What is the time complexity of Make String a Subsequence Using Cyclic Increments?

The optimal solution for Make String a Subsequence Using Cyclic Increments runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse each string once, making it linear in relation to the length of str1.

What is the brute force solution for Make String a Subsequence Using Cyclic Increments?

The brute force approach for Make String a Subsequence Using Cyclic Increments is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of characters in str1 that can be incremented to match str2. This method is straightforward but inefficient, as it may involve many redundant checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Make String a Subsequence Using Cyclic Increments?

Make String a Subsequence Using Cyclic Increments uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Matching.

What are the interview tips for Make String a Subsequence Using Cyclic Increments?

Always clarify the problem requirements before diving into coding. Think about edge cases, such as when str2 is longer than str1. Practice explaining your thought process as you code.

What are common mistakes when solving Make String a Subsequence Using Cyclic Increments?

Not considering the cyclic nature of character increments. Failing to correctly manage the indices while checking matches.

#2825

Make String a Subsequence Using Cyclic Increments

Medium

Two PointersStringTwo PointersString Matching

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a two-pointer technique to efficiently check if str2 can be formed as a subsequence of str1 after at most one increment operation. This reduces unnecessary checks and improves performance.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one for str1 and one for str2.
2Step 2: Iterate through both strings, comparing characters at the current pointers.
3Step 3: If characters match, move both pointers forward. If they don't, check if str1's character can be incremented to match str2's character.
4Step 4: If all characters in str2 are matched by the end of the iteration, return true. Otherwise, return false.

solution.py11 lines

1# Full working Python code
2
3def canMakeSubsequence(str1, str2):
4    i, j = 0, 0
5    while i < len(str1) and j < len(str2):
6        if str1[i] == str2[j]:
7            j += 1
8        elif (ord(str1[i]) + 1 - ord('a')) % 26 == (ord(str2[j]) - ord('a')):
9            j += 1
10        i += 1
11    return j == len(str2)

ℹ

Complexity note: This complexity is achieved because we only traverse each string once, making it linear in relation to the length of str1.

1Understanding cyclic increments is crucial to solving this problem.
2Two pointers can efficiently manage the matching process between the two strings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.