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Problem solution · C++

Data Stream as Disjoint Intervals

Data Stream as Disjoint Intervals: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
54 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

Binary search

For Data Stream as Disjoint Intervals, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 54 lines of C++ from the credited upstream file 352.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 1 loop block detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeData Stream as Disjoint Intervals · C++C++
Use this to learn the idea, then write your own version.
class SummaryRanges { public:  void addNum(int val) {    if (intervals.contains(val))      return;     const int lo = lowerKey(val);    const int hi = higherKey(val);     // {lo, intervals[lo][1]} + val + {hi, intervals[hi][1]} = {lo,    // intervals[hi][1]}    if (lo >= 0 && hi >= 0 && intervals[lo][1] + 1 == val && val + 1 == hi) {      intervals[lo][1] = intervals[hi][1];      intervals.erase(hi);      // {lo, intervals[lo][1]} + val = {lo, val}      // Prevent adding duplicate entry by using '>=' instead of '=='.    } else if (lo >= 0 && intervals[lo][1] + 1 >= val) {      intervals[lo][1] = max(intervals[lo][1], val);    } else if (hi >= 0 && val + 1 == hi) {      // val + {hi, intervals[hi][1]} = {val, intervals[hi][1]}      intervals[val] = {val, intervals[hi][1]};      intervals.erase(hi);    } else {      intervals[val] = {val, val};    }  }   vector<vector<int>> getIntervals() {    vector<vector<int>> res;    for (const auto& [_, interval] : intervals)      res.push_back(interval);    return res;  }  private:  map<int, vector<int>> intervals;  // {start: (start, end)}   // Returns the maximum key in `intervals` < `key`.  int lowerKey(int key) {    auto it = intervals.lower_bound(key);    if (it == intervals.begin())      return -1;    return (--it)->first;  }   // Returns the minimum key in `intervals` > `key`.  int higherKey(int key) {    const auto it = intervals.upper_bound(key);    if (it == intervals.cend())      return -1;    return it->first;  }};

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