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  1. Frequency Queries
  2. Discussions

Frequency Queries

  • dehaka
     + 0 comments

    approximately O(q) using unordered_map, which are hash objects

    each query takes constant time if there's no hash collison

    vector<int> freqQuery(const vector<vector<int>>& queries) {
    unordered_map<int, int> freq, freqCount;
    vector<int> answer;
    for (int i = 0; i < queries.size(); ++i) {
        if (queries[i][0] == 1) {
            auto it = freq.find(queries[i][1]);
            if (it != freq.end()) --freqCount[it->second];
            int t = ++freq[queries[i][1]];
            ++freqCount[t];
        }
        else if (queries[i][0] == 2) {
            auto it = freq.find(queries[i][1]);
            if (it != freq.end() and it->second != 0) {
                --freqCount[it->second];
                --it->second;
                ++freqCount[it->second];
            }
        }
        else if (queries[i][0] == 3) {
            auto it = freqCount.find(queries[i][1]);
            if (it != freqCount.end() and it->second != 0) answer.push_back(1);
            else answer.push_back(0);
        }
    }
    return answer;
}