Intersection between vectors [closed] - c++

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I have data in the following form.
vector<pair<unsigned,unsigned> > vecA; //here first value denotes reolution and second value denotes value. All values are of 4 bits
vecA.push_back(make_pair(2,2)); vecA.push_back(make_pair(2,3)); vecA.push_back(make_pair(3,6)); vecA.push_back(make_pair(3,7)); vecA.push_back(make_pair(4,5));
(2,2)-> signifies that the first 2 bits of value(a 4 bit number) are 10. i.e. the value could be "1000,1001,1010,1011" in binary
(2,3)-> signifies that the first 2 bits of value(a 4 bit number) are 11 i.e. the value could be "1100,1101,1110, 1011" in binary
(3,6)-> signifies that the first 3 bits of value(a 4 bit number) are 110 i.e., the value could be "1100,1101" in binary
(3,7)-> signifies that the first 3 bits of value(a 4 bit number) are 111 i.e., the value could be "1110,1111" in binary
(4,5)-> signifies that the first 4 bits of value(a 4 bit number) are 0101 i.e., the value is "0101" in binary
I have another vector containing the following:
vector<unsigned> vecB; //vecB has a by default resolution of 4. Here too the values are of 4 bits
vecB.push_back(10); vecB.push_back(6); vecB.push_back(13); vecB.push_back(12); vecB.push_back(15); vecB.push_back(5); vecB.push_back(7);
10-> signifies that the 4 bit number is: "1010"
6-> signifies that the 4 bit number is: "0110"
13-> signifies that the 4 bit number is: "1101"
12-> signifies that the 4 bit number is: "1100"
15-> signifies that the 4 bit number is: "1111", etc.
Now the intersection between vecA and vecB should perform a bit level comparison i.e. for 2 bit resolution of vecA just the first two bits of vecB should be seen.
i.e. (2,2) of vecA matches with "10" of vecB
(2,3) of vecA matches with "13,12,15" of vecB
(3,6) of vecA matches with "12,13" of vecB
(3,7) of vecA matches with "15" of vecB
(4,5) matches with "5" of vecB
The intersection should only return the matching values from vecB. i.e. the intersection should return "10,13,12,15,5" as the result.
How can I perform this intersection efficiently in c++?
vector<unsigned> ans;
for(vector<pair<unsigned,unsigned> >::iterator i1=vecA.begin(), l1=vecA.end(); i1!=l1;++i1)
{
for(vector<unsigned>::iterator i2=vecB.begin(),l2=vecB.end();i2!=l2;++i2)
{
if(((*i2)>>(*i1).first)==(*i1).second)
ans.push_back((*i1).second);
}
}

Answer 1

(2,2) represents 10??, where we don't care what ?? are. This is the half-open range 1000 through 1100, aka [2 << 2, (2+1)<<2).
So, produce a set of ranges from the LHS. Anything that overlaps, fuze. You'll have a set of start/finish intervals.
Now sort the RHS. Next, walk through it, keeping track when you enter/exit the LHS intervals. Those that are in the LHS intervals are in the intersection.
The RHS sorting takes O(|RHS| lg |RHS|). The walking takes O(|RHS| + |LHS|).
Making the LHS intervals takes O(|LHS| lg |LHS|) time (including time to sort by start-of-interval). Merging them is a single pass, also O(|LHS|).
So the end result is O(|RHS| lg |RHS| + |LHS| lg |LHS|) time to calculate intersection, instead of O(|RHS| * |LHS|) of your solution above.