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

  • Difficulty Level : Medium
  • Last Updated : 03 Jun, 2021

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

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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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true if
// s[i...j] + s[k...l] + s[p...q]
// is a palindrome
function isPalin(i, j, k, l, p, q, s)
{
    var 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
function countSubStr(s)
{
    // To store the count of
    // required sub-strings
    var count = 0;
    var n = s.length;
 
    // For choosing the first sub-string
    for (var i = 0; i < n - 2; i++) {
        for (var j = i; j < n - 2; j++) {
 
            // For choosing the second sub-string
            for (var k = j + 1; k < n - 1; k++) {
                for (var l = k; l < n - 1; l++) {
 
                    // For choosing the third sub-string
                    for (var p = l + 1; p < n; p++) {
                        for (var 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
var s = "abca";
document.write( countSubStr(s));
 
// This code is contributed by famously.
</script>
Output: 
2

 




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