计算两个列表共有但价格不同的商品

2021年3月11日17:06:57 发表评论 688 次浏览

本文概述

给出两个列表

列表1

list2

包含

ñ

项目。每个项目都与两个字段关联:名称和价格。问题是要计算两个列表中价格不同的商品

例子:

Input : list1[] = {{"apple", 60}, {"bread", 20}, {"wheat", 50}, {"oil", 30}}
    list2[] = {{"milk", 20}, {"bread", 15}, {"wheat", 40}, {"apple", 60}}
Output : 2
bread and wheat are the two items common to both the
lists but with different prices.

资源: 认知面试经验|设置5。

推荐:请尝试以下方法

{IDE}

首先, 在继续解决方案之前。

方法1(天真的方法):使用两个嵌套循环比较列表1与的所有项目list2。如果找到价格不同的匹配项, 则递增计数.

C ++

// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
 
using namespace std;
 
// details of an item
struct item
{
     string name;
     int price;
};
 
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m, item list2[], int n)
{
     int count = 0;
     
     // for each item of 'list1' check if it is in 'list2'
     // but with a different price
     for ( int i = 0; i < m; i++)   
         for ( int j = 0; j < n; j++)
 
             if ((list1[i].name.compare(list2[j].name) == 0) &&
                  (list1[i].price != list2[j].price))
                 count++;       
     
     // required count of items
     return count;
}
 
// Driver program to test above
int main()
{
     item list1[] = {{ "apple" , 60}, { "bread" , 20}, { "wheat" , 50}, { "oil" , 30}};
     item list2[] = {{ "milk" , 20}, { "bread" , 15}, { "wheat" , 40}, { "apple" , 60}};
     
     int m = sizeof (list1) / sizeof (list1[0]);
     int n = sizeof (list2) / sizeof (list2[0]);   
     
     cout << "Count = "   
          << countItems(list1, m, list2, n);
          
     return 0;    
}

Java

// Java implementation to count items common to both
// the lists but with different prices
class GFG{
 
// details of an item
static class item
{
     String name;
     int price;
     public item(String name, int price) {
         this .name = name;
         this .price = price;
     }
};
 
// function to count items common to both
// the lists but with different prices
static int countItems(item list1[], int m, item list2[], int n)
{
     int count = 0 ;
     
     // for each item of 'list1' check if it is in 'list2'
     // but with a different price
     for ( int i = 0 ; i < m; i++)
         for ( int j = 0 ; j < n; j++)
 
             if ((list1[i].name.compareTo(list2[j].name) == 0 ) &&
                 (list1[i].price != list2[j].price))
                 count++;    
     
     // required count of items
     return count;
}
 
// Driver code
public static void main(String[] args)
{
     item list1[] = { new item( "apple" , 60 ), new item( "bread" , 20 ), new item( "wheat" , 50 ), new item( "oil" , 30 )};
         item list2[] = { new item( "milk" , 20 ), new item( "bread" , 15 ), new item( "wheat" , 40 ), new item( "apple" , 60 )};
     
     int m = list1.length;
     int n = list2.length;
     
     System.out.print( "Count = "
         + countItems(list1, m, list2, n));
         
}    
}
 
// This code is contributed by 29AjayKumar

Python3

# Python implementation to
# count items common to both
# the lists but with different
# prices
 
# function to count items
# common to both
# the lists but with different prices
def countItems(list1, list2):
     count = 0
     
     # for each item of 'list1'
     # check if it is in 'list2'
     # but with a different price
     for i in list1:
         for j in list2:
 
             if i[ 0 ] = = j[ 0 ] and i[ 1 ] ! = j[ 1 ]:
                 count + = 1
     
     # required count of items
     return count
 
# Driver program to test above
list1 = [( "apple" , 60 ), ( "bread" , 20 ), ( "wheat" , 50 ), ( "oil" , 30 )]
list2 = [( "milk" , 20 ), ( "bread" , 15 ), ( "wheat" , 40 ), ( "apple" , 60 )]
     
print ( "Count = " , countItems(list1, list2))
     
# This code is contributed by Ansu Kumari.

C#

// C# implementation to count items common to both
// the lists but with different prices
using System;
 
class GFG{
  
// details of an item
class item
{
     public String name;
     public int price;
     public item(String name, int price) {
         this .name = name;
         this .price = price;
     }
};
  
// function to count items common to both
// the lists but with different prices
static int countItems(item []list1, int m, item []list2, int n)
{
     int count = 0;
      
     // for each item of 'list1' check if it is in 'list2'
     // but with a different price
     for ( int i = 0; i < m; i++)
         for ( int j = 0; j < n; j++)
  
             if ((list1[i].name.CompareTo(list2[j].name) == 0) &&
                 (list1[i].price != list2[j].price))
                 count++;    
      
     // required count of items
     return count;
}
  
// Driver code
public static void Main(String[] args)
{
     item []list1 = { new item( "apple" , 60), new item( "bread" , 20), new item( "wheat" , 50), new item( "oil" , 30)};
         item []list2 = { new item( "milk" , 20), new item( "bread" , 15), new item( "wheat" , 40), new item( "apple" , 60)};
      
     int m = list1.Length;
     int n = list2.Length;
      
     Console.Write( "Count = "
         + countItems(list1, m, list2, n));
          
}    
}
// This code is contributed by PrinciRaj1992

输出如下:

Count = 2

时间复杂度:O(m * n)。

辅助空间:O(1)。

方法2(二分搜索):

排序

list2

按项目名称的字母顺序排列。现在, 对于

列表1

检查它是否存在

list2

使用二进制搜索技术并从中获取价格

list2

。如果价格不同, 则增加

计数

.

CPP

// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
 
using namespace std;
 
// details of an item
struct item
{
     string name;
     int price;
};
 
// comparator function used for sorting
bool compare( struct item a, struct item b)
{
     return (a.name.compare(b.name) <= 0);       
}
 
// function to search 'str' in 'list2[]'. If it exists then
// price associated with 'str' in 'list2[]' is being
// returned else -1 is returned. Here binary serach
// trechnique is being applied for searching
int binary_search(item list2[], int low, int high, string str)
{
     while (low <= high)
     {
         int mid = (low + high) / 2;
         
         // if true the item 'str' is in 'list2'        
         if (list2[mid].name.compare(str) == 0)
             return list2[mid].price;
         
         else if (list2[mid].name.compare(str) < 0)   
             low = mid + 1;
         
         else
             high = mid - 1;   
     }
     
     // item 'str' is not in 'list2'        
     return -1;
}
 
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m, item list2[], int n)
{
     // sort 'list2' in alphabetcal order of
     // items name
     sort(list2, list2 + n, compare);
     
     // initial count
     int count = 0;
     
     for ( int i = 0; i < m; i++) {
 
         // get the price of item 'list1[i]' from 'list2'
         // if item in not present in second list then
         // -1 is being obtained
         int r = binary_search(list2, 0, n-1, list1[i].name);
         
         // if item is present in list2 with a
         // different price
         if ((r != -1) && (r != list1[i].price))
             count++;
     }
     
     // required count of items
     return count;
}
 
// Driver program to test above
int main()
{
     item list1[] = {{ "apple" , 60}, { "bread" , 20}, { "wheat" , 50}, { "oil" , 30}};
     item list2[] = {{ "milk" , 20}, { "bread" , 15}, { "wheat" , 40}, { "apple" , 60}};
     
     int m = sizeof (list1) / sizeof (list1[0]);
     int n = sizeof (list2) / sizeof (list2[0]);   
     
     cout << "Count = "   
          << countItems(list1, m, list2, n);
          
     return 0;    
}

输出如下:

Count = 2

时间复杂度:(m * log

2

n)。

辅助空间:O(1)。

为了提高效率, 应对元素数量最大的列表进行排序, 并用于二进制搜索。

方法3(有效方法):

使用创建哈希表

(核心价值)

元组为

(商品名称, 价格)

。插入的元素

列表1

在哈希表中。现在, 对于

list2

检查它是否是哈希表。如果存在, 则检查其价格是否与哈希表中的值不同。如果是这样, 则增加

计数

.

C ++

// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
 
using namespace std;
 
// details of an item
struct item
{
     string name;
     int price;
};
 
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m, item list2[], int n)
{
     // 'um' implemented as hash table that contains
     // item name as the key and price as the value
     // associated with the key
     unordered_map<string, int > um;
     int count = 0;
     
     // insert elements of 'list1' in 'um'
     for ( int i = 0; i < m; i++)
         um[list1[i].name] = list1[i].price;
     
     // for each element of 'list2' check if it is
     // present in 'um' with a different price
     // value
     for ( int i = 0; i < n; i++)   
         if ((um.find(list2[i].name) != um.end()) &&
             (um[list2[i].name] != list2[i].price))
             count++;
     
     // required count of items       
     return count;       
}
 
// Driver program to test above
int main()
{
     item list1[] = {{ "apple" , 60}, { "bread" , 20}, { "wheat" , 50}, { "oil" , 30}};
     item list2[] = {{ "milk" , 20}, { "bread" , 15}, { "wheat" , 40}, { "apple" , 60}};
     
     int m = sizeof (list1) / sizeof (list1[0]);
     int n = sizeof (list2) / sizeof (list2[0]);   
     
     cout << "Count = "   
          << countItems(list1, m, list2, n);
          
     return 0;    
}

Java

// Java implementation to count
// items common to both the lists
// but with different prices
import java.util.*;
class GFG{
 
// details of an item
static class item
{
   String name;
   int price;
   public item(String name, int price)
   {
     this .name = name;
     this .price = price;
   }
};
 
// function to count items common to both
// the lists but with different prices
static int countItems(item list1[], int m, item list2[], int n)
{
   // 'um' implemented as hash table that contains
   // item name as the key and price as the value
   // associated with the key
   HashMap<String, Integer> um = new HashMap<>();
   int count = 0 ;
 
   // insert elements of 'list1' in 'um'
   for ( int i = 0 ; i < m; i++)
     um.put(list1[i].name, list1[i].price);
 
   // for each element of 'list2' check if it is
   // present in 'um' with a different price
   // value
   for ( int i = 0 ; i < n; i++)   
     if ((um.containsKey(list2[i].name)) &&
         (um.get(list2[i].name) != list2[i].price))
       count++;
 
   // required count of items       
   return count;       
}
 
// Driver program to test above
public static void main(String[] args)
{
   item list1[] = { new item( "apple" , 60 ), new item( "bread" , 20 ), new item( "wheat" , 50 ), new item( "oil" , 30 )};
   item list2[] = { new item( "milk" , 20 ), new item( "bread" , 15 ), new item( "wheat" , 40 ), new item( "apple" , 60 )};
 
   int m = list1.length;
   int n = list2.length;
 
   System.out.print( "Count = " +
                     countItems(list1, m, clist2, n));
}    
}
 
// This code is contributed by gauravrajput1

C#

// C# implementation to count
// items common to both the lists
// but with different prices
using System;
using System.Collections.Generic;
class GFG{
 
// Details of an item
public class item
{
   public String name;
   public int price;
   public item(String name, int price)
   {
     this .name = name;
     this .price = price;
   }
};
 
// Function to count items common to
// both the lists but with different prices
static int countItems(item []list1, int m, item []list2, int n)
{
   // 'um' implemented as hash table
   // that contains item name as the
   // key and price as the value
   // associated with the key
   Dictionary<String, int > um = new Dictionary<String, int >();
   int count = 0;
 
   // Insert elements of 'list1'
   // in 'um'
   for ( int i = 0; i < m; i++)
     um.Add(list1[i].name, list1[i].price);
 
   // For each element of 'list2'
   // check if it is present in
   // 'um' with a different price
   // value
   for ( int i = 0; i < n; i++)   
     if ((um.ContainsKey(list2[i].name)) &&
         (um[list2[i].name] != list2[i].price))
       count++;
 
   // Required count of items       
   return count;       
}
 
// Driver code
public static void Main(String[] args)
{
   item []list1 = { new item( "apple" , 60), new item( "bread" , 20), new item( "wheat" , 50), new item( "oil" , 30)};
   item []list2 = { new item( "milk" , 20), new item( "bread" , 15), new item( "wheat" , 40), new item( "apple" , 60)};
   int m = list1.Length;
   int n = list2.Length;
   Console.Write( "Count = " +
                 countItems(list1, m, list2, n));
}    
}
 
// This code is contributed by shikhasingrajput

输出如下:

Count = 2

时间复杂度:O(m + n)。

辅助空间:O(m)。

为了提高效率, 应在哈希表中插入元素数量最少的列表。


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