LeetCode "Palindrome Partition II" - Double DP

A so juicy and interesting problem! Recommend to everyone!

A final solution may not jump into your mind immediately, of course, DP is not a trivial topic. So let's go step by step - it is more interesting that how you get to destination than destination itself!

Since it is palindrome, so we need to know all palindromes - to avoid duplicated calculation, we need DP.

We made a step further but there's still a gap between us and AC: the min cut. "Minimum, maximum, shortest, longest.." etc. It means DP. So, we need a second DP. Since it is a linear value space, an ideal DP would be 1D - and we can. It is all matter of scanning. If s[0..i] is not palindrome, we pick min(dp2[j] + 1, when s[j+1..i] is palindrome). And you can get AC then :)

Reference: http://yucoding.blogspot.com/2013/08/leetcode-question-133-palindrome.html

class Solution {
public:
int minCut(string s) {
size_t len = s.length();
if (len <= 1) return 0;

//    Init
vector<vector<bool>> dp;
dp.resize(len);
for (int i = 0; i < len; i++)
{
dp[i].resize(len);
std::fill(dp[i].begin(), dp[i].end(), false);
dp[i][i] = true;
}

//    Go DP 1 - all palindromes
int min = std::numeric_limits<int>::max();
for (size_t ilen = 2; ilen <= len; ilen += 1)    // current substr len
for (size_t i = 0; i <= len - ilen; i ++)    // current starting inx
{
char c0 = s[i];
char c1 = s[i + ilen - 1];
bool bEq = c0 == c1;
if (ilen == 2)
dp[i][i + 1] = bEq;
else
dp[i][i + ilen - 1] = bEq && dp[i + 1][i + ilen - 2];
}

vector<int> dp2;
dp2.resize(len);
std::fill(dp2.begin(), dp2.end(), 0);

//    Go DP 2
dp2[0] = 0;
for (int i = 1; i < len; i++)
{
if (dp[0][i]) dp2[i] = 0;    // palin already?
else
{
//    find min num of palin
int min = len;
for (int j = 0; j < i; j++)
if (dp[j + 1][i])
{
min = std::min(min, dp2[j] + 1);
}
dp2[i] = min;
}
}
return dp2[len - 1];
}
};