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Problem solution · C++

Next Closest Time

Next Closest Time: 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
29 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 Next Closest Time, 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

  • 29 lines of C++ from the credited upstream file 681.cpp.
  • The implementation visibly relies on ordered lookup.
  • No explicit loop blocks 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 codeNext Closest Time · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string nextClosestTime(string time) {    const set<char> digitsSet{time[0], time[1], time[3], time[4]};    string ans = time;     ans[4] = nextClosest(digitsSet, ans[4], '9');    if (time[4] < ans[4])      return ans;     ans[3] = nextClosest(digitsSet, ans[3], '5');    if (time[3] < ans[3])      return ans;     ans[1] = nextClosest(digitsSet, ans[1], ans[0] == '2' ? '3' : '9');    if (time[1] < ans[1])      return ans;     ans[0] = nextClosest(digitsSet, ans[0], '2');    return ans;  }  private:  char nextClosest(const set<char>& digitsSet, char digit, char limit) {    auto it = digitsSet.upper_bound(digit);    return it == digitsSet.end() || *it > limit ? *digitsSet.begin() : *it;  }};

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