
Sponsored
Sponsored
We use a 2D DP table where `dp[i][j]` represents the number of distinct subsequences of `s[0..i-1]` which forms `t[0..j-1]`. The size of the table is `(m+1)x(n+1)` where `m` is the length of `s` and `n` is the length of `t`.
Initialize `dp[0][0]` to 1 because an empty string is a subsequence of itself, and `dp[i][0]` to 1 for all `i`, because an empty `t` is a subsequence of any prefix of `s`. Then, for each character `s[i-1]` and `t[j-1]`, update the DP table as follows:
Time Complexity: O(m * n), where `m` is the length of `s` and `n` is the length of `t`.
Space Complexity: O(m * n) for the DP table.
1function numDistinct(s, t) {
2 const m = s.length, n = t.length;
3 const dp = Array.from({ length
This JavaScript solution employs an array of arrays `dp` to store the subsequence counts. Loops iterate over characters in `s` and `t`, filling up `dp` as the given conditions dictate. The result is the value at `dp[m][n]`.
We can optimize the space complexity by observing that to calculate `dp[i][j]`, we only need values from the previous row. Thus, we can use a 2-row approach, maintaining only `previous` and `current` arrays to save space. This reduces the space complexity to O(n), where `n` is the length of `t`.
The transition is similar to the 2D approach but only updates the necessary parts of the `current` array based on the `previous` row.
Time Complexity: O(m * n), where `m` is the length of `s` and `n` is the length of `t`.
Space Complexity: O(n), using only two rows for DP storage.
This JavaScript approach uses two 1D arrays `prev` and `cur`. The main logic involves updating only necessary states per iteration, switching arrays at each outer loop step to retain recent results efficiently.