给定一个整数数组,删除所有出现在数组中严格超过 k 次的元素。
例子:
Input : arr[] = {1, 2, 2, 3, 2, 3, 4}
k = 2
Output : 1 3 3 4
Input : arr[] = {2, 5, 5, 7}
k = 1
Output : 2 7
方法:
- 取一个哈希映射,它将存储数组中所有元素的频率。
- 现在,再次穿越。
- 打印出现次数小于或等于 k 次的元素。
C++
// C++ program to remove the elements which
// appear more than k times from the array.
#include "iostream"
#include "unordered_map"
using namespace std;
void RemoveElements(int arr[], int n, int k)
{
// Hash map which will store the
// frequency of the elements of the array.
unordered_map mp;
for (int i = 0; i < n; ++i) {
// Incrementing the frequency
// of the element by 1.
mp[arr[i]]++;
}
for (int i = 0; i < n; ++i) {
// Print the element which appear
// less than or equal to k times.
if (mp[arr[i]] <= k) {
cout << arr[i] << " ";
}
}
}
int main(int argc, char const* argv[])
{
int arr[] = { 1, 2, 2, 3, 2, 3, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
RemoveElements(arr, n, k);
return 0;
}
Java
// Java program to remove the elements which
// appear more than k times from the array.
import java.util.HashMap;
import java.util.Map;
class GFG
{
static void RemoveElements(int arr[], int n, int k)
{
// Hash map which will store the
// frequency of the elements of the array.
Map mp = new HashMap<>();
for (int i = 0; i < n; ++i)
{
// Incrementing the frequency
// of the element by 1.
mp.put(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1);
}
for (int i = 0; i < n; ++i)
{
// Print the element which appear
// less than or equal to k times.
if (mp.containsKey(arr[i]) && mp.get(arr[i]) <= k)
{
System.out.print(arr[i] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 2, 3, 2, 3, 4 };
int n = arr.length;
int k = 2;
RemoveElements(arr, n, k);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python 3 program to remove the elements which
# appear more than k times from the array.
def RemoveElements(arr, n, k):
# Hash map which will store the
# frequency of the elements of the array.
mp = {i:0 for i in range(len(arr))}
for i in range(n):
# Incrementing the frequency
# of the element by 1.
mp[arr[i]] += 1
for i in range(n):
# Print the element which appear
# less than or equal to k times.
if (mp[arr[i]] <= k):
print(arr[i], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [1, 2, 2, 3, 2, 3, 4]
n = len(arr)
k = 2
RemoveElements(arr, n, k)
# This code is contributed by
# Sahil_Shelangia
C#
// C# program to remove the elements which
// appear more than k times from the array.
using System;
using System.Collections.Generic;
class GFG
{
static void RemoveElements(int [] arr,
int n, int k)
{
// Hash map which will store the
// frequency of the elements of the array.
Dictionary mp = new Dictionary();
for (int i = 0; i < n; ++i)
{
// Incrementing the frequency
// of the element by 1.
if(mp.ContainsKey(arr[i]))
mp[arr[i]]++;
else
mp[arr[i]] = 1;
}
for (int i = 0; i < n; ++i)
{
// Print the element which appear
// less than or equal to k times.
if (mp.ContainsKey(arr[i]) && mp[arr[i]] <= k)
{
Console.Write(arr[i] + " ");
}
}
}
// Driver code
static public void Main()
{
int [] arr = { 1, 2, 2, 3, 2, 3, 4 };
int n = arr.Length;
int k = 2;
RemoveElements(arr, n, k);
}
}
// This code is contributed by Mohit kumar 29
Javascript
Python3
# Python3 program to remove the elements which
# appear strictly less than k times from the array.
from collections import Counter
def removeElements(arr, n, k):
# Calculating frequencies
# using Counter function
freq = Counter(arr)
for i in range(n):
# Print the element which appear
# more than or equal to k times.
if (freq[arr[i]] <= k):
print(arr[i], end=" ")
# Driver Code
arr = [1, 2, 2, 3, 2, 3, 4]
n = len(arr)
k = 2
removeElements(arr, n, k)
# This code is contributed by vikkycirus
输出:
1 3 3 4
时间复杂度– O(N)
方法 #2:使用内置Python函数:
- 使用Counter函数计算每个元素的频率
- 遍历数组。
- 打印出现次数小于或等于 k 次的元素。
下面是上述方法的实现:
蟒蛇3
# Python3 program to remove the elements which
# appear strictly less than k times from the array.
from collections import Counter
def removeElements(arr, n, k):
# Calculating frequencies
# using Counter function
freq = Counter(arr)
for i in range(n):
# Print the element which appear
# more than or equal to k times.
if (freq[arr[i]] <= k):
print(arr[i], end=" ")
# Driver Code
arr = [1, 2, 2, 3, 2, 3, 4]
n = len(arr)
k = 2
removeElements(arr, n, k)
# This code is contributed by vikkycirus
输出:
1 3 3 4