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：
Input: arr[] = {1, 2, 2, 3, 3}
Output: 2
Frequency(1) = 1
Frequency(2) = 2
Frequency(3) = 2, frequency can’t be increased
So, delete every occurrence of 3.
Input: arr[] = {2, 3, 2, 3, 4, 4, 4, 4, 5}
Output: 3
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 AnkitRai01C#
// 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)