Given Q queries of type 1, 2, 3 and 4 as described below.
- Type-1: Insert a number to the list.
- Type-2: Delete only one occurrence of a number if exists.
- Type-3: Print the least frequent element, if multiple elements exist then print the greatest among them.
- Type-4: Print the most frequent element, if multiple elements exist then print the smallest among them.
The task is to write a program to perform all the above queries.
Examples:
Input:
Query1: 1 6
Query2: 1 6
Query3: 1 7
Query4: 3
Query5: 1 7
Query6: 2 7
Query7: 1 7
Query8: 3
Query9: 4
Output:
7
7
6
While answering Query4, the frequency of 6 is 2 and that of
7 is 1, hence the least frequent element is 7.
In Query8, the least frequent element is 6 and 7, so print the largest.
In Query9, the most frequent element is 6 and 7, so print the smallest.
A naive approach is to use any Data-Structures(array, vector, ..) and store all the elements. Using a hash-table, the frequency of every element can be stored. While processing the Query of type-2, delete one occurrence of that element from the DS in which the element has been stored. The queries of type-3 and type-4 can be answered by traversing the hash-table. The time complexity will be O(N) per query, where N is the number of elements till then in the DS.
An efficient approach will be to use set container to answer every query. Using two sets, one hash-table the above problem can be solved in O(log n) per query. Two sets s1 and s2 are used, one stores the {num, frequency}, while the other stores the {frequency, number}. A hash-map is used which stores the frequency of each number. Design the set s2 using operator overloading such that it is sorted in ascending order of the first elements. If the first element appears to be same of one or more element, the set will be sorted in descending order of the second elements. The user-defined operating-overloading function thus will be:
bool operator<(pr a, pr b) {
if(a.first == b.first) return a.second > b.second;
return a.first < b.first;
}
Note: Operator overloading only works with user-defined
data-types. pr is a struct which has first and second as two integers.
Below is the algorithm to solve query of every type:
- Type1: Check using hash-table if the element exists. If it does not exist, then mark the number in hash-table. Insert {num, 1} in set s1 and {1, num} in set2 using insert(). If it exists previously, then get the frequency from hash-table and delete {num, frequency} from set1 and {frequency, num} from set2 using find() and erase() function. Insert {num, frequency+1} in set1 and {frequency+1, num} in set2. Also, increase the count in hash-table.
- Type2: Follow the same process as above in query type1. Only difference is to decrease the count in hash-table and insert {num, frequency-1} in set1 and {frequency-1, num} in set2.
- Type3: Print the beginning element which can be obtained using begin() function, as the set has been designed in such a way that begin() returns the least frequent element. If there are more than one, then it returns the largest.
- Type4: Print the last element in the set which can be obtained using rbegin() function in set.
Below is the implementation of the above approach:
CPP
// C++ program for performing // Queries of insert, delete one// occurrence of a number and // print the least and most frequent element#include <bits/stdc++.h>;using namespace std;// user-defined data-typesstruct pr { int first; int second;};// user-defined function to// design a setbool operator<(pr a, pr b){ if (a.first == b.first) return a.second > b.second; return a.first < b.first;}// declare a user-defined setset<pr> s1, s2;// hash map unordered_map<int, int> m;// Function to process the query// of type-1void type1(int num){ // if the element is already there if (m[num]) { // get the frequency of the element int cnt = m[num]; // returns an iterator pointing to // position where the pair is auto it1 = s1.find({ num, cnt }); auto it2 = s2.find({ cnt, num }); // deletes the pair from sets s1.erase(it1); s2.erase(it2); // re-insert the pair by increasing // frequency s1.insert({ num, m[num] + 1 }); s2.insert({ m[num] + 1, num }); } // if the element is not there in the list else { // insert the element with frequency 1 s1.insert({ num, 1 }); s2.insert({ 1, num }); } // increase the count in hash-table m[num] += 1;}// Function to process the query// of type-2void type2(int num){ // if the element exists if (m[num]) { // get the frequency of the element int cnt = m[num]; // returns an iterator pointing to // position where the pair is auto it1 = s1.find({ num, cnt }); auto it2 = s2.find({ cnt, num }); // deletes the pair from sets s1.erase(it1); s2.erase(it2); // re-insert the pair by increasing // frequency s1.insert({ num, m[num] - 1 }); s2.insert({ m[num] - 1, num }); // decrease the count m[num] -= 1; }}// Function to process the query// of type-3int type3(){ // if the set is not empty // return the first element if (!s1.empty()) { auto it = s2.begin(); return it->second; } else return -1;}// Function to process the query// of type-4int type4(){ // if the set is not empty // return the last element if (!s1.empty()) { auto it = s2.rbegin(); return it->second; } else return -1;}// Driver Codeint main(){ // Queries // inserts 6, 6 and 7 type1(6); type1(6); type1(7); // print the answer to query of type3 cout << type3() << endl; // inserts 7 type1(7); // deletes one occurrence of 7 type2(7); // inserts 7 type1(7); // print the answer to query of type3 cout << type3() << endl; // print the answer to query of type4 cout << type4() << endl; return 0;} |
Java
// Java codeimport java.util.*;// user-defined data-typesclass Pr{ int first; int second;}// user-defined function to// design a setclass MyComparator implements Comparator<Pr> { public int compare(Pr a, Pr b){ if (a.first == b.first) return a.second > b.second ? 1 : -1; return a.first < b.first ? -1 : 1; }}// declare a user-defined setTreeSet<Pr> s1 = new TreeSet<Pr>(new MyComparator());TreeSet<Pr> s2 = new TreeSet<Pr>(new MyComparator());// hash map Map<Integer, Integer> m = new HashMap<>();// Function to process the query// of type-1void type1(int num){ // if the element is already there if (m.containsKey(num)) { // get the frequency of the element int cnt = m.get(num); // returns an iterator pointing to // position where the pair is Pr key = new Pr(); key.first = num; key.second = cnt; Pr it1 = s1.ceiling(key); Pr it2 = s2.ceiling(key); // deletes the pair from sets s1.remove(it1); s2.remove(it2); // re-insert the pair by increasing // frequency Pr key1 = new Pr(); key1.first = num; key1.second = m.get(num) + 1; s1.add(key1); Pr key2 = new Pr(); key2.first = m.get(num) + 1; key2.second = num; s2.add(key2); } // if the element is not there in the list else { // insert the element with frequency 1 Pr key1 = new Pr(); key1.first = num; key1.second = 1; s1.add(key1); Pr key2 = new Pr(); key2.first = 1; key2.second = num; s2.add(key2); } // increase the count in hash-table m.put(num, m.getOrDefault(num, 0) + 1);}// Function to process the query// of type-2void type2(int num){ // if the element exists if (m.containsKey(num)) { // get the frequency of the element int cnt = m.get(num); // returns an iterator pointing to // position where the pair is Pr key = new Pr(); key.first = num; key.second = cnt; Pr it1 = s1.ceiling(key); Pr it2 = s2.ceiling(key); // deletes the pair from sets s1.remove(it1); s2.remove(it2); // re-insert the pair by increasing // frequency Pr key1 = new Pr(); key1.first = num; key1.second = m.get(num) - 1; s1.add(key1); Pr key2 = new Pr(); key2.first = m.get(num) - 1; key2.second = num; s2.add(key2); // decrease the count m.put(num, m.get(num) - 1); }}// Function to process the query// of type-3int type3(){ // if the set is not empty // return the first element if (!s1.isEmpty()) { Pr it = s2.first(); return it.second; } else return -1;}// Function to process the query// of type-4int type4(){ // if the set is not empty // return the last element if (!s1.isEmpty()) { Pr it = s2.last(); return it.second; } else return -1;}// Driver Codepublic static void main(String[] args){ // Queries // inserts 6, 6 and 7 type1(6); type1(6); type1(7); // print the answer to query of type3 System.out.println(type3()); // inserts 7 type1(7); // deletes one occurrence of 7 type2(7); // inserts 7 type1(7); // print the answer to query of type3 System.out.println(type3()); // print the answer to query of type4 System.out.println(type4());}// This code is contributed by ishankhandelwals. |
Python
from collections import namedtuplepr = namedtuple('pr', ['first', 'second'])def lt(a, b): if a.first == b.first: return a.second > b.second return a.first < b.firsts1, s2 = set(), set()m = {}def type1(num): if num in m: cnt = m[num] it1 = next(iter(filter(lambda x: x == pr(num, cnt), s1))) it2 = next(iter(filter(lambda x: x == pr(cnt, num), s2))) s1.remove(it1) s2.remove(it2) s1.add(pr(num, m[num] + 1)) s2.add(pr(m[num] + 1, num)) else: s1.add(pr(num, 1)) s2.add(pr(1, num)) m[num] = m.get(num, 0) + 1def type2(num): if num in m: cnt = m[num] it1 = next(iter(filter(lambda x: x == pr(num, cnt), s1))) it2 = next(iter(filter(lambda x: x == pr(cnt, num), s2))) s1.remove(it1) s2.remove(it2) s1.add(pr(num, m[num] - 1)) s2.add(pr(m[num] - 1, num)) m[num] -= 1def type3(): if s1: it = iter(s2) return next(it).second return -1def type4(): if s1: it = iter(s2) return next(reversed(list(it))).second return -1type1(6)type1(6)type1(7)print(type3())type1(7)type2(7)type1(7)print(type3())print(type4()) |
Javascript
// JavaScript Code// user-defined data-typeslet pr = { first: 0, second: 0};// declare a user-defined setlet s1 = new Set();let s2 = new Set();// hash map let m = {};// Function to process the query// of type-1function type1(num) { // if the element is already there if (num in m) { // get the frequency of the element let cnt = m[num]; // returns an iterator pointing to // position where the pair is let it1 = s1.find({ num, cnt }); let it2 = s2.find({ cnt, num }); // deletes the pair from sets s1.delete(it1); s2.delete(it2); // re-insert the pair by increasing // frequency s1.add({ num, m[num] + 1 }); s2.add({ m[num] + 1, num }); } // if the element is not there in the list else { // insert the element with frequency 1 s1.add({ num, 1 }); s2.add({ 1, num }); } // increase the count in hash-table m[num] += 1;}// Function to process the query// of type-2function type2(num) { // if the element exists if (num in m) { // get the frequency of the element let cnt = m[num]; // returns an iterator pointing to // position where the pair is let it1 = s1.find({ num, cnt }); let it2 = s2.find({ cnt, num }); // deletes the pair from sets s1.delete(it1); s2.delete(it2); // re-insert the pair by increasing // frequency s1.add({ num, m[num] - 1 }); s2.add({ m[num] - 1, num }); // decrease the count m[num] -= 1; }}// Function to process the query// of type-3function type3() { // if the set is not empty // return the first element if (!s1.empty()) { let it = s2.first(); return it.second; } else return -1;}// Function to process the query// of type-4function type4() { // if the set is not empty // return the last element if (!s1.empty()) { let it = s2.last(); return it.second; } else return -1;}// Driver Codefunction main() { // Queries // inserts 6, 6 and 7 type1(6); type1(6); type1(7); // print the answer to query of type3 console.log(type3()); // inserts 7 type1(7); // deletes one occurrence of 7 type2(7); // inserts 7 type1(7); // print the answer to query of type3 console.log(type3()); // print the answer to query of type4 console.log(type4());}main();// This code is contributed by ishankhandelwals. |
C#
// C# program for performing // Queries of insert, delete one// occurrence of a number and // print the least and most frequent elementusing System;using System.Collections.Generic;// user-defined data-typespublic struct pr{ public int first; public int second;}// user-defined function to// design a setpublic class Sortbyfirst : IComparer<pr>{ public int Compare(pr a, pr b) { if (a.first == b.first) return a.second > b.second ? 1 : -1; return a.first < b.first ? -1 : 1; }}// declare a user-defined setpublic static SortedSet<pr> s1 = new SortedSet<pr>(new Sortbyfirst());public static SortedSet<pr> s2 = new SortedSet<pr>(new Sortbyfirst());// hash map public static Dictionary<int, int> m = new Dictionary<int, int>();// Function to process the query// of type-1public static void type1(int num){ // if the element is already there if (m.ContainsKey(num)) { // get the frequency of the element int cnt = m[num]; // returns an iterator pointing to // position where the pair is pr target1 = new pr { first = num, second = cnt }; pr target2 = new pr { first = cnt, second = num }; // deletes the pair from sets s1.Remove(target1); s2.Remove(target2); // re-insert the pair by increasing // frequency s1.Add(new pr { first = num, second = m[num] + 1 }); s2.Add(new pr { first = m[num] + 1, second = num }); } // if the element is not there in the list else { // insert the element with frequency 1 s1.Add(new pr { first = num, second = 1 }); s2.Add(new pr { first = 1, second = num }); } // increase the count in hash-table m[num] += 1;}// Function to process the query// of type-2public static void type2(int num){ // if the element exists if (m.ContainsKey(num)) { // get the frequency of the element int cnt = m[num]; // returns an iterator pointing to // position where the pair is pr target1 = new pr { first = num, second = cnt }; pr target2 = new pr { first = cnt, second = num }; // deletes the pair from sets s1.Remove(target1); s2.Remove(target2); // re-insert the pair by increasing // frequency s1.Add(new pr { first = num, second = m[num] - 1 }); s2.Add(new pr { first = m[num] - 1, second = num }); // decrease the count m[num] -= 1; }}// Function to process the query// of type-3public static int type3(){ // if the set is not empty // return the first element if (s1.Count != 0) { return s2.Min.second; } else return -1;}// Function to process the query// of type-4public static int type4(){ // if the set is not empty // return the last element if (s1.Count != 0) { return s2.Max.second; } else return -1;}// Driver Codepublic static void Main(){ // Queries // inserts 6, 6 and 7 type1(6); type1(6); type1(7); // print the answer to query of type3 Console.WriteLine(type3()); // inserts 7 type1(7); // deletes one occurrence of 7 type2(7); // inserts 7 type1(7); // print the answer to query of type3 Console.WriteLine(type3()); // print the answer to query of type4 Console.WriteLine(type4());}// This code is contributed by ishankhandelwals. |
Output:
7 7 6
Time Complexity: O(log N) per query.
Auxiliary Space: O(N)
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