Given a string

Sand a stringT, count the number of distinct subsequences ofTinS.A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie,

`"ACE"`

is a subsequence of`"ABCDE"`

while`"AEC"`

is not).Here is an example:

S=`"rabbbit"`

,T=`"rabbit"`

Return

`3`

.

## Analysis

This is a DP problem. We can use num[i][j] to save the number of distinct subsequences of T(0, j) in S(0, i). We know that for any number i, num[i][0] = 1 because we need to delete all characters from S(0, i) to a empty string.

If the character at position i in S is equal to the character at position j in T, there are two options.

- Delete the character at position i in S. Then the number of distinct subsequences should be the number of distinct subsequences of T(0, j) in S(0, i – 1).
- Remains the character at position i in S. Then the number is the number of distinct subsequences of T(0, j – 1) in S(0, i – 1).

So num[i][j] = num[i – 1][j] + num[i – 1][j – 1].

If the character at position i in S is not equal to the character at position j in T, then we can only delete this character. So num[i][j] = num[i – 1][j].

## Code

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public class Solution { public int numDistinct(String S, String T) { int M = S.length(); int N = T.length(); int[][] num = new int[M + 1][N + 1]; for (int i = 0; i <= N; i++) num[0][i] = 0; for (int i = 0; i <= M; i++) num[i][0] = 1; for (int i = 1; i <= M; i++) { for (int j = 1; j <= N; j++) { if (S.charAt(i - 1) != T.charAt(j - 1)) { num[i][j] = num[i - 1][j]; } else { num[i][j] = num[i - 1][j] + num[i - 1][j - 1]; } } } return num[M][N]; } } |

## Complexity

The complexity is $O(mn)$.