Suppose, we have a lowercase string s that contains letters and parentheses "(" and ")". We have to reverse every string enclosed within parentheses in a recursive manner and return the resultant string.
So, if the input is like s = "back(aps)ce", then the output will be “backspace”.
To solve this, we will follow these steps −
Define a function trav() . This will take s, dir, start, close:= close, ans:= ans
end := "(" if dir is same as −1, otherwise ")"
other := "(" if end is same as ")", otherwise ")"
while start < size of s, and s[start] is not same as end, do
if s[start] is same as other, then
trav(s, -dir, close[other, start] - dir)
start := close[other, start] + dir
otherwise,
insert s[start] at the end of ans
start := start + dir
From the main function, do the following −
ans := a new list
close := a new map containing keys “)” and “(” initially the values are two empty maps
stack := a new list
for each index I and value c in s, do
if c is same as "(", then
push i into stack
otherwise when c is same as ")", then
o := top of stack, then pop from stack
close[")", i] := o
close["(", o] := i
trav(s, 1, 0)
return ans joined with blank string
Let us see the following implementation to get better understanding −
Example
class Solution:
def solve(self, s):
ans = []
close = {")": {}, "(": {}}
stack = []
for i, c in enumerate(s):
if c == "(":
stack.append(i)
elif c == ")":
o = stack.pop()
close[")"][i] = o
close["("][o] = i
def trav(s, dir, start, close=close, ans=ans):
end = "(" if dir == -1 else ")"
other = "(" if end == ")" else ")"
while start < len(s) and s[start] != end:
if s[start] == other:
trav(s, −dir, close[other][start] − dir)
start = close[other][start] + dir
else:
ans.append(s[start])
start += dir
trav(s, 1, 0)
return "".join(ans)
ob = Solution()
print(ob.solve("back(aps)ce"))Input
"back(aps)ce"
Output
backspace