Count items common to both the lists but with different prices

Given two lists list1 and list2 containing m and n items respectively. Each item is associated with two fields: name and price. The problem is to count items that are in both the lists but with different prices.

Examples: 

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.

Source: Cognizant Interview Experience | Set 5.

Method 1 (Naive Approach): Using two nested loops compare each item of list1 with all the items of list2. If a match is found with a different price then increment the count. 

Implementation:

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

Javascript

<script>

// Javascript 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
function countItems(list1, m, list2, n)
{
    var count = 0;

    // for each item of 'list1'
    // check if it is in 'list2'
    // but with a different price
    for (var i = 0; i < m; i++)
        for (var j = 0; j < n; j++)

            if (list1[i][0] === list2[j][0] &&
                 (list1[i][1] != list2[j][1]))
                count++;

    // required count of items
    return count;
}

// Driver program to test above
var list1 = [["apple", 60], ["bread", 20],
                ["wheat", 50], ["oil", 30]];
var list2 = [["milk", 20], ["bread", 15],
               ["wheat", 40], ["apple", 60]];

var m = list1.length;
var n = list2.length;

document.write( "Count = " + countItems(list1, m, list2, n));


</script>

Output

Count = 2

Time Complexity: O(m*n). 
Auxiliary Space: O(1).

Method 2 (Binary Search): Sort the list2 in alphabetical order of its items name. Now, for each item of list1 check whether it in present in list2 using the binary search technique and get its price from list2. If prices are different then increment the count.

Implementation:

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;
};

// 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
// search 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 alphabetical
  // 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<item>
{

  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 alphabetical
  // 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

Python3

# Python3 implementation to count
# items common to both the lists
# but with different prices

# Details of an item
from ast import Str
from functools import cmp_to_key

class item:
    def __init__(self, name, price):
        self.name = name
        self.price = price

# 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
def binary_search(list2, low, high, str):
    while (low <= high):
        mid = ((low + high) // 2)

       # if true the item 'str' is in 'list2'
        # print(list2[mid].name,str)
        if (list2[mid].name == str):
            return list2[mid].price
        elif (list2[mid].name < str):
            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
def custom_logic(a, b):
    return a.name == b.name

def countItems(list1, m, list2, n):

    # sort 'list2' in alphabetical
    # order of items name
    sorted(list2,key=cmp_to_key(custom_logic))

    # initial count
    count = 0

    for i in range(m):

        # get the price of item 'list1[i]'
        # from 'list2' if item in not
        # present in second list then -1
        # is being obtained
        r = binary_search(list2, 0,
                          n - 1,
                          list1[i].name)

        # if item is present in list2
        # with a different price
        if ((r != -1) and (r != list1[i].price)):
            count += 1

  # Required count of items
    return count

# Driver code
list1=[item("apple", 60),
                  item("bread", 20),
                  item("wheat", 50),
                  item("oil", 30)]
list2=[item("milk", 20),
                  item("bread", 15),
                  item("wheat", 40),
                  item("apple", 60)]
m = len(list1)
n = len(list2)
print(f"Count = {countItems(list1, m,list2, n)}")

# This code is contributed by shinjanpatra

Javascript

<script>
// Javascript implementation to count
// items common to both the lists
// but with different prices

// Details of an item
class item
{
    constructor(name,price)
    {
        this.name = name;
        this.price = price;
    }
}

// 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
function binary_search(list2,low,high,str)
{
    while (low <= high)
  {
    let mid = Math.floor((low + high) / 2);

    // if true the item 'str' is in 'list2'
    if (list2[mid].name == (str))
      return list2[mid].price;
    else if (list2[mid].name < (str))
      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
function countItems(list1, m, list2, n)
{

    // sort 'list2' in alphabetical
  // order of items name
  list2.sort(function(a,b){return a.name==b.name;});

  // initial count
  let count = 0;

  for (let 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
    let 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
let list1=[new item("apple", 60),
                  new item("bread", 20),
                  new item("wheat", 50),
                  new item("oil", 30)];
let list2=[new item("milk", 20),
                  new item("bread", 15),
                  new item("wheat", 40),
                  new item("apple", 60)];
let m = list1.length;
let n = list2.length;
document.write("Count = " +
                    countItems(list1, m,
                               list2, n));

// This code is contributed by patel2127
</script>

C#

using System;

class Item
{
    public string name;
    public int price;

    public Item(string name, int price)
    {
        this.name = name;
        this.price = price;
    }
}

class Program
{
    // 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 BinarySearch(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 == str)
            {
                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 alphabetical
        // order of items name
        Array.Sort(list2, (a, b) => a.name.CompareTo(b.name));

        // 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 = BinarySearch(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;
    }

    static void Main(string[] args)
    {
        Item[] list1 = new Item[] {
            new Item("apple", 60),
            new Item("bread", 20),
            new Item("wheat", 50),
            new Item("oil", 30)
        };

        Item[] list2 = new Item[] {
            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.WriteLine("Count = " + CountItems(list1, m, list2, n));
    }
}

Output

Count = 2

Time Complexity: (m*log₂n). 
Auxiliary Space: O(1). 

For efficiency, the list with the maximum number of elements should be sorted and used for binary search.

Method 3 (Efficient Approach): Create a hash table with (key, value) tuple as (item name, price). Insert the elements of list1 in the hash table. Now, for each element of list2 check if it is the hash table or not. If it is present, then check if its price is different from the value from the hash table. If so then increment the count. 

Implementation:

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, list2, n));
    }
}

// This code is contributed by gauravrajput1

Python3

# Python3 implementation to count items common to both
# the lists but with different prices

# details of an item
class item:
    def __init__(self, name, price):
        self.name = name
        self.price = price

# function to count items common to both
# the lists but with different prices
def countItems(list1, m,list2,  n):

    # 'um' implemented as hash table that contains
    # item name as the key and price as the value
    # associated with the key
    um = {}
    count = 0

    # insert elements of 'list1' in 'um'
    for i in range(m):
        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 i in range(n):
        if ((um.get(list2[i].name) != None) and (um[list2[i].name] != list2[i].price)):
            count+=1

    # required count of items
    return count


# Driver program to test above
list1=[item("apple", 60),
                  item("bread", 20),
                  item("wheat", 50),
                  item("oil", 30)]
list2=[item("milk", 20),
                  item("bread", 15),
                  item("wheat", 40),
                  item("apple", 60)]
m = len(list1)
n = len(list2)

print("Count = " ,countItems(list1, m, list2, n))


# This code is contributed by Abhijeet Kumar(abhijeet19403)

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

Javascript

<script>

// JavaScript implementation to count
// items common to both the lists
// but with different prices

// details of an item
class item
{
    constructor(name,price)
    {
        this.name = name;
        this.price = price;
    }
}

// function to count items common to both
// the lists but with different prices
function countItems(list1,m,list2,n)
{
    // 'um' implemented as hash table that contains
  // item name as the key and price as the value
  // associated with the key
  let um = new Map();
  let count = 0;

  // insert elements of 'list1' in 'um'
  for (let i = 0; i < m; i++)
    um.set(list1[i].name, list1[i].price);

  // for each element of 'list2' check if it is
  // present in 'um' with a different price
  // value
  for (let i = 0; i < n; i++)
    if ((um.has(list2[i].name)) &&
        (um.get(list2[i].name) != list2[i].price))
      count++;

  // required count of items
  return count;
}

// Driver program to test above
let list1=[new item("apple", 60),
                  new item("bread", 20),
                  new item("wheat", 50),
                  new item("oil", 30)];
let list2=[new item("milk", 20),
                  new item("bread", 15),
                  new item("wheat", 40),
                  new item("apple", 60)];
let m = list1.length;
let n = list2.length;

document.write("Count = " +
                 countItems(list1, m,
                            list2, n));


// This code is contributed by unknown2108

</script>

Output

Count = 2

Time Complexity: O(m + n). 
Auxiliary Space: O(m). 

For efficiency, the list having a minimum number of elements should be inserted in the hash table.