用数组的总大小和该元素的频率之差替换每个元素

本文概述

• C ++
• Java
• Python3
• C#

给定一个整数数组, 任务是用数组总大小及其频率之差替换每个元素。

例子：

Input: arr[] = { 1, 2, 5, 2, 2, 5, 4 }
Output: 6 4 5 4 4 5 6
Size of the array is 7.
The frequency of 1 is 1. So replace it by 7-1 = 6
The frequency of 2 is 3. So replace it by 7-3 = 4

Input: arr[] = { 4, 5, 4, 5, 6, 6, 6 }
Output: 5 5 5 5 4 4 4

方法：

1. 拿一个哈希图, 它将存储数组中所有元素的频率。
2. 现在, 再次遍历。
3. 现在, 用数组总大小及其频率之差替换所有元素。
4. 打印修改后的数组。

下面是上述方法的实现：

C ++

//C++ program to Replace each element
//by the difference of the total size
//of the array and its frequency
#include <bits/stdc++.h>
using namespace std;
  
//Function to replace the elements
void ReplaceElements( int arr[], int n)
{
     //Hash map which will store the
     //frequency of the elements of the array.
     unordered_map<int , int> mp;
  
     for ( int i = 0; i <n; ++i) 
  
         //Increment the frequency
         //of the element by 1.
         mp[arr[i]]++;
      
  
     //Replace every element by its frequency
     for ( int i = 0; i <n; ++i) 
         arr[i] = n - mp[arr[i]];
  
}
  
//Driver code
int main()
{
     int arr[] = { 1, 2, 5, 2, 2, 5, 4 };
     int n = sizeof (arr) /sizeof (arr[0]);
  
     ReplaceElements(arr, n);
  
     //Print the modified array.
     for ( int i = 0; i <n; ++i)
         cout <<arr[i] <<" " ;
  
     return 0;
}

Java

//Java program to Replace each element
//by the difference of the total size
//of the array and its frequency
import java.util.*;
  
class GFG
{
  
     //Function to replace the elements
     static void ReplaceElements( int arr[], int n) 
     {
         //Hash map which will store the
         //frequency of the elements of the array.
         HashMap<Integer, Integer> mp = new HashMap<>();
  
         for ( int i = 0 ; i <n; i++) 
         {
  
             //Increment the frequency
             //of the element by 1.
             if (!mp.containsKey(arr[i])) 
             {
                 mp.put(arr[i], 1 );
             } 
             else 
             {
                 mp.put(arr[i], mp.get(arr[i]) + 1 );
             }
         }
  
         //Replace every element by its frequency
         for ( int i = 0 ; i <n; ++i) 
         {
             arr[i] = n - mp.get(arr[i]);
         }
  
     }
  
     //Driver code
     public static void main(String[] args) 
     {
         int arr[] = { 1 , 2 , 5 , 2 , 2 , 5 , 4 };
         int n = arr.length;
  
         ReplaceElements(arr, n);
  
         //Print the modified array.
         for ( int i = 0 ; i <n; ++i)
         {
             System.out.print(arr[i] + " " );
         }
     }
}
  
//This code contributed by Rajput-Ji

Python3

# Python3 program to Replace each element
# by the difference of the total size
# of the array and its frequency
  
# Function to replace the elements
def ReplaceElements(arr, n):
      
     # Hash map which will store the
     # frequency of the elements of the array.
     mp = dict ()
  
     for i in range (n):
          
         # Increment the frequency
         # of the element by 1.
         mp[arr[i]] = mp.get(arr[i], 0 ) + 1
  
     # Replace every element by its frequency
     for i in range (n):
         arr[i] = n - mp[arr[i]]
  
# Driver code
arr = [ 1 , 2 , 5 , 2 , 2 , 5 , 4 ]
n = len (arr)
  
ReplaceElements(arr, n)
  
# Print the modified array.
for i in range (n):
     print (arr[i], end = " " )
  
# This code is contributed by mohit kumar

C#

//C# program to Replace each element
//by the difference of the total size
//of the array and its frequency
using System;
using System.Collections.Generic; 
  
class GFG
{
  
     //Function to replace the elements
     static void ReplaceElements( int []arr, int n) 
     {
         //Hash map which will store the
         //frequency of the elements of the array.
         Dictionary<int , int> mp = new Dictionary<int , int>();
  
         for ( int i = 0; i <n; i++) 
         {
  
             //Increment the frequency
             //of the element by 1.
             if (!mp.ContainsKey(arr[i])) 
             {
                 mp.Add(arr[i], 1);
             } 
             else
             {
                 var a = mp[arr[i]] + 1;
                 mp.Remove(arr[i]);
                 mp.Add(arr[i], a);
             }
         }
  
         //Replace every element by its frequency
         for ( int i = 0; i <n; ++i) 
         {
             arr[i] = n - mp[arr[i]];
         }
  
     }
  
     //Driver code
     public static void Main() 
     {
         int []arr = {1, 2, 5, 2, 2, 5, 4};
         int n = arr.Length;
  
         ReplaceElements(arr, n);
  
         //Print the modified array.
         for ( int i = 0; i <n; ++i)
         {
             Console.Write(arr[i] + " " );
         }
     }
}
  
/* This code contributed by PrinciRaj1992 */

输出如下：

6 4 5 4 4 5 6

时间复杂度–O(N)

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