[LeetCode] 940. Distinct Subsequences II

时间：2020-03-18 09:38:23 阅读：52 评论：0 收藏：0 [点我收藏+]

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Given a string S, count the number of distinct, non-empty subsequences of S .

Since the result may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: "abc"
Output: 7
Explanation: The 7 distinct subsequences are "a", "b", "c", "ab", "ac", "bc", and "abc".

Example 2:

Input: "aba"
Output: 6
Explanation: The 6 distinct subsequences are "a", "b", "ab", "ba", "aa" and "aba".

Example 3:

Input: "aaa"
Output: 3
Explanation: The 3 distinct subsequences are "a", "aa" and "aaa".

Note:

S contains only lowercase letters.
1 <= S.length <= 2000

Key Observation: If we ignore the distinct requirement for now, then each new character we takes in doubles the total count of subsequences we have. (Add this new character to the end of all searched subsequences).

Based on the above observation, we derive ssCnt of s[0, i] = 2 * ssCnt of s[0, i - 1], without considering the distinct requirement. If the newly added character has never appear before then we are gold. Otherwise, we need to remove the duplicated subsequences. Let‘s call the previous appearing index j, then we know that all subsequences

that end at s[j] are duplicated by the current index i character. We need to subtract this count.

Dynamic Programming Algorithm O(N) runtime and space

idx[j]: the jth letter‘s previous appearing index, -1 means it has not appeared previously.

dp[i]: the total number of distinct subsequences from s[0, i - 1];

dp[0] = 1, representing the empty subsequence;

dp[i] = dp[i - 1] * 2 + (idx[s[i]] >= 0 ? -dp[idx[s[i]]]: 0);

Answer: dp[N] - 1, -1 to exclude the empty subsequence.

Key note: dp is of length n + 1, so its index represents the current considering substring‘s length, which is 1 ahead of the 0-indexed string scan. So dp[idx[j] + 1] is the total number of distinct subsequences for s[0, j] whereas dp[idx[j]] is the number of distinct subsequences that END at s[j]. When removing duplicates, we only want to remove the ones that are contributed by s[j], i.e, the ones that END at s[j].

class Solution {
    public int distinctSubseqII(String S) {
        int mod = (int)1e9 + 7, n = S.length();
        int[] idx = new int[26];
        long[] dp = new long[n + 1];
        dp[0] = 1;
        Arrays.fill(idx, -1);
        for(int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] * 2 % mod;
            if(idx[S.charAt(i - 1) - ‘a‘] >= 0) {
                dp[i] = (dp[i] + mod - dp[idx[S.charAt(i - 1) - ‘a‘]]) % mod;
            }
            idx[S.charAt(i - 1) - ‘a‘] = i - 1;
        }
        return (int)((dp[n] + mod - 1) % mod);
    }
}

O(1) space optimization

class Solution {
    public int distinctSubseqII(String S) {
        int mod = (int)1e9 + 7, n = S.length();
        //prevOccurCnt[i]: the total number of distinct subsequences that end at s[i]
        long[] prevOccurCnt = new long[26];
        long prevSum = 1, currSum = 0;
        for(int i = 0; i < n; i++) {
            currSum = prevSum * 2 % mod;            
            if(prevOccurCnt[S.charAt(i) - ‘a‘] > 0) {
                currSum = (currSum + mod - prevOccurCnt[S.charAt(i) - ‘a‘]) % mod;
            }
            prevOccurCnt[S.charAt(i) - ‘a‘] = prevSum;
            prevSum = currSum;
        }
        return (int)((currSum + mod - 1) % mod);
    }
}