花花酱 LeetCode 2825. Make String a Subsequence Using Cyclic Increments

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.

Solution: Two pointers

s1[i] and s2[j] can match if
s1[i] == s2[j] or inc(s1[i]) == s2[j]

If matched: ++i; ++j else ++i.

Time complexity: O(n)
Space complexity: O(1)

Iterator version