How to split a string into as few palindromes as possible?

Question

This is an interview question: \"You\'re given a string, and you want to split it into as few strings as possible such that each string is a palindrome\". (I guess a one char st

Answer 1

First find all the palindromes in the string such that L[i][j] represents the length of j-th longest palindrome that ends at S[i]. Lets say S is the input string. This could be done in O(N^2) time by first considering length1 palindromes then then length 2 palindromes and so on. Finding Length i palindromes after you know all length i-2 palindromes is the matter of a single character comparison.

This is a dynamic programming problem after that. Let A[i] represent the smallest number of palindrome that Substring(S,0,i-1) can be decomposed into.

A[i+1] = min_{0 <= j < length(L[i])} A[i - L[i][j]] + 1;

Edit based on Micron's request: Here is the idea behind comuting L[i][j]. I just wrote this up to convey the idea, the code may have problems.

// Every single char is palindrome so L[i][0] = 1;
vector<vector<int> > L(S.length(), vector<int>(1,1));

for (i = 0; i < S.length(); i++) {
 for (j = 2; j < S.length; j++) {
   if (i - j + 1 >= 0 && S[i] == S[i-j + 1]) {
     // See if there was a palindrome of length j - 2 ending at S[i-1]
     bool inner_palindrome = false;
     if (j ==2) {
      inner_palindrome = true;
     } else {
       int k = L[i-1].length;
       if (L[i-1][k-1] == j-2 || (k >= 2 && L[i-1][k-2] == j-2)) {
         inner_palindrome = true;
       }
     }
     if (inner_palindrome) {
       L[i].push_back(j);
     }
   } 
 }
} 

Answer 2

You can do this in O(n^2) time using Rabin-Karp fingerprinting to preprocess the string to find all of the palindromes in O(n^2) time. After the preprocessing, you run code similar to the following:

np(string s) {
  int a[s.size() + 1];
  a[s.size()] = 0;
  for (int i = s.size() - 1; i >= 0; i--) {
    a[i] = s.size() - i;
    for (int j = i + 1; j <= s.size(); j++) {
      if (is_palindrome(substr(s, i, j))) // test costs O(1) after preprocessing
        a[i] = min(a[i], 1 + a[j]);
  }
  return a[0];
}

Answer 3

bool ispalindrome(string inp)
{
    if(inp == "" || inp.length() == 1)
    {
        return true;
    }
    string rev = inp;

    reverse(rev.begin(), rev.end());

    return (rev == inp);
}

int minsplit_count(string inp)
{
    if(ispalindrome(inp))
    {
        return 0;
    }

    int count= inp.length();

    for(int i = 1; i < inp.length(); i++)
    {
        count = min(count, 
                      minsplit_count(inp.substr(0, i))              + 
                      minsplit_count(inp.substr(i, inp.size() - i)) + 
                      1);
    }

    return count;
}

Answer 4

An equivalent problem is that of computing the Snip number of a string.

Suppose you wanted to cut a string using the fewest number of snips, so that each remaining piece was itself a palindrome. The number of such cuts we will call the Snip Number of a string. That is the snip number is always equal to one less than the smallest number of palindromes within a given string. Every string of length n has snip number at most n-1, and each palindrome has snip number 0. Here is working python code.

def snip_number(str):
    n=len(str)
 
 #initialize Opt Table
 # Opt[i,j] = min number of snips in the substring str[i...j]
 
    Opt=[[0 for i in range(n)] for j in range(n) ]
 
 #Opt of single char is 0
    for i in range(n):
     Opt[i][i] = 0
 
 #Opt for adjacent chars is 1 if different, 0 otherwise
    for i in range(n-1):
     Opt[i][i+1]= 1 if str[i]!=str[i+1] else 0
 
 
# we now define sil as (s)substring (i)interval (l) length of the
# interval [i,j] --- sil=(j-i +1) and j = i+sil-1
 
# we compute Opt table entry for each sil length and
# starting index i
 
    for sil in range(3, n+1):
     for i in range(n-sil+1):
       j = i+sil-1
       if (str[i] == str[j] and Opt[i+1][j-1]==0):
         Opt[i][j] = 0
       else:
         snip= min( [(Opt[i][t]+ Opt[t+1][j] + 1 ) for t in range(i,j-1)])
         Opt[i][j] = snip

    return Opt[0][len(str)-1]
#end function snip_number()
mystr=[""for i in range(4)]         
mystr[0]="abc"
mystr[1]="ohiho"
mystr[2]="cabacdbabdc"
mystr[3]="amanaplanacanalpanama aibohphobia "


for i in range(4):
     print mystr[i], "has snip number:", snip_number(mystr[i])
     
# abc has snip number: 2
# ohiho has snip number: 0
# cabacdbabdc has snip number: 2
# amanaplanacanalpanama aibohphobia  has snip number: 1
                                                                    

                                                        
            
            

              
                
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