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Is there a better approach to solve this problem? (recursion, backtracking , possibly dynamic progra

Time:09-25

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 in aa?)
  • 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 char c at position pos 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 tried a, or already tried both a and b);

  • solve has a boolean return value which is true if a solution was found (pos == s.length()), or false 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.

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