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Day-068.cpp
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Day-068.cpp
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class Solution{
bool attacking( pair< int , int> p1 , pair< int , int> p2 ) {
if(p1.first + p1.second == p2.first + p2.second){
return true;
}
if(p1.first - p1.second == p2.first - p2.second){
return true;
}
return false;
}
public:
/*
* else solution can be if we create a map of every diagonal then for every co-ordinated
* we will update the particular key value, at last we can count the pair by simple
* combination trick.
* Time Complexity : O(N*log(N)+N); ( N -> traversing all pair of co-ordinated )
* Space Complexity: O(N)
*
*/
int getAttackingPair( vector< pair< int , int > > &bishop ) { // O(N^2)
int n = static_cast< int >( bishop.size() );
int counter{};
for( int i{}; i < n; ++i ) {
for( int j = i+1; j < n; ++j ) {
if( attacking(bishop[i] , bishop[j]) ) {
++counter;
}
}
}
return counter;
}
};