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104 LeetCode algorithm problem solutions
# #Medium #String #Dynamic_Programming #Big_O_Time_O(n^2)_Space_O(n)
# #2024_08_04_Time_326_ms_(100.00%)_Space_71.3_MB_(100.00%)
defmodule Solution do
@spec count_substrings(s :: String.t()) :: integer
def count_substrings(s) do
tuple_s =
s
|> String.graphemes()
|> List.to_tuple()
get_count(tuple_s, 0, 0)
end
def get_count(tuple_s, count, i) when i >= tuple_size(tuple_s) do
count
end
def get_count(tuple_s, count, i) do
left = right = i
count = count + while(tuple_s, 0, left, right)
left = i
right = i + 1
count = count + while(tuple_s, 0, left, right)
get_count(tuple_s, count, i + 1)
end
def while(tuple_s, acc, left, right)
when left < 0 or
right >= tuple_size(tuple_s) or
elem(tuple_s, left) != elem(tuple_s, right) do
acc
end
def while(tuple_s, acc, left, right) do
while(tuple_s, acc + 1, left - 1, right + 1)
end
end
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