Hash Table implementation in C++ - How to return a key with a particular value
I am trying to solve a programming question using Hash Tables in C++. It is supposed to be fairly simple implementation of hash tables. I am extremely new to Hash Tables and this is my first try at implementation.
The question is that I have been given an array which contains integers. All but one integer repeats itself twice. I have to find an integer that doesn't.
Input: {1,2,1,3,3}
Output: 2
My solution is that I will start putting these keys inside a hash table and if I find a key inside the hash table beforehand, I will delete that key from the hash table.
My code implementation works but I now I wanted to see how I can get back the right value (2 in this case) since even after erasing the key, the keys remain with value 0.
Here is my code:
int main()
{
int num[5] = {1,2,1,3,3};
map <int,int> mymap;
for(int i=0;i<5;i++)
{
if(mymap.find(num[i])!=mymap.end())
{
mymap.erase(num[i]);
}
else
{
mymap[num[i]] = 10; //10 is just a placeholder value.
}
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
}
Output:
0:0
1:0
2:10
3:0
4:0
c++ hash hashtable
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
asked Nov 20 at 19:23
caspian
234
add a comment |
I am trying to solve a programming question using Hash Tables in C++. It is supposed to be fairly simple implementation of hash tables. I am extremely new to Hash Tables and this is my first try at implementation.
The question is that I have been given an array which contains integers. All but one integer repeats itself twice. I have to find an integer that doesn't.
Input: {1,2,1,3,3}
Output: 2
My solution is that I will start putting these keys inside a hash table and if I find a key inside the hash table beforehand, I will delete that key from the hash table.
My code implementation works but I now I wanted to see how I can get back the right value (2 in this case) since even after erasing the key, the keys remain with value 0.
Here is my code:
int main()
{
int num[5] = {1,2,1,3,3};
map <int,int> mymap;
for(int i=0;i<5;i++)
{
if(mymap.find(num[i])!=mymap.end())
{
mymap.erase(num[i]);
}
else
{
mymap[num[i]] = 10; //10 is just a placeholder value.
}
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
}
Output:
0:0
1:0
2:10
3:0
4:0
c++ hash hashtable
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
asked Nov 20 at 19:23
caspian
234
hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29
add a comment |
I am trying to solve a programming question using Hash Tables in C++. It is supposed to be fairly simple implementation of hash tables. I am extremely new to Hash Tables and this is my first try at implementation.
The question is that I have been given an array which contains integers. All but one integer repeats itself twice. I have to find an integer that doesn't.
Input: {1,2,1,3,3}
Output: 2
My solution is that I will start putting these keys inside a hash table and if I find a key inside the hash table beforehand, I will delete that key from the hash table.
My code implementation works but I now I wanted to see how I can get back the right value (2 in this case) since even after erasing the key, the keys remain with value 0.
Here is my code:
int main()
{
int num[5] = {1,2,1,3,3};
map <int,int> mymap;
for(int i=0;i<5;i++)
{
if(mymap.find(num[i])!=mymap.end())
{
mymap.erase(num[i]);
}
else
{
mymap[num[i]] = 10; //10 is just a placeholder value.
}
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
}
Output:
0:0
1:0
2:10
3:0
4:0
c++ hash hashtable
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
asked Nov 20 at 19:23
caspian
234
I am trying to solve a programming question using Hash Tables in C++. It is supposed to be fairly simple implementation of hash tables. I am extremely new to Hash Tables and this is my first try at implementation.
The question is that I have been given an array which contains integers. All but one integer repeats itself twice. I have to find an integer that doesn't.
Input: {1,2,1,3,3}
Output: 2
My solution is that I will start putting these keys inside a hash table and if I find a key inside the hash table beforehand, I will delete that key from the hash table.
My code implementation works but I now I wanted to see how I can get back the right value (2 in this case) since even after erasing the key, the keys remain with value 0.
Here is my code:
int main()
{
int num[5] = {1,2,1,3,3};
map <int,int> mymap;
for(int i=0;i<5;i++)
{
if(mymap.find(num[i])!=mymap.end())
{
mymap.erase(num[i]);
}
else
{
mymap[num[i]] = 10; //10 is just a placeholder value.
}
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
}
Output:
0:0
1:0
2:10
3:0
4:0
c++ hash hashtable
c++ hash hashtable
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
asked Nov 20 at 19:23
caspian
234
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
asked Nov 20 at 19:23
caspian
234
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
edited Nov 20 at 19:33
Matthieu Brucher
11.6k22137
11.6k22137
asked Nov 20 at 19:23
caspian
234
asked Nov 20 at 19:23
caspian
234
asked Nov 20 at 19:23
caspian
234
234
hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29
add a comment |
hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29
hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29
hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29
add a comment |
2 Answers
2
active
oldest
votes
std::map is a tree, not a hash table. For a hash table you want std::unordered_map.
But to answer your question:
You can use the map iterator to get the only remaining value:
if (!mymap.empty()) {
cout << mymap.begin()->first;
}
But beware - when you call cout << mymap[X], it also adds X to the map. So remove all those debugging lines at the end.
And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set).
answered Nov 20 at 19:31
rustyx
28.5k696136
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use->firstinstead (but consider usingset)
– rustyx
Nov 20 at 19:53
add a comment |
Just increment the value in the map, as the integers are default constructed (and thus initialized to 0):
int num[5] = {1,2,1,3,3};
unordered_map <int,int> mymap;
for(int i=0;i<5;i++)
{
++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
std::map is a tree, not a hash table. For a hash table you want std::unordered_map.
But to answer your question:
You can use the map iterator to get the only remaining value:
if (!mymap.empty()) {
cout << mymap.begin()->first;
}
But beware - when you call cout << mymap[X], it also adds X to the map. So remove all those debugging lines at the end.
And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set).
answered Nov 20 at 19:31
rustyx
28.5k696136
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use->firstinstead (but consider usingset)
– rustyx
Nov 20 at 19:53
add a comment |
std::map is a tree, not a hash table. For a hash table you want std::unordered_map.
But to answer your question:
You can use the map iterator to get the only remaining value:
if (!mymap.empty()) {
cout << mymap.begin()->first;
}
But beware - when you call cout << mymap[X], it also adds X to the map. So remove all those debugging lines at the end.
And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set).
answered Nov 20 at 19:31
rustyx
28.5k696136
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use->firstinstead (but consider usingset)
– rustyx
Nov 20 at 19:53
add a comment |
std::map is a tree, not a hash table. For a hash table you want std::unordered_map.
But to answer your question:
You can use the map iterator to get the only remaining value:
if (!mymap.empty()) {
cout << mymap.begin()->first;
}
But beware - when you call cout << mymap[X], it also adds X to the map. So remove all those debugging lines at the end.
And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set).
answered Nov 20 at 19:31
rustyx
28.5k696136
std::map is a tree, not a hash table. For a hash table you want std::unordered_map.
But to answer your question:
You can use the map iterator to get the only remaining value:
if (!mymap.empty()) {
cout << mymap.begin()->first;
}
But beware - when you call cout << mymap[X], it also adds X to the map. So remove all those debugging lines at the end.
And by the way - when you don't have a value, just a key, then you can use a set instead (or unordered_set).
answered Nov 20 at 19:31
rustyx
28.5k696136
edited Nov 20 at 19:51
answered Nov 20 at 19:31
rustyx
28.5k696136
answered Nov 20 at 19:31
rustyx
28.5k696136
answered Nov 20 at 19:31
rustyx
28.5k696136
28.5k696136
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use->firstinstead (but consider usingset)
– rustyx
Nov 20 at 19:53
add a comment |
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use->firstinstead (but consider usingset)
– rustyx
Nov 20 at 19:53
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
For some reason, directly implementing your code did not work, but using iterator from the link that you provided me worked. if (!mymap.empty()) { auto it = mymap.begin(); cout << it->first; }
– caspian
Nov 20 at 19:44
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Also, thank you for ironing out the basics. Your answer was helpful.
– caspian
Nov 20 at 19:45
Ah yes was thinking about the set, answer updated. Should use
->first instead (but consider using set)– rustyx
Nov 20 at 19:53
Ah yes was thinking about the set, answer updated. Should use
->first instead (but consider using set)– rustyx
Nov 20 at 19:53
add a comment |
Just increment the value in the map, as the integers are default constructed (and thus initialized to 0):
int num[5] = {1,2,1,3,3};
unordered_map <int,int> mymap;
for(int i=0;i<5;i++)
{
++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
add a comment |
Just increment the value in the map, as the integers are default constructed (and thus initialized to 0):
int num[5] = {1,2,1,3,3};
unordered_map <int,int> mymap;
for(int i=0;i<5;i++)
{
++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
add a comment |
Just increment the value in the map, as the integers are default constructed (and thus initialized to 0):
int num[5] = {1,2,1,3,3};
unordered_map <int,int> mymap;
for(int i=0;i<5;i++)
{
++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
Just increment the value in the map, as the integers are default constructed (and thus initialized to 0):
int num[5] = {1,2,1,3,3};
unordered_map <int,int> mymap;
for(int i=0;i<5;i++)
{
++mymap[num[i]];
}
cout<<"0:"<<mymap[0]<<endl;
cout<<"1:"<<mymap[1]<<endl;
cout<<"2:"<<mymap[2]<<endl; //Should only find 10 value in 2
cout<<"3:"<<mymap[3]<<endl;
cout<<"4:"<<mymap[4]<<endl;
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
answered Nov 20 at 19:32
Matthieu Brucher
11.6k22137
11.6k22137
add a comment |
add a comment |
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hash table, but you are using a map, not an unordered_map (the former is a standard map, the latter is a hash map).
– Matthieu Brucher
Nov 20 at 19:29