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
.
public class Solution {
public int numDistinct(String S, String T) {
// Start typing your Java solution below
// DO NOT write main() function
HashMap<Character,ArrayList<Integer>> map = new HashMap<Character,ArrayList<Integer>>();
for(int i=0;i<T.length();i++){
if(map.containsKey(T.charAt(i))){
map.get(T.charAt(i)).add(i);
}else{
ArrayList<Integer> temp = new ArrayList<Integer>();
temp.add(i);
map.put(T.charAt(i),temp);
}
}
int[] res = new int[T.length()+1];
res[0]=1;
for(int i=0;i<S.length();i++){
char c = S.charAt(i);
if(map.containsKey(c)){
ArrayList<Integer> temp = map.get(c);
int[] old = new int[temp.size()];
for(int j=0;j<temp.size();j++)
old[j]=res[temp.get(j)];
for(int j=0;j<temp.size();j++)
res[temp.get(j)+1]+=old[j];
}
}
return res[T.length()];
}
}
1 comment:
I have a solution using less space:
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
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