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Head of Career Skills Development & Coaching

*Based on past data of successful IK students

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.

```
{
"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"]
```

- 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.

This is a dynamic programming solution.

O(2^{n-1} * n).

Worst case are strings like `"aaaaaaaaaaaaaaaaaaaa"`

, every substring there is a palindrome.

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(n^{2}).

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

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).

```
/*
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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

Note: Input and Output will already be taken care of.

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.

```
{
"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"]
```

- 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.

This is a dynamic programming solution.

O(2^{n-1} * n).

Worst case are strings like `"aaaaaaaaaaaaaaaaaaaa"`

, every substring there is a palindrome.

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(n^{2}).

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

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).

```
/*
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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

**Attend our free webinar to amp up your career and get the salary you deserve.**

Hosted By

Ryan Valles

Founder, Interview Kickstart