/* 
审题：
由于数字范围为1到100，那我先将三个数组压入三个不同的hash表，
然后遍历1到100，用O(3)的复杂度判断该数字是否在三个hash表中，
把结果累加输出，总复杂度是O(n1 + n2 + n3 + 3 * 100)
*/

vector<int> twoOutOfThree(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3) {
    unordered_map<int, int> numMap1;
    unordered_map<int, int> numMap2;
    unordered_map<int, int> numMap3;
    for (auto i = 0; i < nums1.size(); i++) {
        auto num = nums1[i];
        numMap1[num]++;
    }
    for (auto i = 0; i < nums2.size(); i++) {
        auto num = nums2[i];
        numMap2[num]++;
    }
    for (auto i = 0; i < nums3.size(); i++) {
        auto num = nums3[i];
        numMap3[num]++;
    }
    vector<int> result;
    for (int i = 1; i <= 100; i++) {
        int numCount = 0;
        numCount += static_cast<bool>(numMap1.count(i)) + static_cast<bool>(numMap2.count(i)) + static_cast<bool>(numMap3.count(i));
        if (numCount > 1) {
            result.push_back(i);
        }
    }
    return result;
}

/*
审题：这题不会
看了解析，取中位数，然后遍历所有元素做差看是否能整除x
*/

int minOperations(vector<vector<int>>& grd, int x) {
    int n = grd.size();
    int m = grd[0].size();
    vector<int> cnt(n * m);
    int index = 0;
    for(int i = 0; i < n; i ++){
        for(int j = 0; j < m; j ++){
            cnt[index++] = grd[i][j];
        }
    }
    sort(cnt.begin(), cnt.end());
    int t = cnt.size()/ 2;
    int res = 0;
    for(int i = 0; i < cnt.size(); i ++){
        if(cnt[i] != cnt[t]) {
            int k = abs(cnt[i] - cnt[t]);
            if(k % x  == 0) {
                res += k / x;
            }
            else return -1;
        }
    }
    return res;
}