Let's make this one less mysterious. For 744. Find Smallest Letter Greater Than Target, the solution in this repository is mainly a straightforward implementation solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: straightforward implementation.
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The solution is organized around the main LeetCode entry point and a few local helpers.
Guide
Why?
The point of the implementation is not to make the code longer. It is to avoid doing the same thinking twice.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 12 ms, faster than 93.52% of C++ online submissions for Find Smallest Letter Greater Than Target.
02//Memory Usage: 9 MB, less than 72.73% of C++ online submissions for Find Smallest Letter Greater Than Target.
03
04class Solution {
05public:
06 char nextGreatestLetter(vector<char>& letters, char target) {
07 for(char letter : letters){
08 if(letter > target){
09 return letter;
10 }
11 }
12 return letters[0];
13 }
14};Cost