2825. Make String a Subsequence Using Cyclic Increments - Detailed Explanation

Explanation: The dfs(i, j) recursively explores from position i in str1 and position j in str2. We use lru_cache to memoize results of states (i, j) to avoid redundant calculations. We try skipping characters and matching characters (if possible) and stop when we either find a full match for str2 or exhaust all possibilities. The example tests will print True, True, False for the three sample cases respectively.

Python Implementation – Optimized (Two Pointers)

This Python solution uses the two-pointer greedy method described above. It runs in linear time relative to the input lengths.

This code uses a simple loop to scan through str1 and advances j whenever a character matches (considering the increment rule).

Complexity Analysis:

• Brute Force Approach: The brute force (backtracking) approach in the worst case examines an exponential number of possibilities (each character can be taken or not, leading to roughly O(2^n) combinations in the worst scenario). With memoization, the state space is at most O(n * m) (n = length of str1, m = length of str2), but that can still be extremely large. Time complexity in the worst case is exponential, and even with pruning it can degrade towards O(n*m) in a dense exploration scenario. Space complexity is also high: the recursion stack can go up to O(n) deep, and the memo table can hold up to O(n*m) states in the worst case. This is impractical for large n, m (up to 10^5) and is why this approach is mainly for conceptual understanding or very small inputs.

• Optimized Two-Pointer Approach: This approach runs in linear time relative to the size of the input. We make a single pass through str1 (with pointer i) and at most one pass through str2 (pointer j only moves forward). Each character of str1 and str2 is processed at most once, so the time complexity is about O(n + m), which in big-O terms is O(n) if n is the length of str1 (assuming m <= n, otherwise O(n+m) to be precise). In the worst case, if str1 length = 100k and str2 length = 100k, this will do on the order of 200k comparisons, which is efficient. The space complexity is O(1) extra space, since we only use a few integer counters and a couple of char variables for comparisons. (We are not using any data structures that grow with input size, aside from the input strings themselves.)

Comparison: Clearly, the optimized approach is far superior for large strings, turning a potentially exponential problem into a linear scan by leveraging the greedy strategy. The brute force approach would only be feasible for very short strings due to its complexity.

Step-by-Step Walkthrough (Optimized Solution):

Let's do a detailed walkthrough of the optimized two-pointer solution using an example to see how it works step by step. We’ll use Example 1: str1 = "abc", str2 = "ad" for illustration:

• Initial State:
  str1 = "abc", str2 = "ad".
  Pointers: i = 0 (at str1[0] = 'a'), j = 0 (at str2[0] = 'a').
  We haven’t matched anything yet.

• Step 1: Compare str1[0] with str2[0].

  • str1[i] = 'a', str2[j] = 'a'.
  • They are equal, which means we can match the first character of str2. We will use 'a' as-is (no increment needed).
  • Move j to the next position (j = 1 now, pointing to str2[1] = 'd'). This signifies that the first character of str2 ("a") is matched and we are now looking to match the second character ("d").
  • Move i to the next position in str1 (i = 1, pointing to str1[1] = 'b').

  (So far, we have matched "a" from str2. Current matched subsequence in str1 is "a_ _" where underscore represents what we’ll match next.)

• Step 2: Compare str1[1] with str2[1].

  • str1[i] = 'b', str2[j] = 'd'.
  • They are not equal. Check if incrementing 'b' helps: incrementing 'b' gives 'c' (since 'b' -> 'c'). 'c' is still not equal to 'd'.
  • This means str1[1] (which is 'b') cannot be used to match the current needed character ('d'). So we decide to skip str1[1].
  • We do not change j (still need to match 'd'), but we increment i to move on (i = 2, now pointing to str1[2] = 'c').

  (We skipped 'b'. It was not useful for matching 'd'. The subsequence match progress is still just "a_ _" — only the first character matched so far.)

• Step 3: Compare str1[2] with str2[1].

  • str1[i] = 'c', str2[j] = 'd'.
  • Directly, 'c' != 'd'. Check increment: incrementing 'c' gives 'd'. Yes, 'd' matches the needed character!
  • This means we can use str1[2], provided we increment it in our operation, to match the character 'd' from str2. We take this match.
  • Move j to the next position (j = 2). Now j equals str2.length (which is 2), meaning we've matched all characters of str2.
  • We can stop here, but for completeness, we'd also move i to 3 (i = 3), which is beyond the length of str1.

  (Now we have matched "a" and "d" in sequence. The chosen indices in str1 were 0 (used as 'a') and 2 (used as 'c' incremented to 'd'). The subsequence "ad" is formed successfully.)

• End: Since j reached the end of str2, the algorithm will conclude with j == str2.length and return true. This indicates "ad" can be formed as a subsequence of "abc" with at most one cyclic increment operation (in fact, exactly one increment at index 2).

Visual summary of the operation on str1: We identified that incrementing the character at index 2 was needed. If we actually perform the single allowed operation by choosing index 2, str1 transforms as:

Original str1:  a   b   c
Operation:              ^  (increment 'c' to 'd')
Transformed str1: a   b   d
Subsequence:     ^       ^  (we take 'a' and 'd')

"ad" is clearly a subsequence of the transformed "abd". The two-pointer algorithm found this transformation without explicitly constructing all options.

This step-by-step process can be applied to any input. Essentially, the algorithm is trying to align each character of str2 with a character in str1 or an increment of a character in str1, in order. If at any point a str1 character cannot serve the needed match, it is skipped. If we manage to assign matches to all characters of str2, the answer is true.

Common Mistakes & Edge Cases:

When solving this problem, watch out for these common pitfalls and consider edge cases:

• Mixing up Subsequence vs. Substring: Remember that subsequence characters do not have to be contiguous in str1; they only need to appear in order. A mistake would be to try to match str2 as a contiguous block or to reorder characters. We must preserve the original order of characters from str1. Always use the two-pointer technique or equivalent logic to respect order.

• Forgetting the Wrap-around from 'z' to 'a': The cyclic nature means 'z' is effectively one step before 'a'. For example, if str2 needs an 'a' and you have a 'z' in str1, that 'z' can be incremented to 'a'. Many errors can happen by neglecting this special case. In implementation, be careful to handle 'z' properly. In our checks, we explicitly consider if current_char == 'z' and needed_char == 'a' as a valid match condition.

• Using Multiple Increments on the Same Character: You are only allowed to perform the increment operation once globally (although it can cover multiple characters in that one operation). This means each character in str1 can be incremented at most one time. A common misunderstanding would be thinking you could increment a character twice to reach a target (you cannot do 'a' -> 'c' in one operation; that would require two operations on the same character). Our solution accounts for this by only ever considering a one-step increment. If you find yourself thinking of changing a character by more than one alphabetical step, that path is invalid under the problem's constraints.

• Not Considering the Zero-Operation Scenario: It's possible that str2 is already a subsequence of str1 without any changes. The problem says "at most once", which includes the possibility of zero operations. Make sure your logic doesn’t falsely require at least one increment. If str2 is already a subsequence of str1 (all characters found in order naturally), you should return true immediately. The greedy approach naturally handles this (it would just match all characters without ever using the increment condition), but a mistake can occur if one tries a different method and forgets the no-op case.

• Edge Case: Length Mismatch: If str2 is longer than str1, then it's impossible to form str2 as a subsequence (since even without any operations you can't get more characters than str1 has). Our algorithm implicitly handles this: j can never reach m if m > n. It’s good to check upfront: if str2.length > str1.length, return false immediately as an optimization. Conversely, if str2 is empty (which in our constraints it isn't, but hypothetically), that’s trivially a subsequence of any string (return true).

• Edge Case: All characters require increment: Consider a case like str1 = "zzz" and str2 = "aaa". Here each 'z' needs to wrap to 'a'. The algorithm will handle it by checking each 'z' against 'a' and seeing the increment possibility. If the lengths match, this should return true (increment every character of str1). Make sure your implementation correctly allows 'z' -> 'a'.

• Not advancing the str2 pointer correctly: Some might forget to only advance j when a match is found. Advancing i and j unconditionally or incorrectly will break the logic. Ensure that j only moves forward when a character is matched (considering increment). If you advance j when there's no match, you'll incorrectly drop a needed character from str2. The provided two-pointer logic handles this by only doing j++ in the matching case.

By keeping these points in mind, you can avoid common mistakes. Test edge cases like very short strings, cases where no increment is needed, or cases where only the wrap-around letter can help.

Alternative Variations:

This problem is a twist on a classic subsequence check by adding a constraint of a single cyclic increment operation. There are a few ways this problem (or related problems) could be varied:

• Unlimited or Multiple Operations: If we were allowed to increment characters multiple times (or perform many operations), the problem changes drastically. In fact, if any number of increments on any characters were allowed, any string str1 could be transformed character-by-character into any other string of equal length, so the main limiting factor would only be the length and order. In such a scenario, as long as str2 is not longer than str1, you could always make str2 a subsequence by transforming the appropriate characters in order. The interesting part of this given problem is that each character can only shift by at most one position in the alphabet, which restricts the transformations.

• Single-Character Increment Limit: Another variation could be if you were allowed to increment at most one character in str1 (instead of any subset of characters in one operation). That would mean you could choose one index in str1 to increment by one, and you want to see if str2 is a subsequence after that single-character change. Solving that variation might involve checking each position of str1 as a candidate for increment and seeing if it yields a subsequence match — a potentially more expensive approach (but still manageable by trying at most n possibilities and doing a subsequence check each time).

• Different Operations: Instead of cyclic increments, imagine variations where you could replace a character with any other character (one substitution), or you could delete or insert characters (which moves into edit distance territory). Each of these would create a different type of problem:

  • One substitution to achieve a subsequence might be similar in approach (check if str2 is subsequence allowing one mismatched character to still count via substitution).

  • Allowing deletions from str1 doesn't change anything (deletions only make subsequence easier to form).

  • Insertions into str1 effectively is the reverse of deletions (similar to checking if str2 is a subsequence of str1 because insertions in str1 can always be done to match str2 if needed, limited by number of insertions).

• Subsequence with Shifts: The cyclic increment operation is basically a shift by +1. We could imagine a variation where instead of +1 shifts, you could shift characters by +k positions (mod 26) in one operation. If k is large or if multiple increments are allowed, it could open up more possibilities. The solution approach would need adjustment to check a range of possible matches instead of just the next character.

• Counting or Minimizing Operations: A different problem could ask not just whether it's possible, but how many characters need to be incremented (or what is the minimum number of increments needed) to make str2 a subsequence of str1. That becomes a different challenge – likely a dynamic programming problem – where you try to minimize the modifications. Our current problem restricts it to at most one operation globally, which simplifies the decision to a yes/no.

Many of these variations would require more complex algorithms (often dynamic programming or more involved greedy strategies). However, the fundamental idea of scanning through one string to find a subsequence in another is a recurring theme.

Related Problems for Further Practice:

To strengthen understanding, it’s helpful to practice other problems that involve subsequences or two-pointer techniques:

• 392. Is Subsequence (Easy) – This is the classic subsequence checking problem. You have two strings s and t, and you need to check if s is a subsequence of t. It can be solved with a similar two-pointer approach (without any character transformations). Solving this will solidify the basic idea used in our optimized solution.

• 524. Longest Word in Dictionary through Deleting (Medium) – In this problem, you are given a string and a dictionary of words, and you need to find the longest word in the dictionary that is a subsequence of the given string. This also relies on subsequence checking (often repeatedly checking multiple candidates). It’s a good exercise in efficient subsequence verification.

• 2486. Append Characters to String to Make Subsequence (Medium) – This problem asks for the minimum number of characters that need to be appended to the end of a string s so that another string t becomes a subsequence of s. It uses the subsequence two-pointer idea to see how much of t can be matched and how many characters are left unmatched . It’s a direct application of subsequence logic.

• 792. Number of Matching Subsequences (Medium) – Here you are given one string and a list of many words, and you need to count how many of those words are subsequences of the string. A naive approach would check each word with a two-pointer scan (which could be costly if there are many words), so it requires thinking about optimizing multiple subsequence checks, possibly by pre-processing the positions of characters in the main string. This problem extends the subsequence concept to multiple queries.

• (Advanced) Edit Distance problems (e.g., LeetCode 72) – Though more advanced, problems that involve inserting, deleting, or replacing characters to transform one string into another (like Edit Distance) are related in the sense of string transformations. They use dynamic programming to count minimum operations, which is different from our greedy check, but practicing them can give deeper insight into how allowing edits changes what subsequences/substrings can form.