Given a string s, partition s such that every string of the partition is a palindrome. Return all possible palindrome partitioning of s.
Example :
Input : s = "bcc"
Output : [["b", "c", "c"], ["b", "cc"]]
Input : s = "geeks"
Output : [["g", "e", "e", "k", "s"],
["g", "ee", "k", "s"]]
We have to list the all possible partitions so we will think in the direction of recursion. When we are on index i, we incrementally check all substrings starting from i for being palindromic. If found, we recursively solve the problem for the remaining string and add this in our solution.
Following is the solution-
- We will maintain a 2-dimensional vector for storing all possible partitions and a temporary vector for storing the current partition, new starting index of string to check partitions as we have already checked partitions before this index.
- Now keep on iterating further on string and check if it is palindrome or not.
- If it is a palindrome than add this string in current partitions vector. Recurse on this new string if it is not the end of the string. After coming back again change the current partition vector to the old one as it might have changed in the recursive step.
- If we reach the end of string while iterating than we have our partitions in our temporary vector so we will add it in results.
To check whether it’s a palindrome or not, iterate on string by taking two pointers. Initialize the first to start and other to end of string. If both characters are same increase the first and decrease the last pointer and keep on iterating until first is less than last one.
C++
// C++ program to print all palindromic partitions // of a given string. #include <bits/stdc++.h> using namespace std; // Returns true if str is palindrome, else false bool checkPalindrome(string str) { int len = str.length(); len--; for (int i=0; i<len; i++) { if (str[i] != str[len]) return false; len--; } return true; } void printSolution(vector<vector<string> > partitions) { for (int i = 0; i < partitions.size(); ++i) { for(int j = 0; j < partitions[i].size(); ++j) cout << partitions[i][j] << " "; cout << endl; } return; } // Goes through all indexes and recursively add remaining // partitions if current string is palindrome. void addStrings(vector<vector<string> > &v, string &s, vector<string> &temp, int index) { int len = s.length(); string str; vector<string> current = temp; if (index == 0) temp.clear(); for (int i = index; i < len; ++i) { str = str + s[i]; if (checkPalindrome(str)) { temp.push_back(str); if (i+1 < len) addStrings(v,s,temp,i+1); else v.push_back(temp); temp = current; } } return; } // Generates all palindromic partitions of 's' and // stores the result in 'v'. void partition(string s, vector<vector<string> >&v) { vector<string> temp; addStrings(v, s, temp, 0); printSolution(v); return; } // Driver code int main() { string s = "geeks"; vector<vector<string> > partitions; partition(s, partitions); return 0; } |
Java
// Java program to print all palindromic partitions // of a given string. import java.util.ArrayList; public class GFG { // Returns true if str is palindrome, else false static boolean checkPalindrome(String str) { int len = str.length(); len--; for (int i=0; i<len; i++) { if (str.charAt(i) != str.charAt(len)) return false; len--; } return true; } // Prints the partition list static void printSolution(ArrayList<ArrayList<String>> partitions) { for(ArrayList<String> i: partitions) { for(String j: i) { System.out.print(j+" "); } System.out.println(); } } // Goes through all indexes and recursively add remaining // partitions if current string is palindrome. static ArrayList<ArrayList<String>> addStrings(ArrayList< ArrayList<String>> v, String s, ArrayList<String> temp, int index) { int len = s.length(); String str = ""; ArrayList<String> current = new ArrayList<>(temp); if (index == 0) temp.clear(); // Iterate over the string for (int i = index; i < len; ++i) { str = str + s.charAt(i); // check whether the substring is // palindromic or not if (checkPalindrome(str)) { // if palindrome add it to temp list temp.add(str); if (i + 1 < len) { // recurr to get all the palindromic // partitions for the substrings v = addStrings(v,s,temp,i+1); } else { // if end of the string is reached // add temp to v v.add(temp); } // temp is reinitialize with the // current i. temp = new ArrayList<>(current); } } return v; } // Generates all palindromic partitions of 's' and // stores the result in 'v'. static void partition(String s, ArrayList<ArrayList< String>> v) { // temporary ArrayList to store each // palindromic string ArrayList<String> temp = new ArrayList<>(); // calling addString method it adds all // the palindromic partitions to v v = addStrings(v, s, temp, 0); // printing the solution printSolution(v); } // Driver code public static void main(String args[]) { String s = "geeks"; ArrayList<ArrayList<String>> partitions = new ArrayList<>(); partition(s, partitions); } } // This code is contributed by Sumit Ghosh |
Python3
# Python3 program to print all palindromic # partitions of a given string. def checkPalindrome(string): # Returns true if str is palindrome, # else false length = len(string) length -= 1 for i in range(length): if string[i] != string[length]: return False length -= 1 return True def printSolution(partitions): for i in range(len(partitions)): for j in range(len(partitions[i])): print(partitions[i][j], end = " ") print() def addStrings(v, s, temp, index): # Goes through all indexes and # recursively add remaining partitions # if current string is palindrome. length = len(s) string = "" current = temp[:] if index == 0: temp = [] for i in range(index, length): string += s[i] if checkPalindrome(string): temp.append(string) if i + 1 < length: addStrings(v, s, temp[:], i + 1) else: v.append(temp) temp = current def partition(s, v): # Generates all palindromic partitions # of 's' and stores the result in 'v'. temp = [] addStrings(v, s, temp[:], 0) printSolution(v) # Driver Code if __name__ == "__main__": s = "geeks" partitions = [] partition(s, partitions) # This code is contributed by # vibhu4agarwal |
Output :
g e e k s g ee k s
Related Article:
Dynamic Programming | Set 17 (Palindrome Partitioning)
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