26 Minimum Number of insertion to make a string palindrome.cpp

int LCS(string X, string Y, int n, int m) {
	int dp[n + 1][m + 1]; // DP - matrix

	// base case of recursion --> for initialization of dp - matrix
	for (int i = 0; i <= n; i++)
		for (int j = 0; j <= m; j++)
			if (i == 0 || j == 0)
				dp[i][j] = 0;

	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			if (X[i - 1] == Y[j - 1]) // when last character is same
				dp[i][j] = 1 + dp[i - 1][j - 1];
			else // when last character is not same -> pick max
				dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);

	return dp[n][m];
}

int LPS(string X, int n) {
	string rev_X = X;
	reverse(rev_X.begin(), rev_X.end());
	return LCS(X, rev_X, n, n);
}

int MinInsertForPallindrome(string X, int n) {
	return n - LPS(X, n); // substract LPS from the length of string to get Minimum number of insertion to make a string palindrome
}