Wednesday, November 26, 2014

[LeetCode] Distinct Subsequences

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

思路:

一看要求解的个数,然后又是string匹配,而且形式上和Minimum Edit Distance那题很像,基本就是DP题跑不了了。DP题惯例的三步走:定义状态,推导递推公式,确定状态计算方向和起始状态。

1. 状态i, j分别表示T中长度为i的prefix:T[0:i-1],和S中长度为j的prefix:S[0:j-1]。
DP[i][j]:S[0:j-1]中存在T[0:i-1]作为distinct subsequence的个数。显然如果j<i,DP[i][j] = 0。

2. 递推公式:
(a) T[i]!=s[j]:

T = r a b
S = r c a c b c

DP[i+1][j+1] = DP[i+1][j]

(b) T[i] = s[j]: 

T = r a b b
S = r a b b b  - DP[i+1][j] = 1
S = r a b b b  - DP[i][j] = 2
S = r a b b  /

DP[i+1][j+1] = DP[i][j] + DP[i+1][j]

公式总结:
S[j-1]!= T[i-1]:DP[i][j] = DP[i][j-1]
S[j-1]==T[i-1]:DP[i][j] = DP[i-1][j-1] + DP[i][j-1]


3. 计算方向和起始状态:
DP[i][j]
DP[i+1][j]   DP[i+1][j+1]

所以从上向下,从左到右的顺序可以计算。

根据计算顺序,需要先设置第0行、0列的值。
第0列:DP[i][0] = 0,i>0。因为T的长度大于S的长度,不可能成为S的subsequence。
第0行:DP[0][j] = 1,j>=0。这是为了保证第1行的计算正确:

T = r
S = r a r b r c

      r a  r b r  c  
  1 1 1 1 1 1 1
r 0 1 1 2 2 3 3


4. 计算优化:用滚动数组减少内存消耗。



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class Solution {
public:
    int numDistinct(string S, string T) {
        int n = S.size(), m = T.size();
        vector<int> dp(n+1, 1);
        
        for(int i=1; i<=m; i++) {
            int upLeft = dp[0];
            dp[0] = 0;
            for(int j=1; j<=n; j++) {
                int temp = dp[j];
                dp[j] = dp[j-1];
                if(S[j-1]==T[i-1]) dp[j] += upLeft;
                upLeft = temp;
            }
        }
        
        return dp[n];
    }
};

3 comments:

  1. My same way of space optimization:

    public class Solution {
    public int numDistinct(String s, String t) {
    if(s == null || t == null || t.length() == 0) return 0;
    int[] dp = new int[t.length()];

    for(int i = 0; i=0; j–){
    if(c == t.charAt(j)){
    dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
    }
    }
    }
    return dp[t.length()-1];
    }
    }
    URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html

    ReplyDelete
  2. My same way of space optimization:

    public class Solution {
    public int numDistinct(String s, String t) {
    if(s == null || t == null || t.length() == 0) return 0;
    int[] dp = new int[t.length()];

    for(int i = 0; i=0; j–){
    if(c == t.charAt(j)){
    dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
    }
    }
    }
    return dp[t.length()-1];
    }
    }
    URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html

    ReplyDelete