给定一个由 N 个整数组成的数组 arr[]。任务是找到频率大于或等于数组中该元素的元素的总和。
例子：
Input: arr[] = {2, 1, 1, 2, 1, 6}
Output: 3
The elements in the array are {2, 1, 6}
Where,
2 appear 2 times which is greater than equal to 2 itself.
1 appear 3 times which is greater than 1 itself.
But 6 appears 1 time which is not greater than or equals to 6.
So, sum = 2 + 1 = 3.
Input: arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}
Output: 6方法：
- 遍历数组并将所有元素的频率存储在 C++ 中的 unordered_map 或任何其他编程语言中的等效数据结构中。
- 计算频率大于或等于该元素的元素的总和。
下面是上述方法的实现：
C++
// C++ program to find sum of elements
// in an array having frequency greater
// than or equal to that element
#include
using namespace std;
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
int sumOfElements(int arr[], int n)
{
bool prime[n + 1];
int i, j;
// Map is used to store
// element frequencies
unordered_map m;
for (i = 0; i < n; i++)
m[arr[i]]++;
int sum = 0;
// Traverse the map using iterators
for (auto it = m.begin(); it != m.end(); it++) {
// Calculate the sum of elements
// having frequencies greater than
// or equal to the element itself
if ((it->second) >= (it->first)) {
sum += (it->first);
}
}
return sum;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << sumOfElements(arr, n);
return 0;
} Java
// Java program to find sum of elements
// in an array having frequency greater
// than or equal to that element
import java.util.*;
class Solution
{
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
static int sumOfElements(int arr[], int n)
{
boolean prime[] = new boolean[n + 1];
int i, j;
// Map is used to store
// element frequencies
HashMap m= new HashMap();
for (i = 0; i < n; i++)
{
if(m.get(arr[i])==null)
m.put(arr[i],1);
else
m.put(arr[i],m.get(arr[i])+1);
}
int sum = 0;
// Getting an iterator
Iterator hmIterator = m.entrySet().iterator();
// Traverse the map using iterators
while (hmIterator.hasNext()) {
Map.Entry mapElement = (Map.Entry)hmIterator.next();
// Calculate the sum of elements
// having frequencies greater than
// or equal to the element itself
if (((int)mapElement.getValue()) >= ((int)mapElement.getKey())) {
sum += ((int)mapElement.getKey());
}
}
return sum;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
int n = arr.length;
System.out.println(sumOfElements(arr, n));
}
}
//contributed by Arnab Kundu Python3
# Python3 program to find sum of elements
# in an array having frequency greater
# than or equal to that element
# Function to return the sum of elements
# in an array having frequency greater
# than or equal to that element
def sumOfElements(arr, n) :
# dictionary is used to store
# element frequencies
m = dict.fromkeys(arr, 0)
for i in range(n) :
m[arr[i]] += 1
sum = 0
# traverse the dictionary
for key,value in m.items() :
# Calculate the sum of elements
# having frequencies greater than
# or equal to the element itself
if value >= key :
sum += key
return sum
# Driver code
if __name__ == "__main__" :
arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
n = len(arr)
print(sumOfElements(arr, n))
# This code is contributed by RyugaC#
// C# program to find sum of elements
// in an array having frequency greater
// than or equal to that element
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
static int sumOfElements(int []arr, int n)
{
bool []prime = new bool[n + 1];
int i;
// Map is used to store
// element frequencies
Dictionary m= new Dictionary();
for (i = 0; i < n; i++)
{
if(!m.ContainsKey(arr[i]))
m.Add(arr[i],1);
else
{
var val = m[arr[i]];
m.Remove(arr[i]);
m.Add(arr[i], val + 1);
}
}
int sum = 0;
// Calculate the sum of elements
// having frequencies greater than
// or equal to the element itself
foreach(KeyValuePair entry in m)
{
if(entry.Value >= entry.Key)
{
sum+=entry.Key;
}
}
return sum;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
int n = arr.Length;
Console.WriteLine(sumOfElements(arr, n));
}
}
// This code has been contributed by 29AjayKumar Javascript
Python3
# Python program for the above approach
from collections import Counter
# Function to return the sum of elements
# in an array having frequency greater
# than or equal to that element
def sumOfElements(arr, n):
# Counter function is used to
# calculate frequency of elements of array
m = Counter(arr)
sum = 0
# traverse the dictionary
for key, value in m.items():
# Calculate the sum of elements
# having frequencies greater than
# or equal to the element itself
if value >= key:
sum += key
return sum
# Driver code
if __name__ == "__main__":
arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
n = len(arr)
print(sumOfElements(arr, n))
# This code is contributed by vikkycirus输出
6时间复杂度： O(n)
方法#2：使用内置Python函数：
方法：
- 使用Counter()函数计算频率
- 计算频率大于或等于该元素的元素的总和。
蟒蛇3
# Python program for the above approach
from collections import Counter
# Function to return the sum of elements
# in an array having frequency greater
# than or equal to that element
def sumOfElements(arr, n):
# Counter function is used to
# calculate frequency of elements of array
m = Counter(arr)
sum = 0
# traverse the dictionary
for key, value in m.items():
# Calculate the sum of elements
# having frequencies greater than
# or equal to the element itself
if value >= key:
sum += key
return sum
# Driver code
if __name__ == "__main__":
arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
n = len(arr)
print(sumOfElements(arr, n))
# This code is contributed by vikkycirus
输出
6时间复杂度： O(n)
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