647. Palindromic Substrings
Tags: ‘String’, ‘Dynamic Programming’
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.
Solution 1 DP
$O(n^2)$ time $O(n^2)$ space 23% 45%
Refer to 5.Longest Parlindromic Substring
public int countSubstrings(String s) {
if (s == null) return 0;
int n = s.length(), count = 0;
boolean[][] dp = new boolean[n][n];
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
if (j - i <= 2) {
dp[i][j] = s.charAt(i) == s.charAt(j);
} else {
dp[i][j] = s.charAt(i) == s.charAt(j) && dp[i+1][j-1];
}
if (dp[i][j]) count++;
}
}
return count;
}
Solution 2 Expand around center
public int countSubstrings(String s) {
int count = 0;
for(int i=0; i<s.length(); i++){
count += extractPalindrome(s,i,i);//odd length
count + =extractPalindrome(s,i,i+1);//even length
}
return count;
}
public int extractPalindrome(String s, int left, int right){
int count=0;
while(left >= 0 && right < s.length() && (s.charAt(left) == s.charAt(right))){
left--;
right++;
count++;
}
return count;
}