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Missing Numbers

  • VictoriaB
     + 6 comments

    Following the wonderful observation by @baran_jana, we can make do with only a frequency array of size 201:

        int n;
    cin >> n;

    int pivot, value;
    cin >> pivot;

    int freq[200 + 1] = { 0 };
    freq[100] = 1;
    
    for (int i = 1; i < n; ++i) {
        cin >> value;
        freq[100 + (value - pivot)]++;
    }

    int m;
    cin >> m;

    for (int i = 0; i < m; ++i) {
        cin >> value;
        freq[100 + (value - pivot)]--;
    }

    for (int i = -100; i <= 100; ++i) {
        if (freq[100 + i] != 0)
            cout << (pivot + i) << " ";
    }

    return 0;