从数组中删除其频率在[l，r]范围内的元素

给定一个整数数组，请从该数组中删除其频率在[l，r]范围内的元素。

例子：

Input : arr[] = { 1, 2, 3, 3, 2, 2, 5 }
l = 2, r = 3
Output : 1 5
We remove all those elements with frequencies from 2 to 5.

Input : arr[] = { 1, 2, 3, 3, 3, 3 }
l = 2, r = 3
Output : 3 3 3 3

方法 ：

• 拿一个哈希图，它将存储数组中所有元素的频率。
• 现在，再次遍历。
• 删除频率在[l，r]之间的元素。
• 否则，打印它。

C++

// C++ program to remove the elements whose
// frequency appears in the range [l, r]
#include "iostream"
#include "unordered_map"
using namespace std;
  
void removeElements(int arr[], int n, int l, int r)
{
    // 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 whose frequency
        // is not in the range [l, r]
        if (mp[arr[i]] < l or mp[arr[i] > r]) {
            cout << arr[i] << " ";
        }
    }
}
  
int main()
{
    int arr[] = { 1, 2, 3, 3, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int l = 2, r = 3;
    removeElements(arr, n, l, r);
    return 0;
}

Java

// Java program to remove the elements whose
// frequency appears in the range [l, r]
import java.util.HashMap;
  
class GFG {
  
    static void removeElements(int arr[], int n, int l, int r)
    {
        // Hash map which will store the
        // frequency of the elements of the array.
        HashMap mp = new HashMap<>();
  
        for (int i = 0; i < n; ++i) {
  
            // Incrementing the frequency
            // of the element by 1.
            // mp[arr[i]]++;
            int val = 0;
            if (mp.get(arr[i]) == null) {
                val = 1;
            }
            else {
                val = mp.get(arr[i]) + 1;
            }
            // System.out.println("--"+mp.get(arr[i]));
            mp.put(arr[i], val);
        }
  
        for (int i = 0; i < n; ++i) {
  
            // Print the element whose frequency
            // is not in the range [l, r]
            if (mp.get(arr[i]) < l || mp.get(arr[i]) > r) {
                System.out.print(arr[i] + " ");
            }
        }
    }
  
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 3, 2, 2, 5 };
        int n = arr.length;
        int l = 2, r = 3;
        removeElements(arr, n, l, r);
    }
}
  
// This code is contributed by 29AjayKumar

Python3

# Python 3 program to remove the elements 
# whose frequency appears in the range [l, r]
  
def removeElements(arr, n, l, r):
      
    # 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 whose frequency
        # is not in the range [l, r]
        if (mp[arr[i]] < l or mp[arr[i] > r]):
            print(arr[i], end = " ")
      
# Driver Code
if __name__ == '__main__':
    arr = [1, 2, 3, 3, 2, 2, 5]
    n = len(arr)
    l = 2
    r = 3
    removeElements(arr, n, l, r);
      
# This code is contributed by
# Sahil_Shelangia

C#

// C# program to remove the elements whose
// frequency appears in the range [l, r]
using System;
using System.Collections.Generic;
  
class GFG {
  
    static void removeElements(int[] arr, int n, int l, int r)
    {
        // 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.
            // mp[arr[i]]++;
            int val = 0;
            if (!mp.ContainsKey(arr[i])) {
                val = 1;
            }
            else {
                val = mp[arr[i]] + 1;
            }
            if (!mp.ContainsKey(arr[i]))
                mp.Add(arr[i], val);
            else {
                mp.Remove(arr[i]);
                mp.Add(arr[i], val);
            }
        }
  
        for (int i = 0; i < n; ++i) {
  
            // Print the element whose frequency
            // is not in the range [l, r]
            if (mp[arr[i]] < l || mp[arr[i]] > r) {
                Console.Write(arr[i] + " ");
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 1, 2, 3, 3, 2, 2, 5 };
        int n = arr.Length;
        int l = 2, r = 3;
        removeElements(arr, n, l, r);
    }
}
  
// This code contributed by Rajput-Ji

输出：

1 5

时间复杂度– O(N)

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