Intersection of two std::unordered_map

Question

I have two std::unordered_map

std::unordered_map<int, int> mp1;
std::unordered_map<int, int> mp2;

I need to find the intersection of key-value pairs and store it in another map of the form.

std::unordered_map<int, int> mp;

How can i do this??

Answer 1

You could use std::set_intersection to fill a new container containing the key, value pairs that exists in both maps. set_intersection needs the ranges to be sorted (which is exactly what you won't get from an unordered_map) so either, replace the unordered_maps with map or create temporary maps (or temporary std::set<std::pair<int, int>>s) before using set_intersection.

I would recommend replacing your original unordered_maps with ordered maps for efficiency if you need intersections often:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <map>
#include <unordered_map>
#include <vector>

int main() {
    std::map<int, int> mp1 {{1,0}, {2,0}, {3,0}};
    std::map<int, int> mp2 {{0,0}, {2,0}, {3,0}};

    // this can be unordered unless you plan to use it in an intersection later:
    std::unordered_map<int, int> mp;

    std::set_intersection(
        mp1.begin(), mp1.end(),
        mp2.begin(), mp2.end(), 
        std::inserter(mp, mp.begin())
    );

    for(auto[key, val] : mp) {
        std::cout << key << ',' << val << '\n';
    }
}

Possible output:

3,0
2,0

If you want to stay with unordered_maps and to not have to create temporary sets or maps, you can just replace set_intersection above with a manual filler:

    const auto& [min, max] = std::minmax(mp1, mp2,
                                         [](auto& a, auto& b) {
                                             return a.size() < b.size();
                                         });
    for(auto& [key, value] : min) {               // iterate over the smallest map
        auto fi = max.find(key);                  // find in the bigger map
        if(fi != max.end() && fi->second == value)
            mp.emplace(key, value);               // add the pair if you got a hit
    }

The reason for iterating over the smallest map is to keep the number of find operations down to a minimum. Consider a case where one map contains 1 element and the other 1000000 elements. You then want 1 lookup and not 1000000.

A more generic solution could be making a function template out of it:

template<
    class Key,
    class T,
    class Hash = std::hash<Key>,
    class KeyEqual = std::equal_to<Key>,
    class Allocator = std::allocator< std::pair<const Key, T> >
>
auto unordered_map_intersection(
    const std::unordered_map<Key,T,Hash,KeyEqual,Allocator>& mp1,
    const std::unordered_map<Key,T,Hash,KeyEqual,Allocator>& mp2)
{
    std::unordered_map<Key,T,Hash,KeyEqual,Allocator> mp;

    const auto& [min, max] = std::minmax(mp1, mp2,
                                         [](auto& a, auto& b) {
                                             return a.size() < b.size();
                                         });
    for(auto& [key, value] : min) {               // iterate over the smallest map
        auto fi = max.find(key);                  // find in the bigger map
        if(fi != max.end() && fi->second == value)
            mp.emplace(key, value);               // add the pair if you got a hit
    }
    return mp;
}