all_palindrome_substring.cpp

// C++ program to find palindromic substrings of a string 
// Returns total number of palindrome substring of 
// length greater then equal to 2 
int CountPS(char str[], int n) 
{ 
	// create empty 2-D matrix that counts all palindrome 
	// substring. dp[i][j] stores counts of palindromic 
	// substrings in st[i..j] 
	int dp[n][n]; 
	memset(dp, 0, sizeof(dp)); 

	// P[i][j] = true if substring str[i..j] is palindrome, 
	// else false 
	bool P[n][n]; 
	memset(P, false , sizeof(P)); 

	// palindrome of single length 
	for (int i= 0; i< n; i++) 
		P[i][i] = true; 

	// palindrome of length 2 
	for (int i=0; i<n-1; i++) 
	{ 
		if (str[i] == str[i+1]) 
		{ 
			P[i][i+1] = true; 
			dp[i][i+1] = 1 ; 
		} 
	} 

	// Palindromes of length more than 2. This loop is similar 
	// to Matrix Chain Multiplication. We start with a gap of 
	// length 2 and fill the DP table in a way that gap between 
	// starting and ending indexes increases one by one by 
	// outer loop. 
	for (int gap=2 ; gap<n; gap++) 
	{ 
		// Pick starting point for current gap 
		for (int i=0; i<n-gap; i++) 
		{ 
			// Set ending point 
			int j = gap + i; 

			// If current string is palindrome 
			if (str[i] == str[j] && P[i+1][j-1] ) 
				P[i][j] = true; 

			// Add current palindrome substring ( + 1) 
			// and rest palindrome substring (dp[i][j-1] + dp[i+1][j]) 
			// remove common palindrome substrings (- dp[i+1][j-1]) 
			if (P[i][j] == true) 
				dp[i][j] = dp[i][j-1] + dp[i+1][j] + 1 - dp[i+1][j-1]; 
			else
				dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1]; 
		} 
	} 

	// return total palindromic substrings 
	return dp[0][n-1]; 
}