Palindromic Decomposition Of A String Problem Statement

Find all palindromic decompositions of a given string s.

A palindromic decomposition of string is a decomposition of the string into substrings, such that all those substrings are valid palindromes.

Example

{
"s": "abracadabra"
}

Output:

["a|b|r|a|c|ada|b|r|a", "a|b|r|aca|d|a|b|r|a", "a|b|r|a|c|a|d|a|b|r|a"]

Notes

• Any string is its own substring.
• Output should include ALL possible palindromic decompositions of the given string.
• Order of decompositions in the output does not matter.
• To separate substrings in the decomposed string, use | as a separator.
• Order of characters in a decomposition must remain the same as in the given string. For example, for s = "ab", return ["a|b"] and not ["b|a"].
• Strings in the output must not contain whitespace. For example, ["a |b"] or ["a| b"] is incorrect.

Constraints:

• 1 <= length of s <= 20
• s only contains lowercase English letters.

We have provided two solutions:

1. Recursive solution: other_solution.cpp.
2. Dynamic programming solution: Palindromic Decomposition Of A String Solution: Optimal.

Try to solve the problem using both approaches.

In the dynamic programming solution we have pre-calculated is_palindrome array. In the recursive solution we have not done that to make code more readable. Do that there too.

optimal_solution.cpp

This is a dynamic programming solution.

Time Complexity

O(2^n-1 * n).

Worst case are strings like "aaaaaaaaaaaaaaaaaaaa", every substring there is a palindrome.

Auxiliary Space Used

O(2^n-1 * n).

Answer array stores 2^n-1 palindromic decompositions (in the worst case anyway) of length O(n).

Also is_palindrome array is O(n²).

O(2^n-1 * n) + O(n²) = O(2^n-1 * n).

Space Complexity

O(2^n-1 * n).

Auxiliary space used is O(2^n-1 * n) and input size is O(n).

O(2^n-1 * n) + O(n) = O(2^n-1 * n).

Code For Palindromic Decomposition Of A String Solution: Optimal

/*
Asymptotic complexity in terms of the length of `s` `n`:
* Time: O(2^(n-1) * n).
* Auxiliary space: O(2^(n-1) * n).
* Total space: O(2^(n-1) * n).
*/

// Find is_palindrome[start][stop] for all possible substrings in O(n^2) time.
void find_is_palindrome_for_all_substrings(vector<vector<bool>> &is_palindrome, int n, string &s)
{
    for (int len = 1; len <= n; len++)
    {
        for (int start = 0; start < n; start++)
        {
            int stop = start + len - 1;
            // We want to find if s[start, stop] is a palindrome or not.
            if (stop >= n)
            {
                break;
            }
            is_palindrome[start][stop] = false;
            if (len <= 2)
            {
                is_palindrome[start][stop] = (s[start] == s[stop]);
            }
            else if (s[start] == s[stop])
            {
                /*
                When first and last characters are same then whether string is a palindrome or
                not, depends on the inner string.
                For example:
                1) In "abcba", 'a' = 'a' so string is a palindrome or not will depend on "bcb".
                1) In "abcca", 'a' = 'a' so string is a palindrome or not will depend on "bcc".
                */
                is_palindrome[start][stop] = is_palindrome[start + 1][stop - 1];
            }
        }
    }
}

vector<string> generate_palindromic_decompositions(string &s)
{
    int n = s.length();
    /*
    For all substrings of s, we will pre-calculate if it is a palindrome or not.
    is_palindrome[i][j] (where i <= j and 0 <= i, j < n) will contain whether s[i, j] is a
    palindrome or not.
    */
    vector<vector<bool>> is_palindrome (n, vector<bool> (n, false));
    find_is_palindrome_for_all_substrings(is_palindrome, n, s);
    /*
    decompositions_container[i] will contain all possible palindromic decompositions of s[0, i].
    So our answer will be decompositions_container[0, n - 1].
    We will build our solution like:
    decompositions_container[0, 0] -> decompositions_container[0, 1] ->  ... ->
    decompositions_container[0, n - 1].
    Why we are storing these values? -> To avoid recalculation.
    */
    vector<vector<string>> decompositions_container(n, vector<string>(0));
    /*
    Loop to find decompositions_container[i] (i.e. all possible palindromic decompositions of
    s[0, i].)
    */
    for (int i = 0; i < n; i++)
    {
        // If s[0, i] is a palindrome then add it.
        if (is_palindrome[0][i])
        {
            decompositions_container[i].push_back(s.substr(0, i + 1));
        }
        /*
        Loop to find other palindromic decompositions of s[0, i] using already calculated
        palindromic decompositions of s[0, 0], ..., s[0, n - 1].
        */
        for (int j = 0; j < i; j++)
        {
            if (is_palindrome[j + 1][i])
            {
                /*
                If s[j + 1][i] is a palindromic substring, then we can join palindromic
                decompositions (that we have already found) of s[0, j] with it to form the
                palindromic decomposition of s[0, i].
                */
                string cur_sub_str = '|' + s.substr(j + 1, i - j);
                int len = decompositions_container[j].size();
                for (int k = 0; k < len; k++)
                {
                    /*
                    Here we are directly using previously calculated values, but in recursion we
                    recalculate them. So that is why this solution is faster than recursive
                    solution.
                    */
                    decompositions_container[i].push_back(decompositions_container[j][k] +
                        cur_sub_str);
                }
            }
        }
    }
    return decompositions_container[n - 1];
}

We hope that these solutions to palindrome partitioning problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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