Question

Given a list of Strings (that contains both palindromes and nonpalindromes), write a method that returns a list of all the palindrome
strings.
TIP: Use static method references of Functional Interfaces
Solve it in Java

Answer 1

Answer:

// C++ program to find palindromic substrings of a string

#include<bits/stdc++.h>

using namespace std;

// 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];

}

// Driver program

int main()

{

char str[] = "abaab";

int n = strlen(str);

cout << CountPS(str, n) << endl;

return 0;

}