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: 4Output:
7
7
6While 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 b.second; return a.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. Ony 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:
// 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-types struct pr {
int first;
int second;
}; // user-defined function to // design a set bool operator<(pr a, pr b)
{ if (a.first == b.first)
return a.second > b.second;
return a.first < b.first;
} // declare a user-defined set set<pr> s1, s2; // hash map unordered_map< int , int > m;
// Function to process the query // of type-1 void 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-2 void 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-3 int 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-4 int 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 Code int 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;
} |
7 7 6
Time Complexity: O(log N) per query.
Auxiliary Space: O(N)
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