Leetcode 940: Distinct Subsequences II

Given a string s, return the number of distinct non-empty subsequences of s. Since the answer may be very large, return it modulo 10^9 + 7.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not.

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

• 1 <= s.length <= 2000
• s consists of lowercase English letters.

Problem Analysis:

We can use dynamic programming to solve this problem. When we iterate the string, we record number of distinct subsequences which end with ith character.

When i = 0, there is only one subsequence with the first character.

If i != 0, here we assume the number of distinct subsequences in i-1 case is sum(i-1), for these subsequences, we add ith character in the end of each subsequence to form new subsequences of i case, so sum(i) = sum(i-1).

But here we still need to add 1, as we can regard the ith character as a subsequence.

We add all number of distinct subsequences in i case, then we get the answer.

Solution

class Solution {
public:
    int distinctSubseqII(string s) {
        const int mod = 1e9 + 7;
        vector<long long> ends(26);
        for (const auto ch : s)
            ends[ch - 'a'] = accumulate(begin(ends), end(ends), 1LL) % mod;
        
        return accumulate(begin(ends), end(ends), 0LL) % mod;
    }
};

Time complexity is O(n)

Space complexity is O(1)