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).
Example
Given S = "rabbbit"
, T = "rabbit"
, return 3
.
Challenge
Do it in O(n2) time and O(n) memory.
O(n2) memory is also acceptable if you do not know how to optimize memory.
Solution: Dynamic Programming
Version 1. O(m*n)time O(m*n)space
f[i][j] means the number of different ways to select first j chars of T from first i chars from S, 即从S的前i个字符中挑出T的前j个字符 有多少种方案
f[i][j] = f[i-1][j-1] + f[i-1][j] //S[i]=T[j]
= f[i-1][j] //S[i]!=T[j]
public int numDistinct(String S, String T) { //f[i][j] means the number of different ways to select first j chars of T from first i chars from S //从S的前i个字符中挑出T的前j个字符 有多少种方案 //f[i][j] = f[i-1][j-1] + f[i-1][j] //S[i]=T[j] // = f[i-1][j] //S[i]!=T[j] if (S == null || T == null || S.length() < T.length()) { return 0; } int[][] numDistinct = new int[S.length() + 1][T.length() + 1]; for (int i = 0; i <= S.length(); i++) { numDistinct[i][0] = 1; } for (int i = 1; i <= S.length(); i++) { for (int j = 1; j <= i && j <= T.length(); j++) { if (S.charAt(i - 1) == T.charAt(j - 1)) { numDistinct[i][j] = numDistinct[i - 1][j - 1] + numDistinct[i - 1][j]; } else { numDistinct[i][j] = numDistinct[i - 1][j]; } } } return numDistinct[S.length()][T.length()]; }
Version 2. O(m*n) time, O(n)space
just mod 2 for the row index. Rolling array.
public int numDistinct(String S, String T) { if (S == null || T == null || S.length() < T.length()) { return 0; } int[][] numDistinct = new int[2][T.length() + 1]; for (int i = 0; i < 2; i++) { numDistinct[i][0] = 1; } for (int i = 1; i <= S.length(); i++) { for (int j = 1; j <= i && j <= T.length(); j++) { if (S.charAt(i - 1) == T.charAt(j - 1)) { numDistinct[i % 2][j] = numDistinct[(i - 1) % 2][j - 1] + numDistinct[(i - 1) % 2][j]; } else { numDistinct[i % 2][j] = numDistinct[(i - 1) % 2][j]; } } } return numDistinct[S.length() % 2][T.length()]; }