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Assignment: interval_map — implement assign Use std::map. Goal: Implement the member function assign for the class template interval_map. The class holds a piecewise-constant mapping from K to V. The internal data structure is: - V m_valBegin; // value for all keys before the first change - std::map m_map; // stores changes of value at certain keys (key -> value starting there) Example representation: For interval_map: M.m_valBegin == 'A' M.m_map == { (1,'B'), (3,'A') } This means the mapping is: key: … -2 -1 0 1 2 3 4 5 … val: A A A B B A A A I.e., values are ‘A’ up to key 1, ‘B’ on [1,3), then back to ‘A’ from 3 onward. You must keep m_map as compact as possible (no redundant entries like ..., (3,'A'), (5,'A'), ... that don’t represent real changes). Constraints: - K must be usable as a key in std::map (requires a strict weak ordering via operator<). - V must be equality-comparable (operator==). Skeleton: #include #include #include template class interval_map { friend void IntervalMapTest(); V m_valBegin; std::map m_map; public: // constructor associates whole range of K with val template interval_map(V_forward&& val) : m_valBegin(std::forward(val)) {} // Assign value val to interval [keyBegin, keyEnd). // Overwrite previous values in this interval. // Conforming to the C++ Standard Library conventions, the interval // includes keyBegin, but excludes keyEnd. // If !(keyBegin < keyEnd), this designates an empty interval, // and assign must do nothing. template void assign(K const& keyBegin, K const& keyEnd, V_forward&& val) requires (std::is_same, V>::value) { // TODO: implement } // look-up of the value associated with key V const& operator[](K const& key) const { auto it = m_map.upper_bound(key); if (it == m_map.begin()) { return m_valBegin; } else { return std::prev(it)->second; } } };
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C++ Developer

Interviewed at think-cell

4.4★
Aug 17, 2025

Assignment: interval_map — implement assign Use std::map. Goal: Implement the member function assign for the class template interval_map. The class holds a piecewise-constant mapping from K to V. The internal data structure is: - V m_valBegin; // value for all keys before the first change - std::map m_map; // stores changes of value at certain keys (key -> value starting there) Example representation: For interval_map: M.m_valBegin == 'A' M.m_map == { (1,'B'), (3,'A') } This means the mapping is: key: … -2 -1 0 1 2 3 4 5 … val: A A A B B A A A I.e., values are ‘A’ up to key 1, ‘B’ on [1,3), then back to ‘A’ from 3 onward. You must keep m_map as compact as possible (no redundant entries like ..., (3,'A'), (5,'A'), ... that don’t represent real changes). Constraints: - K must be usable as a key in std::map (requires a strict weak ordering via operator<). - V must be equality-comparable (operator==). Skeleton: #include #include #include template class interval_map { friend void IntervalMapTest(); V m_valBegin; std::map m_map; public: // constructor associates whole range of K with val template interval_map(V_forward&& val) : m_valBegin(std::forward(val)) {} // Assign value val to interval [keyBegin, keyEnd). // Overwrite previous values in this interval. // Conforming to the C++ Standard Library conventions, the interval // includes keyBegin, but excludes keyEnd. // If !(keyBegin < keyEnd), this designates an empty interval, // and assign must do nothing. template void assign(K const& keyBegin, K const& keyEnd, V_forward&& val) requires (std::is_same, V>::value) { // TODO: implement } // look-up of the value associated with key V const& operator[](K const& key) const { auto it = m_map.upper_bound(key); if (it == m_map.begin()) { return m_valBegin; } else { return std::prev(it)->second; } } };

Problem was not very challenging, it was of average complexity. I can not disclose what the problem was but nine hours time was given to write a single functionality (a function). Problem statement explains how functionality should work and what the acceptable status of underlying data structure is after operation. It seems, problem is very black & white and easy to implement but when you observe it closely you will find few gray areas and that is what you have to think as developer to implement functionality correctly. Test allows submitting solution one more time if first submitted solution does not work. The hardest part is, this test is not similar to the test conducted on Hackerrank type of platforms. There are no test cases in this assignment, so you cannot compare you output with expected output.
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C++ Developer

Interviewed at think-cell

4.4★
May 3, 2024

Problem was not very challenging, it was of average complexity. I can not disclose what the problem was but nine hours time was given to write a single functionality (a function). Problem statement explains how functionality should work and what the acceptable status of underlying data structure is after operation. It seems, problem is very black & white and easy to implement but when you observe it closely you will find few gray areas and that is what you have to think as developer to implement functionality correctly. Test allows submitting solution one more time if first submitted solution does not work. The hardest part is, this test is not similar to the test conducted on Hackerrank type of platforms. There are no test cases in this assignment, so you cannot compare you output with expected output.

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