Asked in: Walmart
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All Testcases Passed ✔
Solution
s = input().strip().lower()
n = len(s)
j = 0
ans = 0
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting the number of continuous subarrays consisting only of vowels efficiently, it’s important to understand the properties of the problem and leverage mathematical observations rather than brute force enumeration.
Step 1: Understand the Problem and Constraints
- You have an array of length N, each element being a single English alphabet letter.
- A “good” subarray is a continuous sequence of elements where every character is a vowel (A, E, I, O, U).
- You want to count all such subarrays.
- The naive approach, checking every possible subarray, has a time complexity of O(N^2), which is too slow for large inputs.
Step 2: Key Insight — Consecutive Vowel Segments
- Instead of looking at every subarray, focus on segments of the array that contain only vowels.
- When you identify a maximal consecutive segment of vowels, all subarrays within this segment are automatically “good.”
- For example, if you have a segment of length L consisting solely of vowels, the number of subarrays within this segment is given by the formula: L * (L + 1) / 2.
- This formula comes from summing up the counts of subarrays of lengths 1 through L.
Step 3: Breaking the Array into Vowel Segments
- Iterate through the array once.
- Each time you find a vowel, you extend the current vowel segment.
- When you hit a consonant or the end of the array, the current vowel segment ends.
- Calculate the number of good subarrays for that segment using the formula above, then reset the segment length counter.
Step 4: Efficiency Considerations
- This approach only requires a single pass through the array, resulting in O(N) time complexity.
- No need for nested loops or checking all possible subarrays.
- The memory usage is minimal since you only need counters and no additional complex data structures.
Step 5: Implementation Details to Keep in Mind
- Define a helper method or logic to identify vowels quickly. This can be done by checking if a character is in the set {A, E, I, O, U}.
- Handle the edge case when the array ends with a vowel segment; make sure to add the last segment’s count to the total.
- The result is the sum of all subarray counts from all vowel segments.
Step 6: Alternative Approaches (and why to prefer the above)
- Sliding window or prefix sums are common in subarray problems, but here the simplicity of counting consecutive vowels is more direct and efficient.
- Sliding window can be useful if the problem involved constraints like maximum length or additional conditions.
- Prefix sums usually assist in problems involving sums or counts with varying conditions, but here counting continuous vowel segments is simpler.
Step 7: Final Thoughts
- The crucial observation is that continuous vowel segments fully define the set of good subarrays.
- By reducing the problem to counting the lengths of these segments and applying the triangular number formula, you avoid exponential computation.
- This approach scales well with large input sizes.
- It highlights the importance of problem decomposition and leveraging mathematical properties to optimize solutions.
Summary:
1. Identify continuous segments of vowels.
2. Calculate number of subarrays for each segment using L * (L + 1) / 2.
3. Sum these counts to get the total number of good subarrays.
4. Return or output the total count.
This method efficiently counts all vowel-only subarrays with just a single pass and simple arithmetic.
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