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])为0 。
例子:
Input: arr[] = {2, -2, 1, 1}
Output : 2
Explanation:
(arr[0] + arr[1]) * (arr[0] – arr[1]) = 0
(arr[3] + arr[4]) * (arr[3] – arr[4]) = 0
Input: arr[] = {5, 9, -9, -9}
Output : 3
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)