easyStringsPattern: Two Pointers
Verify Nickname Subsequence Solution
Problem Statement
Given two strings, fullName and nickname, determine if nickname is a valid subsequence of fullName. A nickname is considered a valid subsequence if it can be derived from fullName by deleting zero or more characters without changing the relative order of the remaining characters. The comparison must be case-insensitive.
Examples
Example 1:
Input:fullName = "Benjamin", nickname = "Bnj"
Output:true
Explanation: The characters 'B', 'n', and 'j' appear in "Benjamin" in the same relative order (B-e-n-j-a-m-i-n) when ignoring case.
Example 2:
Input:fullName = "Sophia", nickname = "Osa"
Output:false
Explanation: Although 'o', 's', and 'a' exist in "Sophia", their relative order is not preserved ('s' comes before 'o' in "Sophia", whereas 'O' comes before 's' in "Osa").
Constraints
- 1 <= fullName.length <= 10^4
- 1 <= nickname.length <= 10^4
- fullName and nickname consist only of uppercase and lowercase English letters.
Time: O(n + m) Space: O(1)
The optimal approach is to use two pointers, one for the fullName string and one for the nickname string, to compare characters in order. This approach allows us to find a subsequence in linear time.
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