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Leetcode115-distinctSubsequences

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example

Example 1:

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Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:
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Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

Solution

We first keep in mind that s is the longer string [0, i-1], t is the shorter substring [0, j-1]. We can assume t is fixed, and s is increasing in character one by one (this is the key point).

For example:
t : ab—> ab —> ab —> ab
s: a —> ac —> acb —> acbb

The first case is easy to catch. When the new character in s, s[i-1], is not equal with the head char in t, t[j-1], we can no longer increment the number of distinct subsequences, it is the same as the situation before incrementing the s, so dp[i][j] = dp[i-1][j].

However, when the new incrementing character in s, s[i-1] is equal with t[j-1], which contains two case:

We don’t match those two characters, which means that it still has original number of distinct subsequences, so dp[i][j] = dp[i-1][j].
We match those two characters, in this way. dp[i][j] = dp[i-1][j-1];
Thus, including both two case, dp[i][j] = dp[i-1][j] + dp[i-1][j-1]

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class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();
        int[][] dp = new int[m+1][n+1]; 
        for (int i = 0; i <= m; i++) dp[i][0] = 1;
        for (int j = 1; j <= n; j++){
            for (int i = 1; i <= m; i++){
                if (s.charAt(i-1) == t.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                else dp[i][j] = dp[i-1][j];
            }
        }
        
        return dp[m][n];
    }
}