easyStringsPattern: Slow/Fast Pointers
Shift Hash Characters Solution
Problem Statement
You are given a string s. Rearrange the characters of s such that all hash characters ('#') are moved to the end of the string, while the relative order of all other characters remains unchanged. Return the modified string.
Examples
Example 1:
Input:s = "a#b#c"
Output:"abc##"
Explanation: The non-hash characters 'a', 'b', and 'c' maintain their relative order, while the two '#' characters are shifted to the end.
Example 2:
Input:s = "###code"
Output:"code###"
Explanation: All three '#' characters at the beginning are shifted to the end of the string, keeping 'code' intact at the front.
Constraints
- 1 <= s.length <= 10^5
- s consists only of lowercase English letters and '#' characters.
Time: O(N) Space: O(N)
Convert the string to a mutable array and use a two-pointer read/write approach. The write pointer tracks the placement of non-hash characters, which are copied forward sequentially, and the rest of the array is backfilled with hashes in a single pass.
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