LWC 59:730. Count Different Palindromic Subsequences
2017-11-19 21:45
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LWC 59:730. Count Different Palindromic Subsequences
传送门:730. Count Different Palindromic SubsequencesProblem:
Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.
A subsequence of a string S is obtained by deleting 0 or more characters from S.
A sequence is palindromic if it is equal to the sequence reversed.
Two sequences A_1, A_2, … and B_1, B_2, … are different if there is some i for which A_i != B_i.
Example 1:
Input:
S = ‘bccb’
Output: 6
Explanation:
The 6 different non-empty palindromic subsequences are ‘b’, ‘c’, ‘bb’, ‘cc’, ‘bcb’, ‘bccb’.
Note that ‘bcb’ is counted only once, even though it occurs twice.
Example 2:
Input:
S = ‘abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba’
Output: 104860361
Explanation:
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.
Note:
The length of S will be in the range [1, 1000].
Each character S[i] will be in the set {‘a’, ‘b’, ‘c’, ‘d’}.
思路:
难点在于如何划分子问题,才能保证更新dp时没有重复,其中需要解决重复元素子串的表达。为了保证每个子问题的回文在原问题中没有出现过,定义如下规则:子问题求出的回文串必须套上一层外壳,即子问题中的回文串集合Set = {s | s 为回文}, 有新的回文 s’ = “a” + s + “a” or “b” + s + “b”,….
定义函数如下f(i, j) 表示当前对应S[i,…j]的不重复回文串个数,于是有:
初始化: ans = 0 1. 子问题的回文串批层外衣,有 ans += f(i + 1, j - 1) , 其中S[i] == S[j] 2. 考虑"a_..._a", "_..._"表示子问题的回文串,抽出a'= a...a,其中"..."表示x个a,那么有新的回文串aa...a 和 aa...aa,有ans += 2
代码如下:
public int countPalindromicSubsequences(String S) { int n = S.length(); int[][] next = new int[4][1010]; int[][] prev = new int[4][1010]; char[] cs = S.toCharArray(); for (int i = 0; i < 4; ++i) Arrays.fill(next[i], n); for (int i = n - 1; i >= 0; --i) { int c = cs[i] - 'a'; for (int j = 0; j < 4; ++j) next[j][i] = i + 1 == n ? n : next[j][i + 1]; next[c][i] = i; } for (int i = 0; i < 4; ++i) Arrays.fill(prev[i], -1); for (int i = 0; i < n; ++i) { int c = cs[i] - 'a'; for (int j = 0; j < 4; ++j) prev[j][i] = i - 1 == -1 ? -1 : prev[j][i - 1]; prev[c][i] = i; } dp = new int[1010][1010]; return f(cs, next, prev, 0, n - 1); } int mod = 1000000000 + 7; int[][] dp; int f(char[] cs, int[][] next, int[][] prev, int s, int e) { if (s > e) return 0; if (dp[s][e] > 0) return dp[s][e]; long ans = 0; for (int i = 0; i < 4; ++i) { int ns = next[i][s]; int ne = prev[i][e]; if (ns > ne) continue; if (ns != ne) ans += 1; ans ++; ans += f(cs, next, prev, ns + 1, ne - 1); } dp[s][e] = (int)(ans % mod); return dp[s][e]; }
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