Given a string S of length N, return a string that is the result of replacing each '?'
in the string S with an 'a'
or a 'b'
character and does not contain three identical consecutive letters (in other words, neither 'aaa'
not 'bbb'
may occur in the processed string).
Examples:
Given S="a?bb", output= "aabb"
Given S="??abb", output= "ababb" or "bbabb" or "baabb"
Given S="a?b?aa", output= "aabbaa"
1<=n<= 500000
I solved the problem using backtracking, but my solution is slow and won't work for greater N values, is there a better approach?
My solution:
#include <bits/stdc .h>
using namespace std;
bool isValid(string &s, int pos)
{
if (pos < 2)
return true;
for (int i = 1; i < pos; i )
{
if (s[i - 1] == s[i] && s[i] == s[i 1])
{
return false;
}
}
return true;
}
void solve(string &s, int pos, string &ans)
{
if (pos == s.length())
{
ans = s;
return;
}
if (!isValid(s, pos))
return;
if (s[pos] == '?')
{
s[pos] = 'a';
if (isValid(s, pos))
solve(s, pos 1, ans);
s[pos] = 'b';
if (isValid(s, pos))
{
solve(s, pos 1, ans);
}
s[pos] = '?';
}
else
solve(s, pos 1, ans);
}
int main()
{
string s;
cin >> s;
string ans;
solve(s, 0, ans);
cout << ans;
return 0;
}
CodePudding user response:
first of all x??y
with more than 1? would always pass
a??b => abab
(produce no increase of length on both side, *no impact* at following)a??a => abba
(no impact)a???????????????????b => ab?????????????????ab
so you only need to consider the case with single ?
aa? => aab
(no other possibility)a?a => aba
(no impact)
so the only problem comes to a?b
aa?b => aabb
(as describe inaa?
)ba?b => baab
(no impact)^a?b
=>^aab
(only for start, no impact)
You did at most 2 look-back and 2 look-ahead so this is a O(n)
solution,
If it can contain invalid input, simply run another O(n)
check
CodePudding user response:
The O(n)
dynamic program has four states: a character is either a_first
, a_second
, b_first
or b_second
. Let dp[i]
represent the possibility of creating a valid string up to index i
. Then:
dp[i][a_first] = True
dp[i][a_second] = True
dp[i][b_first] = True
dp[i][b_second] = True
where i < 0
if s[i] == '?':
dp[i][a_first] = dp[i-1][b_first] or dp[i-1][b_second]
dp[i][a_second] = dp[i-1][a_first]
dp[i][b_first] = dp[i-1][a_first] or dp[i-1][a_second]
dp[i][b_second] = dp[i-1][b_first]
else if s[i] == 'a':
dp[i][a_first] = dp[i-1][b_first] or dp[i-1][b_second]
dp[i][a_second] = dp[i-1][a_first]
dp[i][b_first] = False
dp[i][b_second] = False
else:
dp[i][a_first] = False
dp[i][a_second] = False
dp[i][b_first] = dp[i-1][a_first] or dp[i-1][a_second]
dp[i][b_second] = dp[i-1][b_first]
where i ≥ 0
To get the strings, we can follow the paths backwards that maintain True throughout and keep the consecutive character constraint.
Python code:
def f(s):
dp = [[None] * 4 for _ in range(len(s))]
def get_dp(i, j):
return True if i < 0 else dp[i][j]
for i in range(len(s)):
if s[i] == '?':
dp[i][0] = get_dp(i-1, 2) or get_dp(i-1, 3)
dp[i][1] = get_dp(i-1, 0)
dp[i][2] = get_dp(i-1, 0) or get_dp(i-1, 1)
dp[i][3] = get_dp(i-1, 2)
elif s[i] == 'a':
dp[i][0] = get_dp(i-1, 2) or get_dp(i-1, 3)
dp[i][1] = get_dp(i-1, 0)
dp[i][2] = False
dp[i][3] = False
else:
dp[i][0] = False
dp[i][1] = False
dp[i][2] = get_dp(i-1, 0) or get_dp(i-1, 1)
dp[i][3] = get_dp(i-1, 2)
# Build the output
result = []
i = len(s) - 1
need = None
while i >= 0:
if dp[i][0] and need != 'b':
result.append('a')
need = None if (len(result) == 1 or result[-2] == 'b') else 'b'
i-= 1
elif dp[i][1] and need != 'b':
result.extend(['a', 'a'])
need = 'b'
i -= 2
elif dp[i][2] and need != 'a':
result.append('b')
need = None if (len(result) == 1 or result[-2] == 'a') else 'a'
i -= 1
elif dp[i][3] and need != 'a':
result.extend(['b', 'b'])
need = 'a'
i -= 2
else:
return ""
return "".join(reversed(result))
Output:
strs = [
"a?bb", # "aabb"
"??abb", # "ababb" or "bbabb" or "baabb"
"a?b?aa", # "aabbaa"
"?bb?",
"aa?bb", # NO SOLUTION
"aa??aa", # "aabbaa"
"ab???bb?",
"????",
"??",
"?????",
"??????"
]
for s in strs:
print("%s\n%s\n" % (s, f(s)))
"""
a?bb
aabb
??abb
baabb
a?b?aa
aabbaa
?bb?
abba
aa?bb
aa??aa
aabbaa
ab???bb?
abbaabba
????
abaa
??
aa
?????
aabaa
??????
baabaa
"""
CodePudding user response:
An iterative backtracking solution.
isValid
only checks whether inserting a charc
at positionpos
would create an issue, without iterating over the whole string;maintain a variable
int dir = -1
to know whether we're moving forward or backwards in the string (this is only important so that we know in which direction to move when skipping over a non-?
character in the original string);forest of
if/else if/else
to decide where we're at (non-?
to skip, or fresh?
going forward, or already trieda
, or already tried botha
andb
);solve
has a boolean return value which istrue
if a solution was found (pos == s.length()
), orfalse
if no solution was found (pos == (unsigned int) -1
).
#include <iostream>
#include <vector>
bool isValid(char c, std::string const &s, unsigned int pos)
{
return ((pos < 2 || s[pos - 2] != c || s[pos - 1] != c)
&& (pos < 1 || pos 1 >= s.length() || s[pos - 1] != c || s[pos 1] != c)
&& (pos 2 >= s.length() || s[pos 1] != c || s[pos 2] != c));
}
bool solve(std::string const &in, std::string &out)
{
unsigned int pos = 0;
int dir = 1;
out = in;
while (pos < in.size())
{
if (in[pos] != '?') // nothing to do: keep moving (forward or backward)
{
pos = dir;
}
else if (out[pos] == '?') // going forward, will try 'a' then 'b'
{
dir = 1;
if (isValid('a', out, pos))
{
out[pos] = 'a';
pos = dir;
}
else if (isValid('b', out, pos))
{
out[pos] = 'b';
pos = dir;
}
else
{
dir = -1;
pos = dir;
}
}
else if (out[pos] == 'a' && isValid('b', out, pos)) // going backward, already tried 'a'
{
out[pos] = 'b';
dir = 1;
pos = dir;
}
else // already tried everything: backtrack
{
dir = -1;
out[pos] = '?';
pos = dir;
}
}
return (pos == in.size());
}
int main(void)
{
std::vector<std::string> ins = {"a?bb", "??abb", "a?b?aa", "aa?bb"};
std::vector<std::string> outs = {"", "", "", "", ""};
for (unsigned int i = 0; i < ins.size(); i)
{
if (solve(ins[i], outs[i]))
std::cout << ins[i] << " --> " << outs[i] << std::endl;
else
std::cout << ins[i] << " --> NO SOLUTION" << std::endl;
}
return 0;
}
Output:
a?bb --> aabb
??abb --> ababb
a?b?aa --> aabbaa
aa?bb --> NO SOLUTION
CodePudding user response:
You can try the algorithm below. Let's start changing sequences of ?
into a
or b
from left to right.
For a single ?
a?a -> aba
a?b -> try aab, then (in case of aa?b) try abb
b?a -> try bba, then (in case of bb?a) try baa
b?b -> bab
For a long (more then one ?
) sequence start adding alternating a
and b
from both ends:
a?????????b -> ab???????ab -> aba?????bab -> abab???abab -> ababa?babab -> ababaababab
We can have 4 possible pattern all being valid
a?????????a -> ababa .... ababa, so we'll have either abababa or ababbaba
a?????????b -> ababa .... babab, so we'll have either abababab or ababbabab
b?????????a -> babab .... ababa, so we'll have either babababa or babaababa
b?????????b -> babab .... babab, so we'll have either bababab or babaabab
CodePudding user response:
I quickly tried out the solution I proposed in the comments:
#include <iostream>
#include <string>
int main(void)
{
std::string s = "?????aaa??bbaa????";
std::cout << s <<'\n';
for(char &c:s)
if(c=='?')
c='a';
std::cout << s <<'\n';
for(int i =0; i<s.size()-2; i )
{
if(s[i] == 'a'
&& s[i 1] == 'a'
&& s[i 2] == 'a' )
{
s[i 1] = 'b';
i ;
}
}
std::cout << s <<'\n';
return 0;
}
It seems to be conformant with the given rules. but maybe there's some case to check and fix.