2168. Unique Substrings With Equal Digit Frequency - Detailed Explanation

Time Complexity: Building the prefix table takes O(10 * n) = O(n) time. Then, we iterate over all substrings with two loops (≈ n²/2 substrings) and perform a constant amount of work (checking 10 counts) for each. This yields a time complexity of O(n²). For n = 1000, n² is 1,000,000, and checking 10 digits for each is a 10x multiplier (10,000,000 operations), which is efficient in C++/Java and manageable in Python with optimizations. This is a dramatic improvement over the O(n³) brute force.

Space Complexity: The prefix table uses O(10 * n) space, which is O(n). We also use a set to store seen substrings or their hashes. In the worst case, this set could hold O(n²) substrings (in scenarios where many substrings qualify and are all distinct). Storing actual substrings can be memory heavy (as discussed before). However, using a hash (integer) representation for each substring can reduce per-entry space. In terms of Big-O, the worst-case space remains O(n²) due to potentially many valid substrings. Typically, the distribution of digits will limit how many substrings qualify, and using hashing can mitigate memory usage.

Note: In a real interview or contest setting, you might implement further optimizations:

• Use rolling hash or a tuple of frequency counts as the key in the set to avoid storing large strings.
• Prune some checks by noticing that the length of a valid substring must be divisible by the number of distinct digits in it (since each distinct digit's frequency times the number of distinct digits gives the substring length). This can skip checking some substrings that cannot possibly have equal frequencies. However, such pruning can be complex to implement and the above solution is already efficient enough for the given constraints.

Edge Cases and Common Mistakes

• Single Character String: If s has length 1 (e.g., "5"), the result should be 1, since the only substring is the string itself and trivially all (single) digit frequencies are equal. Ensure the solution handles n=1 correctly.
• All Characters Same: If s consists of the same digit repeated (e.g., "1111" or "000"), then every substring is valid because any substring will have just that one digit. However, many substrings will be duplicates in terms of content. For example, s = "1111" has substrings "1", "11", "111", "1111" as the unique ones (any substring of length 2 is "11", etc.). A common mistake is to count duplicates separately if not using a set. Always ensure uniqueness by using a set or equivalent.
• Ignoring Absent Digits: Only consider digits that appear in the substring when checking frequency equality. Digits that are not present can be ignored. A mistake would be to include zeros in the equality check. For instance, in substring "11", we have two 1s and zero of every other digit. We consider it valid because the only digit present (1) appears with equal frequency (there’s just one distinct digit). We do not require that 0,2,3,...,9 also appear two times (that would be impossible).
• Duplicate Substrings: Ensure that if the same qualifying substring appears in multiple places, it is counted once. This is why using a set of seen substrings is crucial. Forgetting this will overcount results. For example, in "1212", the substring "12" appears at indices [0..1] and [2..3]; without a uniqueness check, one might count it twice.
• Performance Pitfalls: A common pitfall is to attempt a brute force solution without any optimization on frequency counting. Computing frequency from scratch for each substring (O(n) per substring) will time out on larger inputs. Using the prefix sum method or maintaining a running count is important. Another performance mistake can be not breaking out of loops early when a inequality is found in frequencies (in our code, we break out as soon as we detect two different counts).
• Off-by-One Errors: When using prefix sum arrays or slicing, it’s easy to get indices wrong. Be careful with the inclusive/exclusive nature of indices. For example, we used prefix[j+1] - prefix[i] to get counts for substring s[i..j]. Forgetting the +1 or using prefix[j] instead would lead to incorrect counts.

Alternative Variations

• Letters Instead of Digits: The problem could be generalized to a string with alphabetic characters (e.g., lowercase letters 'a'-'z') instead of digits. The condition would then be that every letter in the substring appears the same number of times. The solution approach would be similar: you would use a frequency array of size 26 for letters (or size 52 if considering both lowercase and uppercase), and the prefix sum technique would still apply. The complexity might increase because there are more possible characters, but for moderate string lengths it would still be manageable.
• Mixed Character Types: If the string can contain any characters (letters, digits, symbols), you can extend the approach by using a hashmap or a large frequency array to count characters. The concept of checking equal frequencies remains the same. Using a dynamic structure like a hashmap for frequencies might slow things down a bit, but the overall approach of using prefix sums (perhaps with an index mapping for characters) would still work.
• Exact k Frequency (Fixed count): A twist on this problem is if you were asked to find substrings where each unique character appears exactly k times (for a given k). This is actually a known variant (related to “equal count substrings”). In that case, you would look for substrings of length m * k where m distinct characters each appear k times. The approach could involve fixing k and using a sliding window of length that is a multiple of k to count how many valid substrings meet that criterion.
• Almost Equal Frequency: Another variation could allow a tolerance, for example substrings where each distinct digit's frequency is either x or x+1. This is a more complex variant, but it shows up in problems like reorganizing strings or making frequencies almost equal. Solving that would require careful counting logic beyond exact equality.
• Subsequence vs Substring: This problem specifically deals with contiguous substrings. If one were to ask for subsequences (not necessarily contiguous) with equal digit frequency, the problem becomes much harder, as combinatorial explosion of subsequences would need different techniques (usually not feasible to brute force for large strings).

In summary, the idea of ensuring equal frequency has broad applications. The approach of counting frequencies and using prefix sums or hashing can be adapted to many of these variations with appropriate adjustments.

Related Problems

• Remove All Occurrences of a Substring

• Find Longest Special Substring That Occurs Thrice I

• Maximum Number of Occurrences of a Substring