LeetCode 940: Distinct Subsequences II Solution

Master LeetCode problem 940 (Distinct Subsequences II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

940. Distinct Subsequences II

Hard

String Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. Let dp[i] represent the number of distinct subsequences that can be formed using the first i characters of the string s. We can update dp[i] based on the following conditions:

If s[i] has not occurred before, dp[i] = 2 * dp[i-1] + 1.
If s[i] has occurred before, we need to subtract the last occurrence of s[i] from the total, so dp[i] = 2 * dp[i-1] - dp[last_occurrence].

We also need to keep track of the last occurrence of each character to calculate dp[i] efficiently.

Time complexity: O(n) Space complexity: O(n)

Solution Code

class Solution {
    public int distinctSubseqII(String s) {
        int mod = (int)1e9 + 7;
        int n = s.length();
        int[] dp = new int[n + 1];
        int[] lastOccurrence = new int[26];
        Arrays.fill(lastOccurrence, -1);
        dp[0] = 1;

        for (int i = 1; i <= n; i++) {
            dp[i] = (2 * dp[i - 1]) % mod;
            int idx = s.charAt(i - 1) - 'a';
            if (lastOccurrence[idx] != -1) {
                dp[i] = (dp[i] - dp[lastOccurrence[idx]] + mod) % mod;
            }
            lastOccurrence[idx] = i - 1;
        }

        return (dp[n] - 1 + mod) % mod;
    }
}

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