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Why using unordered_map gives TLE and how to spot it?
By noobnocap, history, 4 weeks ago, In English

Unorded_map lets us access any element in O(1). So unordered_map is very handy for some problems. But sometimes it gives TLE as the worst case time is o(n). How to spot it beforehand and know the hash function will generate heavy collisions?

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4 weeks ago, # |
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You need to know:

  1. The test cases
  2. The hash function you are using

See if the test cases will generate heavy collisions.

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4 weeks ago, # |
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just don't use unordered map simple

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4 weeks ago, # |
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who said that unordered_map lets us access any element in O(1), this is a misconception.

Worst case: O(n) Best case: O(1)

So just use map instead, cuz map works O(log(n)) in any case

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    4 weeks ago, # ^ |
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    Actually, you can still use unordered_map with custom hash. Please refer to this blog if you want to use a safe unordered_map. Link. If you are feeling lazy to open the blog, here is the source code:

    struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

    Via this safe custom hash, you can use unordered_map like this:

    unordered_map<long long, int, custom_hash> safe_map;
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