Wednesday, January 6, 2016

Find all distinct palindromic sub-strings of a given string - GeeksforGeeks


Find all distinct palindromic sub-strings of a given string - GeeksforGeeks
Given a string of lowercase ASCII characters, find all distinct continuous palindromic sub-strings of it.
Input: str = "abaaa"
Output:  Below are 5 palindrome sub-strings
a
aa
aaa
aba
b
Step 1: Finding all palindromes using modified Manacher’s algorithm:
Considering each character as a pivot, expand on both sides to find the length of both even and odd length palindromes centered at the pivot character under consideration and store the length in the 2 arrays (odd & even).
Time complexity for this step is O(n^2)
Step 2: Inserting all the found palindromes in a HashMap:
Insert all the palindromes found from the previous step into a HashMap. Also insert all the individual characters from the string into the HashMap (to generate distinct single letter palindromic sub-strings).
Time complexity of this step is O(n^3) assuming that the hash insert search takes O(1) time. Note that there can be at most O(n^2) palindrome sub-strings of a string. In below C++ code ordered hashmap is used where the time complexity of insert and search is O(Logn). In C++, ordered hashmap is implemented using Red Black Tree.
Step 3: Printing the distinct palindromes and number of such distinct palindromes:
void palindromeSubStrs(string s)
{
    map<string, int> m;
    int n = s.size();
    // table for storing results (2 rows for odd-
    // and even-length palindromes
    int R[2][n+1];
    // Find all sub-string palindromes from the given input
    // string insert 'guards' to iterate easily over s
    s = "@" + s + "#";
    for (int j = 0; j <= 1; j++)
    {
        int rp = 0;   // length of 'palindrome radius'
        R[j][0] = 0;
        int i = 1;
        while (i <= n)
        {
            //  Attempt to expand palindrome centered at i
            while (s[i - rp - 1] == s[i + j + rp])
                rp++;  // Incrementing the length of palindromic
                       // radius as and when we find vaid palindrome
            // Assigning the found palindromic length to odd/even
            // length array
            R[j][i] = rp;
            int k = 1;
            while ((R[j][i - k] != rp - k) && (k < rp))
            {
                R[j][i + k] = min(R[j][i - k],rp - k);
                k++;
            }
            rp = max(rp - k,0);
            i += k;
        }
    }
    // remove 'guards'
    s = s.substr(1, n);
    // Put all obtained palindromes in a hash map to
    // find only distinct palindromess
    m[string(1, s[0])]=1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= 1; j++)
            for (int rp = R[j][i]; rp > 0; rp--)
               m[s.substr(i - rp - 1, 2 * rp + j)]=1;
        m[string(1, s[i])]=1;
    }
    //printing all distinct palindromes from hash map
   cout << "Below are " << m.size()-1
        << " palindrome sub-strings";
   map<string, int>::iterator ii;
   for (ii = m.begin(); ii!=m.end(); ++ii)
      cout << (*ii).first << endl;
}
http://www.techiedelight.com/find-possible-palindromic-substrings-string/
We can solve this problem in O(n2) time and O(1) space. The idea is inspired from Longest Palindromic Substring problem. For each character in the given string, we consider it as mid point of a palindrome and expand in both directions to find all palindromes that have it as mid-point. For even length palindrome, we consider every adjacent pair of characters as mid point. We use a set to store all unique palindromic substrings.
void expand(string str, int low, int high, auto &set)
{
    // run till str[low.high] is a palindrome
    while (low >= 0 && high < str.length()
            && str[low] == str[high])
    {
        // push all palindromes into the set
        set.insert(str.substr(low, high - low + 1));
 
        // expand in both directions
        low--, high++;
    }
}
 
// Function to find all unique palindromic substrings of given string
void allPalindromicSubstrings(string str)
{
    // create an empty set to store all unique palindromic substrings
    unordered_set<string> set;
 
    for (int i = 0; i < str.length(); i++)
    {
        // find all odd length palindrome with str[i] as mid point
        expand(str, i, i, set);
        
        // find all even length palindrome with str[i] and str[i+1] as
        // its mid points
        expand(str, i, i + 1, set);
    }
    
    // print all unique palindromic substrings
    for (auto i : set)
        cout << i << " ";
}


http://siyang2notleetcode.blogspot.com/2015/02/finding-all-palindromes-in-string.html

https://discuss.leetcode.com/topic/90/count-palindromes
Count the number of possible palindrome substrings in a string. A palindrome is a word that reads the same way spelled backwards.
Example:
input: lasagna.
Possible palindromes are asa, l,a,s,a,g,n,a.
output: count is 8.
input:hellolle
ellolle,lloll,lol,ll,ll,h,e,l,l,o,l,l,e.
output:13
https://discuss.leetcode.com/topic/35753/print-all-the-palindromic-substrings-of-a-given-string

http://www.geeksforgeeks.org/count-palindrome-sub-strings-string/
Given a string, the task is to count all palindrome substring in a given string. Length of palindrome substring is greater then or equal to 2.

    static int CountPS(char str[], int n)
    {
        // creat empty 2-D matrix that counts all palindrome
        // substring. dp[i][j] stores counts of palindromic
        // substrings in st[i..j]
        int dp[][] = new int[n][n];
      
        // P[i][j] = true if substring str[i..j] is palindrome,
        // else false
        boolean P[][] = new boolean[n][n];
      
        // palindrome of single lenght
        for (int i= 0; i< n; i++)
            P[i][i] = true;
      
        // palindrome of length 2
        for (int i=0; i<n-1; i++)
        {
            if (str[i] == str[i+1])
            {
                P[i][i+1] = true;
                dp[i][i+1] = 1 ;
            }
        }
      
        // Palindromes of length more then 2. This loop is similar
        // to Matrix Chain Multiplication. We start with a gap of
        // length 2 and fill DP table in a way that gap between
        // starting and ending indexes increases one by one by
        // outer loop.
        for (int gap=2 ; gap<n; gap++)
        {
            // Pick starting point for current gap
            for (int i=0; i<n-gap; i++)
            {
                // Set ending point
                int j = gap + i;
      
                // If current string is palindrome
                if (str[i] == str[j] && P[i+1][j-1] )
                    P[i][j] = true;
      
                // Add current palindrome substring ( + 1)
                // and rest palinrome substring (dp[i][j-1] + dp[i+1][j])
                // remove common palinrome substrings (- dp[i+1][j-1])
                if (P[i][j] == true)
                    dp[i][j] = dp[i][j-1] + dp[i+1][j] + 1 - dp[i+1][j-1];
                else
                    dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
            }
        }
      
        // return total palindromic substrings
        return dp[0][n-1];
    }
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