
Problem Statement
In this problem, two 0-indexed strings str1
and str2
are provided. The objective is to determine if str2
can be made a subsequence of str1
by performing a specific operation at most once on str1
. This operation involves selecting a set of indices in str1
and incrementing each character at these indices to its next character cyclically, where after 'z'
, the cycle continues back to 'a'
. For instance, if a character at an index is 'a'
, it will become 'b'
and so forth. A subsequence is defined as a derivative string formed without rearranging the order but potentially removing some characters from the original string. The task is to return true
if the transformation allows str2
to be a subsequence of str1
; otherwise, return false
.
Examples
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
andstr2
consist of only lowercase English letters.
Approach and Intuition
The challenge essentially involves checking whether we can transform str1
into a form where str2
can be derived as its subsequence with at most one pass of transformations on specified indices. We can break down the approach as follows:
Cyclic Character Transformation:
- For each character in
str1
, assess the potential character it could transform into if the character cycles once. This involves a simple mapping where each letter moves to the next, and 'z' returns to 'a'.
- For each character in
Subsequence Check:
- Using the transformed versions of
str1
from the cyclic operation, verify ifstr2
can be a subsequence. This subsequence check should ideally move sequentially through both strings, trying to match characters ofstr2
with those ofstr1
.
- Using the transformed versions of
Edge Cases:
- Only a single operation is allowed; hence all necessary index changes for transformation should be kept within this limit.
- Consider cases where
str2
is inherently already a subsequence without any operations. This requires a direct comparison before the expensive operation. - Address scenarios where transformations can lead to potential sequences forming in parts of the strings that do not require modification or only small sections do.
From the given examples:
- In Example 3,
str2
is "d" which cannot be formed from "ab" since neither 'a' nor 'b' can become 'd' with a single cyclic increment, resulting in "false". - Examples 1 and 2 demonstrate scenarios where strategic increments at selected indices successfully allow
str2
to be a subsequence post-transformation.
In essence, the solution needs to evaluate the potential transformation outcomes against the subsequence condition efficiently, adhering to the provided constraints.
Solutions
- C++
- Java
- Python
class Solution {
public:
bool validateSubsequence(string s1, string s2) {
int indexS2 = 0;
int len1 = s1.length(), len2 = s2.length();
for (int indexS1 = 0; indexS1 < len1 && indexS2 < len2; ++indexS1) {
if (s1[indexS1] == s2[indexS2] ||
(s1[indexS1] + 1 == s2[indexS2]) ||
(s1[indexS1] - 25 == s2[indexS2])) {
indexS2++;
}
}
return indexS2 == len2;
}
};
The provided C++ solution focuses on verifying if string s2
can be transformed into a subsequence of string s1
by using cyclic increments. To achieve this, the function validateSubsequence
loops through each character of s1
to check if there is a match with the current character of s2
under three conditions:
- Direct character match between
s1[indexS1]
ands2[indexS2]
. - The character in
s1
is one less than that ins2
, allowing for an increment to match. - The character in
s1
is the alphabetically last (i.e., 'z'), and the character ins2
is the alphabetically first (i.e., 'a'), treating the relationship cyclically with a +1 increment.
The incrementing is deemed cyclic because on reaching 'z', it loops back to 'a'. If any of these conditions are satisfied, the pointer for s2
moves to the next character (incremented by 1).
The function accelerates the process by immediately moving forward in the search as soon as a suitable condition is met, ensuring efficient mapping through possible cyclic increments. It returns true
if it's possible to resequence s2
into a subsequence of s1
following these incremental rules, indicated by traversing all characters in s2
(indexS2 == len2
). Otherwise, it returns false indicating that s2
cannot be transformed into a valid subsequence of s1
under these conditions.
class Solution {
public boolean isSubsequencePossible(String s1, String s2) {
int indexS2 = 0;
int lenS1 = s1.length(), lenS2 = s2.length();
for (
int indexS1 = 0;
indexS1 < lenS1 && indexS2 < lenS2;
++indexS1
) {
if (
s1.charAt(indexS1) == s2.charAt(indexS2) ||
(s1.charAt(indexS1) + 1 == s2.charAt(indexS2)) ||
(s1.charAt(indexS1) - 25 == s2.charAt(indexS2))
) {
indexS2++;
}
}
return indexS2 == lenS2;
}
}
The Java function isSubsequencePossible(String s1, String s2)
checks whether the string s2
can be made a subsequence of s1
via cyclic increments. A cyclic increment involves modifying a character by either incrementing it by one or treating 'z' as a cyclic increment to 'a'.
- First, initialize an index for
s2
,indexS2
, starting at zero. - Use the lengths of
s1
ands2
, denoted aslenS1
andlenS2
, for determining loop termination. - Iterate through each character of
s1
. For each character ofs1
:- If the current character of
s1
equals the corresponding character ins2
or represents a cyclic increment (either directly by one or cycling through from 'z' to 'a'), increment thes2
indexindexS2
by one.
- If the current character of
- After the loop, determine if
s2
is a subsequence by checking ifindexS2
equalslenS2
.
The method returns true
if s2
can successfully be a subsequence of s1
under these constraints, false
otherwise.
class Solution:
def isSubsequencePossible(self, sequence1: str, sequence2: str) -> bool:
index_seq2 = 0
len_seq1, len_seq2 = len(sequence1), len(sequence2)
# Iterate over characters of sequence1
for char_index in range(len_seq1):
if index_seq2 < len_seq2 and (
sequence1[char_index] == sequence2[index_seq2]
or ord(sequence1[char_index]) + 1 == ord(sequence2[index_seq2])
or ord(sequence1[char_index]) - 25 == ord(sequence2[index_seq2])
):
# Increment index for sequence2 upon finding a matching condition
index_seq2 += 1
# If index_seq2 reached length of sequence2, all characters have been matched
return index_seq2 == len_seq2
The provided code defines a Python class, Solution
, which encompasses a method isSubsequencePossible
to determine if one string (sequence1
) can be transformed into a subsequence of another string (sequence2
) through specific character modifications. The method employs cyclic increment logic, allowing character transformations by incrementing characters by 1 or wrapping from 'z' to 'a', to match characters in sequence2
.
Implement the following steps to understand the provided functionality:
- Initialize an index (
index_seq2
) to keep track of the position insequence2
. - Calculate the lengths of both input strings
sequence1
andsequence2
. - Loop through each character in
sequence1
. Within this loop, check if the current character ofsequence1
matches the corresponding character ofsequence2
or can be transformed to do so by either incrementing by 1 or by cyclically incrementing 'z' to 'a'. - If a match or valid transformation is found, advance the
index_seq2
to check against the next character insequence2
. - After the loop, determine if all characters of
sequence2
have been matched by comparing ifindex_seq2
is equal to the length ofsequence2
.
The method returns True
if sequence2
has been fully matched (i.e., it can be turned into a subsequence of sequence1
using the described transformations) and False
otherwise. This solution supports checking sequences directly and transformation via specific cyclic character increments.
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