元素数量，它与X的和/差也存在于数组中
中
给定数组arr[]和整数X，任务是对数组中的元素进行计数，以使它们在数组中存在元素arr[i] - X或arr[i] + X。
示例：
输入：
arr[] = {3, 4, 2, 5}, X = 2输出：4
说明：
在上述示例中，有 4 个此类数字：
对于元素 3：数组中可能存在 1、5，存在元素 5，对于元素 4：可能的数字为 2、6，存在元素 2，对于元素 2：可能的数字为 0、4，4 存在于数组中，对于数字 4：可能的数字为 3、7，而在数组中存在 3。 计数为 4。
输入：
arr[] = {2, 2, 4, 5, 6}, X = 3输出：3
说明：
在上述示例中，有 3 个这样的数字
{2, 2, 5}。
方法：这个想法是使用哈希映射检查元素是否在O(1)时间出现在哈希映射中。 然后，遍历数组的元素，并针对每个元素检查数组中是否存在arr[i] - X或arr[i] + X。 如果是，则将此类元素的计数加 1。
下面是上述方法的实现：
C++
// C++ implementation to count of
// elements such that its sum/difference
// with X also exists in the Array
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of
// elements in the array such that
// element at the difference at X
// is present in the array
void findAns(int arr[], int n, int x)
{
int ans;
unordered_set<int> s;
// Loop to insert the elements
// of the array into the set
for (int i = 0; i < n; i++)
s.insert(arr[i]);
ans = 0;
// Loop to iterate over the array
for (int i = 0; i < n; i++) {
// if any of the elements are there
// then increse the count variable
if (s.find(arr[i] + x) != s.end() || s.find(arr[i] - x) != s.end())
ans++;
}
cout << ans;
return;
}
// Driver Code
int main()
{
int arr[] = { 2, 2, 4, 5, 6 };
int n = sizeof(arr) / sizeof(int);
int x = 3;
findAns(arr, n, x);
return 0;
}
Java
// Java implementation to count of
// elements such that its sum/difference
// with X also exists in the Array
import java.util.*;
class GFG{
// Function to find the count of
// elements in the array such that
// element at the difference at X
// is present in the array
static void findAns(int arr[],
int n, int x)
{
int ans;
HashSet<Integer> s = new HashSet<Integer>();
// Loop to insert the elements
// of the array into the set
for (int i = 0; i < n; i++)
s.add(arr[i]);
ans = 0;
// Loop to iterate over the array
for (int i = 0; i < n; i++)
{
// if any of the elements are there
// then increse the count variable
if (s.contains(arr[i] + x) ||
s.contains(arr[i] - x))
ans++;
}
System.out.print(ans);
return;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 2, 4, 5, 6 };
int n = arr.length;
int x = 3;
findAns(arr, n, x);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation to count of
# elements such that its sum/difference
# with X also exists in the array
# Function to find the count of
# elements in the array such that
# element at the difference at X
# is present in the array
def findAns(arr, n, x):
s = set()
# Loop to insert the elements
# of the array into the set
for i in range(n):
s.add(arr[i])
ans = 0
# Loop to iterate over the array
for i in range(n):
# If any of the elements are there
# then increase the count variable
if arr[i] + x in s or arr[i] - x in s:
ans = ans + 1
print(ans)
# Driver Code
arr = [ 2, 2, 4, 5, 6 ]
n = len(arr)
x = 3
# Function call
findAns(arr, n, x)
# This code is contributed by ishayadav181
C
// C# implementation to count of
// elements such that its sum/difference
// with X also exists in the Array
using System;
using System.Collections.Generic;
class GFG{
// Function to find the count of
// elements in the array such that
// element at the difference at X
// is present in the array
static void findAns(int[] arr,
int n, int x)
{
int ans;
HashSet<int> s = new HashSet<int>();
// Loop to insert the elements
// of the array into the set
for(int i = 0; i < n; i++)
s.Add(arr[i]);
ans = 0;
// Loop to iterate over the array
for(int i = 0; i < n; i++)
{
// if any of the elements are there
// then increse the count variable
if (s.Contains(arr[i] + x) ||
s.Contains(arr[i] - x))
ans++;
}
Console.Write(ans);
return;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 2, 2, 4, 5, 6 };
int n = arr.Length;
int x = 3;
findAns(arr, n, x);
}
}
// This code is contributed by ShubhamCoder
输出：
3
效果分析：
-
时间复杂度：
O(n)。 -
辅助空间复杂度：
O(n)。
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