Find all possible Palindromic Partitions of the given String in C++

October 14, 2022

All Possible Palindromic Partitions of the given String in C++

Today in this article we will discuss the program to Find all possible Palindromic partitions of the given String in C++ Programming Language. We will discuss both approach and code in this page.

All Possible Palindromic Partitions of the given String in C++

Algorithm:

  • We 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.
Palindromic Partitions of given string in C++

Code in C++

Run 
#include<bits/stdc++.h>
using namespace std;

bool isPalindrome(string str, int low, int high)
{
    while (low < high)
    {
        if (str[low] != str[high])
            return false;
        low++;
        high--;
    }
    return true;
}

void allPalPartUtil(vector<vector<string>>&allPart, vector<string> &currPart, int start, int n, string str)
{
    if (start >= n)
    {
        allPart.push_back(currPart);
        return;
    }
 
    for (int i=start; i<n; i++)
    {
        if (isPalindrome(str, start, i))
        {
            currPart.push_back(str.substr(start, i-start+1));
 
            allPalPartUtil(allPart, currPart, i+1, n, str);
          
            currPart.pop_back();
        }
    }
}

void allPalPartitions(string str)
{
    int n = str.length();
 
    vector<vector<string> > allPart;
 
    vector<string> currPart;
 
    allPalPartUtil(allPart, currPart, 0, n, str);
 
    for (int i=0; i<allPart.size(); i++ )
    {
        for (int j=0; j<allPart[i].size(); j++)
            cout << allPart[i][j] << " ";
        cout << "\n";
    }
}

int main()
{
    string str = "nitin";
    allPalPartitions(str);
    return 0;
}

Output :

n i t i n 
n iti n 
nitin
While loop in C

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One comment on “Find all possible Palindromic Partitions of the given String in C++”


  • amanbajpai5734

    C++ code
    #include
    using namespace std;
    int main()
    {
    string str= “NITIN”;
    for(int i=0;i<str.length();i++)
    {
    for(int j=i;j<str.length();j++)
    {
    if(i==j)
    cout<<str[i]<<endl;
    else{
    string sub1= str.substr(i,j+1);
    string sub2= sub1;
    reverse(sub1.begin(),sub1.end());
    //cout<<sub1<<" "<<sub2<<endl;
    if(sub1==sub2)
    cout<<sub1<<endl;
    }
    }
    }

    return 0;
    }

    1