Write a function for retrieving the total number of substring palindromes.
For example the input is 'abba' then the possible palindromes= a, b, b, a, bb, abba
So the result is 6.
Updated at 11/11/2013:
After the interview I got know that the O(n^3) solution is not enough to go to the next round. It would have been better to know before starting implementing the solution unnecessarily ...
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Similar to leetcode Longest Palindromic Substring Part II in my blog, the code is like:
#include <iostream>
#include <map>
#include <algorithm>
#include <limits.h>
#include <assert.h>
#include <string.h>
#include <vector>
using namespace std;
string preprocess(string s) {
string res = "^#";
for (int i = 0; i < s.length(); ++i) {
res += s[i];
res += '#';
}
res += '$';
return res;
}
int getPalindromeNum(string s) {
string str = preprocess(s);
int i, j, len = str.length(), C = 0, R = 0, res = 0, ii;
vector<int> T(len + 1, 0), P(len + 1, 0);
for (i = 1; i < len; ++i) {
ii = 2*C - i;
P[i] = (R - i) > 0 ? min(P[ii], R-i) : 0;
//bug1: P[i] = (R - i) > 0 ? P[i] : 0
while (str[i + P[i] + 1] == str[i - P[i] - 1])
++P[i];
res += (P[i] + 1) / 2;
//bug2: res += P[i];
if (i + P[i] > R) {
C = i;
R = i + P[i];
}
}
return res;
}
int main() {
//string s = "abcba";
string s = "aaaaa";
int res = getPalindromeNum(s);
return 0;
}
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Leetcode collects this problem:
Given a string, your task is to count how many palindromic substrings in this string.
The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
Example 1:
Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".
Example 2:
Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Note:
- The input string length won't exceed 1000.
Accepted
147,409
Submissions
249,379
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Python
class Solution:
def pre_process(self, s):
res = '^#'
for ch in s:
res += ch + '#'
res += '$'
return res
def countSubstrings(self, s):
s1 = self.pre_process(s)
l, res, c, r = len(s1), 0, 0, 0
p = [0 for i in range(l)]
for i in range(1, l-1): #bug1: for i in range(1, l-1)
ii = c * 2 - i
# bug2: p[i] = 0 if ii < 0 else min(p[ii], r)
p[i] = 0 if ii < 0 else min(p[ii], r-i)
while (s1[i + p[i] + 1] == s1[i - p[i] - 1]):
p[i] += 1
if (i + p[i] > r):
r = i + p[i]
c = i
res += (p[i] + 1) >> 1
return res
s = Solution()
print(s.countSubstrings("aaa"))