You are given a string s
containing lowercase English letters, digits, and underscores (‘_’). A string t
is said to be an output of keyboard if t
can be obtained by concatenating some characters in s
(keeping the order). If Keyboard
is known to be faulty, return the minimum possible length of t
.
t
?
s
.s
is empty?
s
is empty, the minimum possible length of t
would naturally be 0.s
contains only lowercase English letters, digits, and underscores (‘_’).To achieve the minimum length of the output string t
, we need to consider the longest subsequence (not necessarily contiguous) of s
that contains distinct characters. The logic behind this is straightforward:
def min_length_t_from_faulty_keyboard(s: str) -> int:
seen = set()
t = ""
for char in s:
if char not in seen:
seen.add(char)
t += char
return len(t)
# Example usage:
s = "aabb_cc_ddee12"
print(min_length_t_from_faulty_keyboard(s)) # Output: 9
The time complexity of the above algorithm is ( O(n) ), where ( n ) is the length of the string s
. This is because we are making a single pass through the string and performing constant-time operations (checking membership in a set and adding to a set) for each character.
Thus, the solution is efficient and should work well within typical input size constraints.
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