LeetCode 2825: Make String a Subsequence Using Cyclic Increments

Problem Description

LeetCode Problem 2825

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. That is ‘a’ becomes ‘b’, ‘b’ becomes ‘c’, and so on, and ‘z’ becomes ‘a’.

Return true if it is possible to make str2 a subsequence of str1 by performing the operation at most once, and false otherwise.

Note: A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

Example 1:

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.

Example 2:

Input: str1 = “zc”, str2 = “ad”
Output: true
Explanation: Select indices 0 and 1 in str1.
Increment str1[0] to become ‘a’.
Increment str1[1] to become ‘d’.
Hence, str1 becomes “ad” and str2 is now a subsequence. Therefore, true is returned.

Example 3:

Input: str1 = “ab”, str2 = “d”
Output: false
Explanation: In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once.
Therefore, false is returned.

Constraints:

• 1 <= str1.length <= 105
• 1 <= str2.length <= 105
• str1 and str2 consist of only lowercase English letters.

Difficulty: Medium

Tags: two pointers, string

Rating: 93.21%

Solution

Here’s my Python solution to this problem:

class Solution:
    def canMakeSubsequence(self, str1: str, str2: str) -> bool:
        j = 0

        for i in range(len(str1)):
            if j < len(str2):
                if str1[i] == str2[j] or chr((ord(str1[i])-97+1)%26+97) == str2[j]:
                    j += 1
        
        return j == len(str2)

Complexity Analysis

The solution has the following complexity characteristics:

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$

Note: This is an automated analysis and may not capture all edge cases or specific algorithmic optimizations.