Count of three non-overlapping sub-strings which on concatenation forms a palindrome

Given a string str, the task is to count the number of ways a palindromic sub-string could be formed by concatenation of three sub-strings x, y and z of the string str such that all of them are non-overlapping i.e. sub-string y occurs after sub-string x and sub-string z occurs after sub-string y.

Examples:

Input: str = “abca”
Output: 2
The two valid pairs are (“a”, “b”, “a”) and (“a”, “c”, “a”)

Input: str = “abba”
Output: 5

Approach: Find all the possible pairs of three non-overlapping sub-strings and for every pairs check whether the string generated by their concatenation is a palindrome or not. If yes then increment the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if
// s[i...j] + s[k...l] + s[p...q]
// is a palindrome
bool isPalin(int i, int j, int k, int l,
             int p, int q, string s)
{
    int start = i, end = q;
    while (start < end) {
        if (s[start] != s[end])
            return false;
  
        start++;
        if (start == j + 1)
            start = k;
        end--;
        if (end == p - 1)
            end = l;
    }
    return true;
}
  
// Function to return the count
// of valid sub-strings
int countSubStr(string s)
{
    // To store the count of
    // required sub-strings
    int count = 0;
    int n = s.size();
  
    // For choosing the first sub-string
    for (int i = 0; i < n - 2; i++) {
        for (int j = i; j < n - 2; j++) {
  
            // For choosing the second sub-string
            for (int k = j + 1; k < n - 1; k++) {
                for (int l = k; l < n - 1; l++) {
  
                    // For choosing the third sub-string
                    for (int p = l + 1; p < n; p++) {
                        for (int q = p; q < n; q++) {
  
                            // Check if the concatenation
                            // is a palindrome
                            if (isPalin(i, j, k, l, p, q, s)) {
                                count++;
                            }
                        }
                    }
                }
            }
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    string s = "abca";
  
    cout << countSubStr(s);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
    // Function that returns true if
    // s[i...j] + s[k...l] + s[p...q]
    // is a palindrome
    static boolean isPalin(int i, int j, int k, int l,
                            int p, int q, String s) 
    {
        int start = i, end = q;
        while (start < end) {
            if (s.charAt(start) != s.charAt(end))
            {
                return false;
            }
              
            start++;
            if (start == j + 1
            {
                start = k;
            }
            end--;
            if (end == p - 1
            {
                end = l;
            }
        }
        return true;
    }
  
    // Function to return the count
    // of valid sub-strings
    static int countSubStr(String s)
    {
        // To store the count of
        // required sub-strings
        int count = 0;
        int n = s.length();
  
        // For choosing the first sub-string
        for (int i = 0; i < n - 2; i++) 
        {
            for (int j = i; j < n - 2; j++) 
            {
  
                // For choosing the second sub-string
                for (int k = j + 1; k < n - 1; k++) 
                {
                    for (int l = k; l < n - 1; l++) 
                    {
  
                        // For choosing the third sub-string
                        for (int p = l + 1; p < n; p++) 
                        {
                            for (int q = p; q < n; q++) 
                            {
  
                                // Check if the concatenation
                                // is a palindrome
                                if (isPalin(i, j, k, l, p, q, s))
                                {
                                    count++;
                                }
                            }
                        }
                    }
                }
            }
        }
          
        return count;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        String s = "abca";
          
        System.out.println(countSubStr(s));
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function that returns true if 
# s[i...j] + s[k...l] + s[p...q] 
# is a palindrome 
def isPalin(i, j, k, l, p, q, s) : 
  
    start = i; end = q; 
    while (start < end) :
          
        if (s[start] != s[end]) :
            return False
  
        start += 1
        if (start == j + 1) :
            start = k; 
              
        end -= 1
        if (end == p - 1) : 
            end = l; 
      
    return True
  
  
# Function to return the count 
# of valid sub-strings 
def countSubStr(s) : 
  
    # To store the count of 
    # required sub-strings 
    count = 0
    n = len(s); 
  
    # For choosing the first sub-string 
    for i in range(n-2) :
          
        for j in range(i, n-2) : 
  
            # For choosing the second sub-string 
            for k in range(j + 1, n-1) : 
                for l in range(k, n-1) : 
  
                    # For choosing the third sub-string 
                    for p in range(l + 1, n) :
                        for q in range(p, n) :
  
                            # Check if the concatenation 
                            # is a palindrome 
                            if (isPalin(i, j, k, l, p, q, s)) :
                                count += 1
              
    return count; 
  
  
# Driver code 
if __name__ == "__main__"
  
    s = "abca"
  
    print(countSubStr(s)); 
  
# This course is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
class GFG 
  
    // Function that returns true if 
    // s[i...j] + s[k...l] + s[p...q] 
    // is a palindrome 
    static bool isPalin(int i, int j, int k, int l, 
                        int p, int q, String s) 
    
        int start = i, end = q; 
        while (start < end) 
        
            if (s[start] != s[end]) 
            
                return false
            
              
            start++; 
            if (start == j + 1) 
            
                start = k; 
            
            end--; 
            if (end == p - 1) 
            
                end = l; 
            
        
        return true
    
  
    // Function to return the count 
    // of valid sub-strings 
    static int countSubStr(String s) 
    
        // To store the count of 
        // required sub-strings 
        int count = 0; 
        int n = s.Length; 
  
        // For choosing the first sub-string 
        for (int i = 0; i < n - 2; i++) 
        
            for (int j = i; j < n - 2; j++) 
            
  
                // For choosing the second sub-string 
                for (int k = j + 1; k < n - 1; k++) 
                
                    for (int l = k; l < n - 1; l++) 
                    
  
                        // For choosing the third sub-string 
                        for (int p = l + 1; p < n; p++) 
                        
                            for (int q = p; q < n; q++) 
                            
  
                                // Check if the concatenation 
                                // is a palindrome 
                                if (isPalin(i, j, k, l, p, q, s)) 
                                
                                    count++; 
                                
                            
                        
                    
                
            
        
          
        return count; 
    
  
    // Driver code 
    public static void Main(String[] args) 
    
        String s = "abca"
          
        Console.WriteLine(countSubStr(s)); 
    
  
// This code is contributed by Princi Singh

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Output:

2


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