Given two strings,searchWord and resultWord:
- Determine the minimum number of characters you need to append to resultWord so that it becomes a subsequence of searchWord.
Example:
Input:
searchWord = "abcz"
resultWord = "azdb"
Output: 2
Explanation: By appending 'd' and 'b' to "azdb", we can transform it into a subsequence of "abcz".
Understanding the Problem:
A subsequence of a string is derived by deleting some or no characters from it without changing the order of the remaining characters. For example:
"ace" is a subsequence of "abcde" because 'a', 'c', and 'e' appear in order.
"aec" is NOT a subsequence of "abcde" because 'e' appears before 'c'.
In this problem:
Input: Two strings, searchWord and resultWord.
Output: The number of characters to append to resultWord so that all characters in resultWord appear in order as a subsequence in searchWord.
Approach to Solve the Problem
This problem can be solved efficiently using the two-pointer technique. Below is a step-by-step explanation of the approach:
Two Pointers:
- Use one pointer (
i
) to traversesearchWord
. - Use another pointer (
j
) to traverseresultWord
.
- Use one pointer (
Matching Characters:
- If the character at
searchWord[i]
matchesresultWord[j]
, move both pointers forward (i.e., progress to the next character in both strings). - If they don’t match, move only the
i
pointer to search for a match in the next character ofsearchWord
.
- If the character at
Stopping Condition:
- The traversal stops when either:
- All characters of
searchWord
have been processed. - All characters of
resultWord
have been matched.
- All characters of
- The traversal stops when either:
Calculate Unmatched Characters:
- After traversal, the pointer
j
indicates how many characters ofresultWord
were matched insearchWord
. - The unmatched characters are given by:
Time and Space Complexity
Time Complexity:
- The algorithm processes both strings in a single traversal.
- Complexity: O(n + m), where
n
is the length ofsearchWord
andm
is the length ofresultWord
.
Space Complexity:
- No extra space is used apart from variables.
- Complexity: O(1).
Try to write the code by yourself if you feel anywhere stuck in this problem then comment it down👇
Code Link: https://shorturl.at/lsFmP
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