计算两个清单共有但价格不同的商品

给定两个分别包含m和n项的列表list1和list2 。每个项目都与两个字段相关联：名称和价格。问题是要对两个列表中但价格不同的商品进行计数。
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.

资料来源：认知面试经验|设置5。

方法1(天真方法)：使用两个嵌套循环将list1的每个项目与list2的所有项目进行比较。如果找到价格不同的匹配项，则增加计数。

C++

// C++ implementation to count items common to both
// the lists but with different prices
#include 
 
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：

Count = 2

时间复杂度：O(m * n)。
辅助空间：O(1)。
方法2(二进制搜索)：按列表2的项名称的字母顺序对其进行排序。现在，使用二进制搜索技术检查list1的每个项目是否存在于list2中，并从list2获得其价格。如果价格不同，则增加计数。

C++

// C++ implementation to count
// items common to both the lists
// but with different prices
#include 
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 technique 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 code
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;
  item(String name,
       int price)
  {
    this.name = name;
    this.price = price;
  }
};
 
// comparator function used for sorting
static class Com implements Comparator
{
 
  public int compare(item a,
                     item b)
  {
    return a.name.compareTo(b.name);
  }
 
}
 
// 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 search technique is being 
// applied for searching
static 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.compareTo(str) == 0)
      return list2[mid].price;
    else if (list2[mid].name.compareTo(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
static int countItems(item list1[], int m,
                      item list2[], int n)
{
  // sort 'list2' in alphabetcal
  // order of items name
  Arrays.sort(list2, new Com());
 
  // 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 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

输出：

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)：创建一个哈希表，其中将(键，值)元组作为(项目名称，价格) 。将list1的元素插入哈希表中。现在，对于list2的每个元素，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 
 
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 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 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 um = new Dictionary();
  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：

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。