range(N):
        hash[abs(arr[i])] += 1
    
    # Calculate desired number
    # of pairs
    for i in range(MAXN):
        desiredPairs += ((hash[i]) *
                         (hash[i] - 1)) // 2
     
    # Print desired pairs
    print (desiredPairs)
 
# Driver Code
if __name__ == "__main__":
   
    # Given arr[]
    arr = [2, -2, 1, 1]
 
    # Size of the array
    N = len(arr)
 
    # Function Call
    countPairs(arr, N)
 
# This code is contributed by Chitranayal

// C# program for the above approach
using System;
 
class GFG{
   
static int MAXN = 100005;
 
// Function to count required
// number of pairs
static void countPairs(int []arr, int N)
{
     
    // Stores count of pairs
    int desiredPairs = 0;
 
    // Initialize hash with 0
    int []hash = new int[MAXN];
 
    // Count frequency of each element
    for(int i = 0; i < N; i++)
    {
        hash[Math.Abs(arr[i])]++;
    }
 
    // Calculate desired number of pairs
    for(int i = 0; i < MAXN; i++)
    {
        desiredPairs += ((hash[i]) *
                         (hash[i] - 1)) / 2;
    }
 
    // Print desired pairs
    Console.Write(desiredPairs);
}  
   
// Driver Code
public static void Main(String[] args)
{
     
    // Given []arr
    int []arr = { 2, -2, 1, 1 };
 
    // Size of the array
    int N = arr.Length;
 
    // Function call
    countPairs(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

<script>
 
// Javascript program for the above approach
 
// Function to count required
// number of pairs
function countPairs(arr, N)
{
    // Stores count of pairs
    let desiredPairs = 0;
 
    // Initialize hash with 0
    let hash = new Uint8Array(9100005);
 
    // Count frequency of each element
    for (let i = 0; i < N; i++) {
        hash[Math.abs(arr[i])]++;
    }
 
    // Calculate desired number of pairs
    for (let i = 0; i < 9100005; i++) {
        desiredPairs
            += ((hash[i]) * (hash[i] - 1)) / 2;
    }
 
    // Print desired pairs
    document.write(desiredPairs);
}
 
// Driver Code
 
    // Given arr[]
    let arr = [2, -2, 1, 1];
 
    // Size of the array
    let N = arr.length;
 
    // Function Call
    countPairs(arr, N);
 
// This code is contributed by Mayank Tyagi
 
</script>

2

Complejidad temporal: O(N)
Espacio auxiliar: O(N)