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Problem solution · C++

Design Most Recently Used Queue

Design Most Recently Used Queue: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Hash-based lookup
Source
Kamyu LeetCode Solutions
Length
116 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Design Most Recently Used Queue, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 116 lines of C++ from the credited upstream file design-most-recently-used-queue.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 8 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDesign Most Recently Used Queue · C++C++
Use this to learn the idea, then write your own version.
// Time:  ctor:  O(n + m), m is the max number of calls// Space: fetch: O(log(n + m)) class MRUQueue {public:    MRUQueue(int n)     : bit_{n}     , curr_{n} {        for (int i = 0; i < n; ++i) {            lookup_[i] = i + 1;        }    }        int fetch(int k) {        int pos  = bit_.binary_lift(k);        int val = lookup_[pos];        lookup_.erase(pos);        bit_.add(pos, -1);        bit_.add(curr_, 1);        lookup_[curr_++] = val;        return val;    } private:    class BIT {    public:        static const int MAX_CALLS = 2000;         BIT(int n) : bit_(n + MAX_CALLS + 1) {  // 0-indexed            for (int i = 1; i < size(bit_); ++i) {                bit_[i] = ((i - 1 < n) ? 1 : 0) + bit_[i - 1];            }            for (int i = size(bit_) - 1; i >= 1; --i) {                int last_i = i - lower_bit(i);                bit_[i] -= bit_[last_i];            }        }                void add(int i, int val) {            ++i;            for (; i < size(bit_); i += lower_bit(i)) {                bit_[i] += val;            }        }         int query(int i) const {            ++i;            int total = 0;            for (; i > 0; i -= lower_bit(i)) {                total += bit_[i];            }            return total;        }                int binary_lift(int k) const {            int total = 0;            int pos = 0;            for (int i = floor_log2_x(size(bit_) - 1); i >= 0; --i) {                if (pos + (1 << i) < size(bit_) && !(total + bit_[pos + (1 << i)] >= k)) {                    total += bit_[pos + (1 << i)];                    pos += (1 << i);                }            }            return (pos + 1) - 1;        }        private:        int lower_bit(int i) const {            return i & -i;        }                int floor_log2_x(int x) const {            return 8 * sizeof(int) - __builtin_clz(x) - 1;        };                vector<int> bit_;    };        BIT bit_;    unordered_map<int, int> lookup_;    int curr_;}; // Time:  ctor:  O(n)// Space: fetch: O(sqrt(n))// sqrt decomposition solutionclass MRUQueue2 {public:    MRUQueue2(int n)     : buckets_(ceil(sqrt(n))) {        for (int i = 0; i < n; ++i) {            buckets_[i / size(buckets_)].emplace_back(i + 1);        }    }        int fetch(int k) {        --k;        int left = k / size(buckets_);        int idx = k % size(buckets_);        auto cit = cbegin(buckets_[left]);        advance(cit, idx);        int val = *cit;        buckets_[left].erase(cit);        buckets_.back().emplace_back(val);        for (int i = size(buckets_) - 2; i >= left; --i) {            int x = buckets_[i + 1].front();            buckets_[i + 1].pop_front();            buckets_[i].emplace_back(x);        }        return val;    }    private:    vector<list<int>> buckets_;};

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