2825. Make String a Subsequence Using Cyclic Increments

Description

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.

Solutions

Solution 1: Two Pointers

This problem actually requires us to determine whether a string $s$ is a subsequence of another string $t$. However, the characters do not have to match exactly. If two characters are the same, or one character is the next character of the other, they can match.

The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of the strings $str1$ and $str2$ respectively. The space complexity is $O(1)$.

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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)

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class Solution {
    public boolean canMakeSubsequence(String str1, String str2) {
        int i = 0, n = str2.length();
        for (char c : str1.toCharArray()) {
            char d = c == 'z' ? 'a' : (char) (c + 1);
            if (i < n && (str2.charAt(i) == c || str2.charAt(i) == d)) {
                ++i;
            }
        }
        return i == n;
    }
}

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class Solution {
public:
    bool canMakeSubsequence(string str1, string str2) {
        int i = 0, n = str2.size();
        for (char c : str1) {
            char d = c == 'z' ? 'a' : c + 1;
            if (i < n && (str2[i] == c || str2[i] == d)) {
                ++i;
            }
        }
        return i == n;
    }
};

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func canMakeSubsequence(str1 string, str2 string) bool {
	i, n := 0, len(str2)
	for _, c := range str1 {
		d := byte('a')
		if c != 'z' {
			d = byte(c + 1)
		}
		if i < n && (str2[i] == byte(c) || str2[i] == d) {
			i++
		}
	}
	return i == n
}

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function canMakeSubsequence(str1: string, str2: string): boolean {
    let i = 0;
    const n = str2.length;
    for (const c of str1) {
        const d = c === 'z' ? 'a' : String.fromCharCode(c.charCodeAt(0) + 1);
        if (i < n && (str2[i] === c || str2[i] === d)) {
            ++i;
        }
    }
    return i === n;
}