27. Palindromic Substrings.cpp

/*
Palindromic Substrings
======================

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:
Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:
Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:
The input string length won't exceed 1000.

Hint #1  
How can we reuse a previously computed palindrome to compute a larger palindrome?

Hint #2  
If “aba” is a palindrome, is “xabax” and palindrome? Similarly is “xabay” a palindrome?

Hint #3  
Complexity based hint:
If we use brute-force and check whether for every start and end position a substring is a palindrome we have O(n^2) start - end pairs and O(n) palindromic checks. Can we reduce the time for palindromic checks to O(1) by reusing some previous computation?
*/

class Solution
{
public:
  int countSubstrings(string s)
  {
    int n = s.size();
    vector<vector<int>> dp(n, vector<int>(n, 0));

    for (int i = 0; i < n; ++i)
      dp[i][i] = 1;
    for (int i = 0; i + 1 < n; ++i)
    {
      if (s[i] == s[i + 1])
        dp[i][i + 1] = 1;
    }

    for (int gap = 2; gap < n; ++gap)
    {
      for (int i = 0; i + gap < n; ++i)
      {
        int j = i + gap;
        if (s[i] == s[j])
          dp[i][j] = dp[i + 1][j - 1];
      }
    }

    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
      for (int j = 0; j < n; ++j)
      {
        ans += dp[i][j];
      }
    }
    return ans;
  }
};