Hackerrank Sets-STL Solution
- Published
- Updated
- Reading time
- 3 min read
- Length
- 525 words
Sets are a part of the C++ STL. Sets are containers that store unique elements following a specific order. Here are some of the frequently used member functions of sets:
Declaration:
set<int>s; //Creates a set of integers.Size:
int length=s.size(); //Gives the size of the set.Insert:
s.insert(x); //Inserts an integer x into the set s.Erasing an element:
s.erase(val); //Erases an integer val from the set s.Finding an element:
set<int>::iterator itr=s.find(val); //Gives the iterator to the element val if it is found otherwise returns s.end() .
Ex: set<int>::iterator itr=s.find(100); //If 100 is not present then it==s.end().To know more about sets click Here. Coming to the problem, you will be given queries. Each query is of one of the following three types:
: Add an element to the set.
: Delete an element from the set. (If the number is not present in the set, then do nothing).
: If the number is present in the set, then print "Yes"(without quotes) else print "No"(without quotes).
Input Format
The first line of the input contains where is the number of queries. The next lines contain query each. Each query consists of two integers and where is the type of the query and is an integer.
Constraints
Output Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}
For queries of type print "Yes"(without quotes) if the number is present in the set and if the number is not present, then print "No"(without quotes).
Each query of type should be printed in a new line.
Sample Input
8
1 9
1 6
1 10
1 4
3 6
3 14
2 6
3 6Sample Output
Yes
No
NoSolution in cpp#
Approach 1.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <unordered_set>
#include <algorithm>
using namespace std;
int main() {
unordered_set<int> s;
int q; cin >> q;
for(int j=0;j<q;++j) {
int y, x; cin >> y >> x;
switch(y) {
case 1: s.insert(x); break;
case 2: s.erase(x); break;
case 3: cout << (s.count(x) > 0 ? "Yes" : "No") << endl;
}
}
return 0;
}Approach 2.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
using namespace std;
int main() {
int q, y, x;
set<int>::iterator it;
set<int> s;
cin >> q;
while (q--)
{
cin >> y >> x;
switch (y)
{
case 3:
it = s.find(x);
(it != s.end()) ? cout << "Yes" : cout << "No";
cout << endl;
break;
case 2:
s.erase(x);
break;
case 1:
s.insert(x);
break;
default:
break;
}
}
return 0;
}Approach 3.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
using namespace std;
int main() {
int q;
cin >> q;
set<int> st;
for(int i = 0; i < q; i++) {
int y, x;
cin >> y >> x;
if(y == 1) {
st.insert(x);
} else if(y == 2) {
st.erase(x);
} else {
auto z = st.find(x);
if(z != st.end()) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}
}
return 0;
}
