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📜  最小删除要求的数量,以使任意数量的X都可以准确出现X次

📅  最后修改于: 2021-05-07 01:08:22             🧑  作者: Mango

给定的阵列ARR [] N个整数的,任务是找到最小缺失阵列对于i的所有可能值中需要使得ARR的频率[i]的准确的常用3 [i]中

例子:

方法:有两种情况:

  • 如果X的频率大于或等于X,则我们删除X的额外频率,以得到值X的X个元素。
  • 如果X的频率小于X,则我们将删除所有出现的X,因为不可能获得值X的额外元素。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum
// deletions required
int MinDeletion(int a[], int n)
{
  
    // To store the frequency of
    // the array elements
    unordered_map map;
  
    // Store frequency of each element
    for (int i = 0; i < n; i++)
        map[a[i]]++;
  
    // To store the minimum deletions required
    int ans = 0;
  
    for (auto i : map) {
  
        // Value
        int x = i.first;
  
        // It's frequency
        int frequency = i.second;
  
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) {
  
            // Delete extra occurrences
            ans += (frequency - x);
        }
  
        // Delete every occurrence of x
        else
            ans += frequency;
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int a[] = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << MinDeletion(a, n);
  
    return 0;
}


Java
// Java Implementation of above approach 
import java.util.*;
  
class GFG 
{
      
// Function to return the minimum
// deletions required
static int MinDeletion(int a[], int n)
{
  
    // To store the frequency of
    // the array elements
    Map mp = new HashMap<>();
  
    // Store frequency of each element
    for (int i = 0 ; i < n; i++)
    {
        if(mp.containsKey(a[i]))
        {
            mp.put(a[i], mp.get(a[i])+1);
        }
        else
        {
            mp.put(a[i], 1);
        }
    }
    // To store the minimum deletions required
    int ans = 0;
  
    for (Map.Entry i : mp.entrySet()) 
    {
  
        // Value
        int x = i.getKey();
  
        // It's frequency
        int frequency = i.getValue();
  
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) 
        {
  
            // Delete extra occurrences
            ans += (frequency - x);
        }
  
        // Delete every occurrence of x
        else
            ans += frequency;
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int a[] = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = a.length;
  
    System.out.println(MinDeletion(a, n));
}
}
  
// This code is contributed by Rajput-Ji


Python3
# Python3 implementation of the approach 
  
# Function to return the minimum 
# deletions required 
def MinDeletion(a, n) :
  
    # To store the frequency of 
    # the array elements 
    map = dict.fromkeys(a, 0); 
  
    # Store frequency of each element 
    for i in range(n) : 
        map[a[i]] += 1; 
  
    # To store the minimum deletions required 
    ans = 0; 
  
    for key,value in map.items() :
  
        # Value 
        x = key; 
  
        # It's frequency 
        frequency = value; 
  
        # If number less than or equal 
        # to it's frequency 
        if (x <= frequency) :
  
            # Delete extra occurrences 
            ans += (frequency - x); 
  
        # Delete every occurrence of x 
        else :
            ans += frequency; 
  
    return ans; 
  
  
# Driver code 
if __name__ == "__main__" : 
  
    a = [ 2, 3, 2, 3, 4, 4, 4, 4, 5 ];
    n = len(a); 
  
    print(MinDeletion(a, n)); 
  
# This code is contributed by AnkitRai01


C#
// C# Implementation of above approach
using System;
using System.Collections.Generic; 
  
class GFG 
{
      
// Function to return the minimum
// deletions required
static int MinDeletion(int []a, int n)
{
  
    // To store the frequency of
    // the array elements
    Dictionary mp = new Dictionary();
  
    // Store frequency of each element
    for (int i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(a[i]))
        {
            var val = mp[a[i]];
            mp.Remove(a[i]);
            mp.Add(a[i], val + 1); 
        }
        else
        {
            mp.Add(a[i], 1);
        }
    }
      
    // To store the minimum deletions required
    int ans = 0;
  
    foreach(KeyValuePair i in mp)
    {
  
        // Value
        int x = i.Key;
  
        // It's frequency
        int frequency = i.Value;
  
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) 
        {
  
            // Delete extra occurrences
            ans += (frequency - x);
        }
  
        // Delete every occurrence of x
        else
            ans += frequency;
    }
  
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int []a = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = a.Length;
  
    Console.WriteLine(MinDeletion(a, n));
}
}
  
// This code is contributed by PrinciRaj1992


输出:
3

时间复杂度: O(N)