Java实习任务：平衡子词计数测试用例（结果28）困惑求助
Hey there! Let's dig into why you might be stuck on this test case—no code dumps, just the core logic and common pitfalls that trip people up when counting balanced substrings.
First, let's re-clarify the key definition to make sure we're aligned:
A balanced substring is non-empty, and every character present in it appears exactly the same number of times. So single characters like "a" count (only one character exists, so its count is consistent), "ab" counts (a and b each once), "aabb" counts (a and b each twice), but "aab" does not (a twice, b once).
Breaking Down the 28 Result
Let's start with how the total 28 is calculated—this will help you spot where you might be missing counts:
- Single-character substrings: Every individual character is balanced. For "aabbabcccba", we have 5 a's, 4 b's, 3 c's → that's 5+4+3 = 12 right there. This is a super common area to undercount if you're focusing only on longer substrings.
- 2-character balanced substrings: These are substrings where exactly two distinct characters appear, each with the same count. Examples include adjacent pairs like "aa", "bb", "ab", "ba", plus longer intervals like positions 0-3 ("aabb": 2a, 2b) or positions 2-5 ("baba": 2a, 2b). There are 14 such substrings here.
- 3-character balanced substrings: These are substrings where all three characters (a, b, c) appear the same number of times. For this test case, that means substrings where each character appears exactly once (since the total counts of a, b, c are 5,4,3—no interval can fit each character appearing twice). There are 2 such substrings here.
Adding those up: 12 +14 +2 =28.
Common Mistakes Throwing Off Your Count
Here are the most frequent missteps that lead to missing or overcounting:
- Forgetting single-character substrings: It's easy to fixate on longer substrings and overlook that every single character is a valid balanced substring. Always start with this count—it's your baseline.
- Only checking adjacent character blocks: Balanced substrings don't have to be made of consecutive identical character groups. For example, "abba" (positions 1-4 of the test string) is balanced (2a,2b) but isn't just two blocks of a and two of b. You need to consider all possible intervals, not just ones aligned with character runs.
- Not using state tracking for frequency differences: For multi-character cases, brute-forcing every substring and counting frequencies is error-prone. Instead, think in terms of relative frequency states:
- For 2 characters (a and b), track the difference between a's count and b's count. If this difference repeats at two indices, the substring between those indices has equal counts of a and b (since the difference didn't change—meaning a and b increased by the same amount).
- For 3 characters (a,b,c), track pairs like
(a-b, a-c). If this pair repeats, the substring between those indices has equal counts of all three characters (since relative differences stayed the same, so each character increased by the same number of times).
- Overlooking small multi-character substrings: Short valid substrings like "abc" or "bca" are easy to miss when scanning for longer balanced intervals.
Key Takeaways to Adjust Your Approach
- Start with the obvious: count all single-character substrings first.
- For multi-character cases, use state-based tracking to efficiently find valid intervals instead of manually enumerating every possible substring (which leads to human error).
- Verify each category of balanced substrings (k=1, k=2, k=3) separately to ensure you don't double-count or miss any.
内容的提问来源于stack exchange，提问作者Bajter
