给定一个数组arr[]和一个整数X ,任务是计算数组的元素,使它们存在一个元素或者在数组中。
例子:
Input: arr[] = {3, 4, 2, 5}, X = 2
Output: 4
Explanation:
In the above-given example, there are 4 such numbers –
For Element 3: Possible numbers are 1, 5, whereas 5 is present in array
For Element 4: Possible numbers are 2, 6, whereas 2 is present in array
For Element 2: Possible numbers are 0, 4, whereas 4 is present in array
For Element 5: Possible numbers are 3, 7, whereas 3 is present in array
Therefore, Total count = 4
Input: arr[] = {2, 2, 4, 5, 6}, X = 3
Output: 3
Explanation:
In the above-given example, there are 3 such numbers {2, 2, 5}
方法:想法是使用哈希映射来检查元素是否在 O(1) 时间内出现在哈希映射中。然后,遍历数组的元素,并为每个元素检查或者存在于数组中。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
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 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 increase 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 s = new HashSet();
// 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 increase 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 s = new HashSet();
// 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 increase 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
Javascript
3
性能分析:
- 时间复杂度: O(N)
- 辅助空间复杂度: O(N)