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
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
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(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。