Given a target K, and an array of distinct elements, the task is to remove all pairs from the array which sums to K.
This is the approach I followed
for(i=0;i<N;i++)
cin>>array[i];
cin>>K;
map<int, int> mp;
for(i=0;i<N;i++)
{
if(mp.find(array[i])==mp.end())
mp[array[i]] = 1;
else
mp[array[i]]++;
}
Logic for deletion
for(i=0;i<N;i++)
{
if(mp[K-array[i]]>0)
{
mp.erase(K-array[i]);
mp.erase(array[i]);
}
}
Print output :
cout<<mp.size();
Input :
array = 6 5 4 2 1 0
K = 6
Program output
4
Expected Output
0
if(mp[K-array[i]]>0)
std::map::operator::[]
will add if element doesn't exists
You will have to find and remove like following:
for(int i=0;i<N;i++)
{
if(mp.find(K-array[i]) != mp.end() &&
K-array[i] != array[i])
//~~~~~~~~~~~~~~~~~~~~~ check required to see k/2 = array[i]
// Note: all numbers are distinct
{
mp.erase(K-array[i]);
mp.erase(array[i]);
}
}
A better strategy would be to use remove elements during scanning using std::unordered_set
std::unordered_set<int> s;
for(const auto& ele: array) {
auto it = s.find( K - ele);
if ( it == s.end() ) {
s.insert(ele);
} else {
s.erase(it);
}
}
std::cout << s.size();
See here