Intersection of Two Arrays II — C++

// If the given array is not sorted and the memory is unlimited:
//   - Time:  O(m + n)
//   - Space: O(min(m, n))
// elif the given array is already sorted:
//   if m << n or m >> n:
//     - Time:  O(min(m, n) * log(max(m, n)))
//     - Space: O(1)
//   else:
//     - Time:  O(m + n)
//     - Soace: O(1)
// else: (the given array is not sorted and the memory is limited)
//     - Time:  O(max(m, n) * log(max(m, n)))
//     - Space: O(1)

// If the given array is not sorted and the memory is unlimited.
// Time:  O(m + n)
// Space: O(min(m, n))
// Hash solution.
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() > nums2.size()) {
            return intersect(nums2, nums1);
        }

        unordered_map<int, int> lookup;
        for (const auto& i : nums1) {
            ++lookup[i];
        }

        vector<int> result;
        for (const auto& i : nums2) {
            if (lookup[i] > 0) {
                result.emplace_back(i);
                --lookup[i];
            }
        }

        return result;
    }
};


// If the given array is already sorted, and the memory is limited, and (m << n or m >> n).
// Time:  O(min(m, n) * log(max(m, n)))
// Space: O(1)
// Binary search solution.
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() > nums2.size()) {
            return intersect(nums2, nums1);
        }

        // Make sure it is sorted, doesn't count in time.
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());

        vector<int> result;
        auto it = nums2.cbegin();
        for (const auto& i : nums1) {
            it = lower_bound(it, nums2.cend(), i);
            if (it != nums2.end() && *it == i) {
                result.emplace_back(*it++);
            }
        }
        
        return result;
    }
};


// If the given array is already sorted, and the memory is limited or m ~ n.
// Time:  O(m + n)
// Soace: O(1)
// Two pointers solution.
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        // Make sure it is sorted, doesn't count in time.
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        auto it1 = nums1.cbegin(), it2 = nums2.cbegin();
        while (it1 != nums1.cend() && it2 != nums2.cend()) {
            if (*it1 < *it2) {
                ++it1;
            } else if (*it1 > *it2) {
                ++it2;
            } else {
                result.emplace_back(*it1);
                ++it1, ++it2;
            }
        }
        return result;
    }
};


// If the given array is not sorted, and the memory is limited.
// Time:  O(max(m, n) * log(max(m, n)))
// Space: O(1)
// Two pointers solution.
class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        // O(max(m, n) * log(max(m, n)))
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        auto it1 = nums1.cbegin(), it2 = nums2.cbegin();
        while (it1 != nums1.cend() && it2 != nums2.cend()) {
            if (*it1 < *it2) {
                ++it1;
            } else if (*it1 > *it2) {
                ++it2;
            } else {
                result.emplace_back(*it1);
                ++it1, ++it2;
            }
        }
        return result;
    }
};

// Time:  O(mlogm + nlogn)
// Space: O(m + n)
// simple solution
class Solution2 {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        sort(begin(nums1), end(nums1)); sort(begin(nums2), end(nums2));
        set_intersection(cbegin(nums1), cend(nums1), cbegin(nums2), cend(nums2), back_inserter(result));
        return result;
    }
};