计算数组的偶对数量,其和与差之积等于 0
给定大小为N
的数组arr[]
,任务是计算可能的成对数组元素arr[i], arr[j]
,使得(arr[i] + arr[j])(arr[i] - arr[j])
为0
。
示例:
输入:
arr[] = {2, -2, 1, 1}
输出:2
说明:
(arr [0] + arr [1]) * (arr [0] – arr [1]) = 0 (arr [3] + arr [4]) * (arr [3] – arr [4]) = 0
输入:
arr[] = {5, 9, -9, -9}
输出:3
方法:可以看出,方程(arr[i] + arr[j])(arr[i] - arr[j]) = 0
可简化为arr[i] ^ 2 = arr[j] ^ 2
。 因此,任务减少到对具有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。 -
通过为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 <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
// 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
输出:
2
时间复杂度:O(n)
。
辅助空间:O(n)
。
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