Minimum No-Repeat Segments After One Character Removal

Given a lowercase English string s, choose exactly one lowercase English character and remove every occurrence of that character from s.
Partition the remaining string into the minimum number of non-empty, disjoint, contiguous segments such that no segment contains a repeated character. Return that minimum number of segments.
If removing the chosen character makes the string empty, return 0.

Function
getNoRepeatSegments(s: String) → int

Examples
Example 1
s = "abcccde"
return = 1
Remove every 'c'. The remaining string is "abde", whose characters are all distinct, so one segment is sufficient.
Example 2
s = "abdaa"
return = 1
Remove every 'a'. The remaining string is "bd", so the minimum is one segment.
Example 3
s = "aaaa"
return = 0
Remove every 'a'. The remaining string is empty, so there are no non-empty segments.

Constraints
1 <= s.length <= 2 * 10^5
s contains only lowercase English letters.

The trick is that there are only 26 candidate characters. For each one, build the filtered string in O(n), then count segments greedily. Walk left to right, keep a set (or a 26-slot boolean array) of letters in the current segment. When the next character is already in the set, start a new segment, clear the set, and add that character. Greedy works because extending a segment as far as possible never hurts. Take the minimum across all 26 choices. Total cost is about 26 * n, which is fine for 2 * 10^5. Pitfalls: forgetting the empty-string case that returns 0, and clearing the set in O(26) too often. Also, only try letters that exist in s, or at least handle absent letters cleanly, since removing nothing is a valid-looking choice but may not be allowed. If you freeze on the greedy reset logic mid-assessment, StealthCoder can hand you the clean loop.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.