# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```2
```

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