Leetcode #2825: Make String a Subsequence Using Cyclic Increments
In this guide, we solve Leetcode #2825 Make String a Subsequence Using Cyclic Increments in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed strings str1 and str2. In an operation, you select a set of indices in str1, and for each index i in the set, increment str1[i] to the next character cyclically.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Two Pointers, String
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: str1 = "abc", str2 = "ad"
Output: true
Explanation: Select index 2 in str1.
Increment str1[2] to become 'd'.
Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned.
Python Solution
class Solution:
def canMakeSubsequence(self, str1: str, str2: str) -> bool:
i = 0
for c in str1:
d = "a" if c == "z" else chr(ord(c) + 1)
if i < len(str2) and str2[i] in (c, d):
i += 1
return i == len(str2)
Complexity
The time complexity is , where and are the lengths of the strings and respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.