Problem

Given two strings s1 and s2, return s1 with every character that appears anywhere in s2 removed. The relative order of the remaining characters is preserved.

Input / Output

• Input: strings s1, s2.
• Output: filtered string.

Examples

Example 1 Input: s1 = "computer", s2 = "cat" Output: "ompuer" Explanation: c, a and t are removed.

Example 2 Input: s1 = "hello", s2 = "" Output: "hello"

Example 3 Input: s1 = "aaaa", s2 = "a" Output: ""

Constraints

• 0 <= |s1|, |s2| <= 10^5

Approach

The naive method checks each character of s1 against the whole of s2: O(|s1| · |s2|). Instead:

1. Put the characters of s2 into a set (or a boolean array of size 128/256 for ASCII) — O(|s2|).
2. Scan s1 once and keep only characters not in the set — O(|s1|).

The set turns membership into O(1), giving linear total time. If the alphabet is lowercase letters, a 26-entry boolean array is the fastest and most memory-frugal version; a bitmask in a single integer (mask |= 1 << (c - 'a')) is the trick to mention.

Complexity

• Time: O(|s1| + |s2|)
• Space: O(|Σ|) — the alphabet size.

Reference solution (Python)

def remove_chars(s1: str, s2: str) -> str:
    banned = set(s2)
    return ''.join(c for c in s1 if c not in banned)

Test cases

Input	Expected
"computer", "cat"	"ompuer"
"hello", ""	"hello"
"aaaa", "a"	""
"abc", "xyz"	"abc"