1007: Minimum-Domino-Rotations-For-Equal-Row
Medium

table of contents

This question is weird. I dislike my solution, but I will post about it anyways because I alksjdaslkjdlaskjd.

Anyways, basically my algo takes the first domino and creates a hashmap using tops[0] and bottoms[0] as keys, where in the [key, value] pair, the value is a pair<int, int> structure with value.first storing the tops count and value.second storing the bottoms count.

It then just iterates through all the dominoes and checks if either keys are present on either side of the domino.

Afterwards,

code

class Solution {
public:
    int minDominoRotations(vector<int>& tops, vector<int>& bottoms) {
        unordered_map<int, pair<int, int>> mp;
        ++mp[tops[0]].first;
        ++mp[bottoms[0]].second;
        for (int i = 1; i < tops.size(); ++i) {
            if (mp.size() == 0) break;
            vector<int> erase;
            for (auto& [key, value] : mp) {
                if (key == tops[i] || key == bottoms[i]) {
                    if (key == tops[i]) {
                        ++value.first;
                    }
                    if (key == bottoms[i]) {
                        ++value.second;
                    }
                } else {
                    erase.push_back(key);
                }
            }
            for (int& n : erase) {
                mp.erase(n);
            }
        }
        if (mp.size() == 0) {
            return -1;
        }
        int ans = INT32_MAX;
        for (auto& [key, value] : mp) {
            ans = min(ans, (int)min(tops.size()-value.first, tops.size()-value.second));
        }
        return ans;
    }
};

complexity