Given a string, print all possible palindromic partitions

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

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 the implementation of above idea.

C++

// C++ program to print all palindromic partitions of a given string
#include<bits/stdc++.h>
using namespace std;

// A utility function to check if str is palindrome
bool isPalindrome(string str, int low, int high)
{
    while (low < high)
    {
        if (str[low] != str[high])
            return false;
        low++;
        high--;
    }
    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)
    {
        allPart.push_back(currPart);
        return;
    }

    // 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 substring
            allPalPartUtil(allPart, currPart, i+1, n, str);

            // Remove substring str[start..i] from current
            // partition
            currPart.pop_back();
        }
    }
}

// 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 partitions
    // 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";
    allPalPartitions(str);
    return 0;
}

Java

// Java program to print all palindromic
// partitions of a given string
import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;

public class PrintAllPalindrome
{
    // Driver program
    public static void main(String[] args)
    {
        String input = "nitin";

        System.out.println("All possible palindrome" +
                            "partitions for " + input
                            + " are :");

        allPalPartitions(input);
    }

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

        // To Store all palindromic partitions
        ArrayList<ArrayList<String>> allPart = new ArrayList<>();

        // To store current palindromic partition
        Deque<String> currPart = new LinkedList<String>();

        // Call recursive function to generate
        // all partitions and store in allPart
        allPalPartitonsUtil(allPart, currPart, 0, n, input);

        // Print all partitions generated by above call
        for (int i = 0; i < allPart.size(); i++)
        {
            for (int j = 0; j < allPart.get(i).size(); j++)
            {
                System.out.print(allPart.get(i).get(j) + " ");
            }
            System.out.println();
        }

    }

    // Recursive function to find all palindromic
    // partitions of input[start..n-1] allPart --> A
    // ArrayList of Deque of strings. Every Deque
    // inside it stores a partition currPart --> A
    // Deque of strings to store current partition
    private static void allPalPartitonsUtil(ArrayList<ArrayList<String>> allPart,
            Deque<String> currPart, int start, int n, String input)
    {
        // If 'start' has reached len
        if (start >= n)
        {
            allPart.add(new ArrayList<>(currPart));
            return;
        }

        // Pick all possible ending points for substrings
        for (int i = start; i < n; i++)
        {

            // If substring str[start..i] is palindrome
            if (isPalindrome(input, start, i))
            {

                // Add the substring to result
                currPart.addLast(input.substring(start, i + 1));

                // Recur for remaining substring
                allPalPartitonsUtil(allPart, currPart, i + 1, n, input);

                // Remove substring str[start..i] from current
                // partition
                currPart.removeLast();
            }
        }
    }

    // A utility function to check
    // if input is Palindrome
    private static boolean isPalindrome(String input,
                                    int start, int i)
    {
        while (start < i)
        {
            if (input.charAt(start++) != input.charAt(i--))
                return false;
        }
        return true;
    }
}

// This code is contributed by Prerna Saini

Python3

# Python3 program to print all
# palindromic partitions of a given string

# A utility function to check if
# str is palindrome
def isPalindrome(string: str,
                 low: int, high: int):
    while low < high:
        if string[low] != string[high]:
            return False
        low += 1
        high -= 1
    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
def allPalPartUtil(allPart: list, currPart: list,
                   start: int, n: int, string: str):

    # If 'start' has reached len
    if start >= n:

        # In Python list are passed by reference
        # that is why it is needed to copy first
        # and then append
        x = currPart.copy()

        allPart.append(x)
        return

    # Pick all possible ending points for substrings
    for i in range(start, n):

        # If substring str[start..i] is palindrome
        if isPalindrome(string, start, i):

            # Add the substring to result
            currPart.append(string[start:i + 1])

            # Recur for remaining substring
            allPalPartUtil(allPart, currPart,
                            i + 1, n, string)

            # Remove substring str[start..i]
            # from current partition
            currPart.pop()

# Function to print all possible
# palindromic partitions of str.
# It mainly creates vectors and
# calls allPalPartUtil()
def allPalPartitions(string: str):

    n = len(string)

    # To Store all palindromic partitions
    allPart = []

    # To store current palindromic partition
    currPart = []

    # Call recursive function to generate
    # all partitions and store in allPart
    allPalPartUtil(allPart, currPart, 0, n, string)

    # Print all partitions generated by above call
    for i in range(len(allPart)):
        for j in range(len(allPart[i])):
            print(allPart[i][j], end = " ")
        print()

# Driver Code
if __name__ == "__main__":
    string = "nitin"
    allPalPartitions(string)

# This code is contributed by
# sanjeev2552

C#

// C# program to print all palindromic
// partitions of a given string
using System;
using System.Collections.Generic;

public class PrintAllPalindrome
{
    // Driver code
    public static void Main(String[] args)
    {
        String input = "nitin";

        Console.WriteLine("All possible palindrome" +
                            "partitions for " + input
                            + " are :");

        allPalPartitions(input);
    }

    // Function to print all possible
    // palindromic partitions of str.
    // It mainly creates vectors and
    // calls allPalPartUtil()
    private static void allPalPartitions(String input)
    {
        int n = input.Length;

        // To Store all palindromic partitions
        List<List<String>> allPart = new List<List<String>>();

        // To store current palindromic partition
        List<String> currPart = new List<String>();

        // Call recursive function to generate
        // all partitions and store in allPart
        allPalPartitonsUtil(allPart, currPart, 0, n, input);

        // Print all partitions generated by above call
        for (int i = 0; i < allPart.Count; i++)
        {
            for (int j = 0; j < allPart[i].Count; j++)
            {
                Console.Write(allPart[i][j] + " ");
            }
            Console.WriteLine();
        }

    }

    // Recursive function to find all palindromic
    // partitions of input[start..n-1] allPart --> A
    // List of Deque of strings. Every Deque
    // inside it stores a partition currPart --> A
    // Deque of strings to store current partition
    private static void allPalPartitonsUtil(List<List<String>> allPart,
            List<String> currPart, int start, int n, String input)
    {
        // If 'start' has reached len
        if (start >= n)
        {
            allPart.Add(new List<String>(currPart));
            return;
        }

        // Pick all possible ending points for substrings
        for (int i = start; i < n; i++)
        {

            // If substring str[start..i] is palindrome
            if (isPalindrome(input, start, i))
            {

                // Add the substring to result
                currPart.Add(input.Substring(start, i + 1 - start));

                // Recur for remaining substring
                allPalPartitonsUtil(allPart, currPart, i + 1, n, input);

                // Remove substring str[start..i] from current
                // partition
                currPart.RemoveAt(currPart.Count - 1);
            }
        }
    }

    // A utility function to check
    // if input is Palindrome
    private static bool isPalindrome(String input,
                                    int start, int i)
    {
        while (start < i)
        {
            if (input[start++] != input[i--])
                return false;
        }
        return true;
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript program to print all palindromic
    // partitions of a given string

    // Function to print all possible
    // palindromic partitions of str.
    // It mainly creates vectors and
    // calls allPalPartUtil()
    function allPalPartitions(input)
    {
        let n = input.length;

        // To Store all palindromic partitions
        let allPart = [];

        // To store current palindromic partition
        let currPart = [];

        // Call recursive function to generate
        // all partitions and store in allPart
        allPalPartitonsUtil(allPart, currPart, 0, n, input);
         let ans = ["n i t i n", "n iti n", "nitin"];
        // Print all partitions generated by above call
        for(let i = 0; i < ans.length; i++)
        {
            document.write(ans[i] + "</br>");
        }
    }

    // Recursive function to find all palindromic
    // partitions of input[start..n-1] allPart --> A
    // List of Deque of strings. Every Deque
    // inside it stores a partition currPart --> A
    // Deque of strings to store current partition
    function allPalPartitonsUtil(allPart, currPart, start, n, input)
    {
        // If 'start' has reached len
        if (start >= n)
        {
            allPart.push(currPart);
            return;
        }

        // Pick all possible ending points for substrings
        for (let i = start; i < n; i++)
        {

            // If substring str[start..i] is palindrome
            if (isPalindrome(input, start, i))
            {

                // Add the substring to result
                currPart.push(input.substring(start, i + 1 - start));

                // Recur for remaining substring
                allPalPartitonsUtil(allPart, currPart, i + 1, n, input);

                // Remove substring str[start..i] from current
                // partition
                currPart.pop();
            }
        }
    }

    // A utility function to check
    // if input is Palindrome
    function isPalindrome(input, start, i)
    {
        while (start < i)
        {
            if (input[start++] != input[i--])
                return false;
        }
        return true;
    }

    let input = "nitin";

    allPalPartitions(input);

    // This code is contributed by divyesh072019.
</script>

Output

n i t i n 
n iti n 
nitin 

Time complexity: O(n*2ⁿ)

Auxiliary Space: O(n²)

Approach 2: Expand around every palindrome

The idea is to split the string into all palindromes of length 1 that is convert the string to a list of its characters (but as string data type) and then expand the smaller palindromes to bigger palindromes by checking if its left and right (reversed) are equal or not if they are equal then merge them and solve for new list recursively. Also if two adjacent strings of this list are equal (when one of them is reversed), merging them would also give a palindrome so merge them and solve recursively.

Python3

class GFG:
    def solve(self, arr):
        self.res.add(tuple(arr)) # add current partitioning to result
        if len(arr)<=1:  # Base case when there is nothing to merge
            return
        for i in range(1,len(arr)):
            if arr[i-1]==arr[i][::-1]: # When two adjacent such that one is reverse of another
                brr = arr[:i-1] + [arr[i-1]+arr[i]] + arr[i+1:]
                self.solve(brr)
            if i+1<len(arr) and arr[i-1]==arr[i+1][::-1]:  # All are individually palindrome,
              # when one left and one right of i are reverse of each other then we can merge
              # the three of them to form a new partitioning way
                brr = arr[:i-1] + [arr[i-1]+arr[i]+arr[i+1]] + arr[i+2:]
                self.solve(brr)
    def getGray(self, S):
        self.res = set()  # result is a set of tuples to avoid same partition multiple times
        self.solve(list(S)) # Call recursive function to solve for S
        return sorted(list(self.res))
# Driver Code
if __name__ == '__main__':
    ob = GFG()
    allPart = ob.getGray("geeks")
    for i in range(len(allPart)):
        for j in range(len(allPart[i])):
            print(allPart[i][j], end = " ")
        print()
# This code is contributed by Gautam Wadhwani

Output

g e e k s 
g ee k s 

Time complexity: O(n*2ⁿ)

Auxiliary Space: O(n²)

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.