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:
- Recursive solution: other_solution.cpp.
- 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(2n-1 * n).
Worst case are strings like "aaaaaaaaaaaaaaaaaaaa", every substring there is a palindrome.
Auxiliary Space Used
O(2n-1 * n).
Answer array stores 2n-1 palindromic decompositions (in the worst case anyway) of length O(n).
Also is_palindrome array is O(n2).
O(2n-1 * n) + O(n2) = O(2n-1 * n).
Space Complexity
O(2n-1 * n).
Auxiliary space used is O(2n-1 * n) and input size is O(n).
O(2n-1 * n) + O(n) = O(2n-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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