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DIFFICULTY

EASY

EXPLANATION

The problem can be solved easily by dynamic programming:

F(i) = CountPal(s') for s' is the subtring of s[1]..s[i]

so F(i) = sum of F(j) with j < i whether the substring of s[j+1]..s[i] is a palindrome.

To verify a substring is a palindrome or not, we can pre-compute to save time of DP.

Complexity: O(|s|^2)