Count items common to both the lists but with different prices - GeeksforGeeks

07 Mar, 2022
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. 
C++
Java
Python3
C#
Javascript
// C++ implementation to count items common to both// the lists but with different prices#include <bits/stdc++.h> using namespace std; // details of an itemstruct item{    string name;    int price;}; // function to count items common to both// the lists but with different pricesint 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 aboveint 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 implementation to count items common to both// the lists but with different pricesclass GFG{ // details of an itemstatic 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 pricesstatic 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 codepublic 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
# 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 pricesdef 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 abovelist1 = [("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# implementation to count items common to both// the lists but with different pricesusing System; class GFG{  // details of an itemclass 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 pricesstatic 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 codepublic 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
<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 pricesfunction 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 abovevar 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.
C++
Java
Python3
Javascript
// C++ implementation to count// items common to both the lists// but with different prices#include <bits/stdc++.h>using namespace std; // Details of an itemstruct item{  string name;  int price;}; // comparator function// used for sortingbool 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 searchingint 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 pricesint 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 codeint 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 implementation to count// items common to both the lists// but with different pricesimport java.util.*;class GFG{ // Details of an itemstatic class item{  String name;  int price;  item(String name,       int price)  {    this.name = name;    this.price = price;  }}; // comparator function used for sortingstatic 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 searchingstatic 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 pricesstatic 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 codepublic 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 implementation to count# items common to both the lists# but with different prices # Details of an itemfrom ast import Strfrom 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 searchingdef 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 pricesdef 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
<script>// Javascript implementation to count// items common to both the lists// but with different prices // Details of an itemclass 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 searchingfunction 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 pricesfunction 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>
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. 
C++
Java
C#
Javascript
// C++ implementation to count items common to both// the lists but with different prices#include <bits/stdc++.h> using namespace std; // details of an itemstruct item{    string name;    int price;}; // function to count items common to both// the lists but with different pricesint 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 aboveint 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 implementation to count// items common to both the lists// but with different pricesimport java.util.*;class GFG{ // details of an itemstatic 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 pricesstatic 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 abovepublic 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# implementation to count// items common to both the lists// but with different pricesusing System;using System.Collections.Generic;class GFG{ // Details of an itempublic 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 pricesstatic 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 codepublic 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
<script> // JavaScript implementation to count// items common to both the lists// but with different prices // details of an itemclass item{    constructor(name,price)    {        this.name = name;        this.price = price;    }} // function to count items common to both// the lists but with different pricesfunction 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 abovelet 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. 
29AjayKumar
princiraj1992
GauravRajput1
shikhasingrajput
itsok
rajeev0719singh
patel2127
unknown2108
ruhelaa48
shinjanpatra
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