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Problem solution · C++

Reverse Words with Same Vowel Count

Reverse Words with Same Vowel Count: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
36 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

Sliding window or two pointers

For Reverse Words with Same Vowel Count, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 36 lines of C++ from the credited upstream file reverse-words-with-same-vowel-count.cpp.
  • The implementation visibly relies on hash lookup.
  • 2 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeReverse Words with Same Vowel Count · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(1) // string, inplaceclass Solution {public:    string reverseWords(string s) {        static const unordered_set<char> VOWELS = {'a', 'e', 'i', 'o', 'u'};        const auto& count = [&](int left, int right) {            int result = 0;            for (int i = left; i <= right; ++i) {                result += VOWELS.count(s[i]);            }            return result;        };         for (int i = 0, l = 0, cnt = -1; i < size(s); ++i) {            if (s[i] == ' ') {                l = 0;                continue;            }            ++l;            if (i + 1 != size(s) && s[i + 1] != ' ') {                continue;            }            const auto& c = count(i - l + 1, i);            if (cnt == -1) {                cnt = c;            } else if (cnt == c) {                reverse(begin(s) + i - l + 1, begin(s) + (i + 1));            }        }        return s;    }};

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