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📜  计算来自其和与差之积等于0的数组的对

📅  最后修改于: 2021-05-17 19:13:05             🧑  作者: Mango

给定大小为N的数组arr [] ,任务是对可能的成对数组元素(arr [i],arr [j])进行计数,使得(arr [i] + arr [j])*(arr [i] – arr [j])0

例子:

方法:可以看出,方程(arr [i] + arr [j])*(arr [i] – arr [j])= 0可以简化为arr [i] 2 = arr [j] 2 。因此,任务减少到对绝对值相等的对进行计数。请按照以下步骤解决问题:

  • 初始化数组hash []以存储每个数组元素的绝对值的频率。
  • 通过对每个数组不同的绝对值相加(hash [x] *(hash [x] – 1))/ 2 ,计算对数。

下面是上述方法的实现:

C++14
// C++ program for the above approach
#include 
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
// 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
# 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#
// 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


Javascript


输出:
2

时间复杂度: O(N)
辅助空间: O(N)