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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.
This C code uses a two-dimensional array `dp` where dp[i][j] contains the number of ways the first `i` characters of `s` can form the first `j` characters of `t`. We initialize the first column (except `dp[0][0]`) to 1 because the empty string `t` is a subsequence of any string `s`. We fill the DP table according to the conditions based on whether the current characters of `s` and `t` match or not, then return the value at dp[m][n], which contains the number of distinct subsequences.
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.
1class Solution {
2 public int numDistinct(String s, String t) {
3 int m = s.length(), n = t.length();
4 int[] prev = new int[n + 1];
5 int[] cur = new int[n + 1];
6 prev[0] = cur[0] = 1;
7
8 for (int i = 1; i <= m; i++) {
9 for (int j = 1; j <= n; j++) {
10 cur[j] = s.charAt(i - 1) == t.charAt(j - 1) ? prev[j - 1] + prev[j] : prev[j];
11 }
12 System.arraycopy(cur, 0, prev, 0, n + 1);
13 }
14 return prev[n];
15 }
16
17 public static void main(String[] args) {
18 Solution sol = new Solution();
19 System.out.println(sol.numDistinct("rabbbit", "rabbit"));
20 }
21}This Java method uses two 1D arrays, `prev` and `cur`, to minimize space usage. After each iteration, the current state `cur` is copied into `prev` for the next iteration using `System.arraycopy`.
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