Given a string, print all possible palindromic partitions

Given a string, find all possible palindromic partitions of given string.


Note that this problem is different from Palindrome Partitioning Problem, there the task was to find the partitioning with minimum cuts in input string. Here we need to print all possible partitions.

The idea is to go through every substring starting from first character, check if it is palindrome. If yes, then add the substring to solution and recur for remaining part. Below is complete algorithm.

Below is C++ implementation of above idea


// C++ program to print all palindromic partitions of a given string
using namespace std;

// A utility function to check if str is palindroem
bool isPalindrome(string str, int low, int high)
	while (low < high)
		if (str[low] != str[high])
			return false;
	return true;

// Recursive function to find all palindromic partitions of str[start..n-1]
// allPart --> A vector of vector of strings. Every vector inside it stores
//			 a partition
// currPart --> A vector of strings to store current partition 
void allPalPartUtil(vector<vector<string> >&allPart, vector<string> &currPart, 
                   int start, int n, string str)
	// If 'start' has reached len
	if (start >= n)

	// Pick all possible ending points for substrings
	for (int i=start; i<n; i++)
		// If substring str[start..i] is palindrome
		if (isPalindrome(str, start, i))
			// Add the substring to result
			currPart.push_back(str.substr(start, i-start+1));

			// Recur for remaining remaining substring
			allPalPartUtil(allPart, currPart, i+1, n, str);
			// Remove substring str[start..i] from current 
			// partition

// Function to print all possible palindromic partitions of
// str. It mainly creates vectors and calls allPalPartUtil()
void allPalPartitions(string str)
	int n = str.length();

	// To Store all palindromic partitions
	vector<vector<string> > allPart;

	// To store current palindromic partition
	vector<string> currPart; 

	// Call recursive function to generate all partiions
	// and store in allPart
	allPalPartUtil(allPart, currPart, 0, n, str);

	// Print all partitions generated by above call
	for (int i=0; i< allPart.size(); i++ )
		for (int j=0; j<allPart[i].size(); j++)
			cout << allPart[i][j] << " ";
		cout << "\n";

// Driver program
int main()
	string str = "nitin";
	return 0;


# A O(n^2) time and O(1) space program to find the 
#longest palindromic substring

# This function prints the longest palindrome substring (LPS)
# of str[]. It also returns the length of the longest palindrome
def longestPalSubstr(string):
    maxLength = 1

    start = 0
    length = len(string)

    low = 0
    high = 0

    # One by one consider every character as center point of 
    # even and length palindromes
    for i in xrange(1, length):
        # Find the longest even length palindrome with center
	# points as i-1 and i.
        low = i - 1
        high = i
        while low >= 0 and high < length and string[low] == string[high]:
            if high - low + 1 > maxLength:
                start = low
                maxLength = high - low + 1
            low -= 1
            high += 1

        # Find the longest odd length palindrome with center 
        # point as i
        low = i - 1
        high = i + 1
        while low >= 0 and high < length and string[low] == string[high]:
            if high - low + 1 > maxLength:
                start = low
                maxLength = high - low + 1
            low -= 1
            high += 1

    print "Longest palindrome substring is:",
    print string[start:start + maxLength]

    return maxLength

# Driver program to test above functions
string = "forgeeksskeegfor"
print "Length is: " + str(longestPalSubstr(string))

# This code is contributed by BHAVYA JAIN


n i t i n
n iti n

This article is contributed by Ekta Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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