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164 lines (150 loc) · 4.44 KB
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// Source : https://oj.leetcode.com/problems/distinct-subsequences/
// Author : Hao Chen
// Date : 2014-07-06
/**********************************************************************************
*
* 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.
*
*
**********************************************************************************/
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include <iostream>
#include <string>
#include <map>
#include <vector>
using namespace std;
//=====================
// Dynamic Programming
//=====================
//
// The idea as below:
//
// Considering m[i][j] means the distance from T[i] to S[j], and add the empty "" case, then,
//
// A) Initialization for empty case: m[0][j] = 1;
//
// B) Calculation
//
// a) Target-len > Source-len cannot found any substring
// i > j : m[i][j] = 0;
//
// b) if not equal, take the value of T[i] => S[j-1] (e.g. ["ra" => "rabb"] =["ra" => "rab"] )
// S[j] != T[i] : m[i][j] = m[i][j-1]
//
// c) if equal. (e.g. ["rab" => "rabb"] = ["rab" =>"rab"] + ["ra" => "rab"] )
// S[j] == T[i] : m[i][j] = m[i][j-1] + m[i-1][j-1]
//
// 1) Initialize a table as below
// "" r a b b b i t
// "" 1 1 1 1 1 1 1 1
// r
// b
// t
//
// 2) Calculation
// "" r a b b b i t
// "" 1 1 1 1 1 1 1 1
// r 0 1 1 1 1 1 1 1
// b 0 0 0 1 2 3 3 3
// t 0 0 0 0 0 0 0 3
//
int numDistinct1(string S, string T) {
vector< vector<int> > m(T.size()+1, vector<int>(S.size()+1));
for (int i=0; i<m.size(); i++){
for (int j=0; j<m[i].size(); j++){
if (i==0){
m[i][j] = 1;
continue;
}
if ( i>j ) {
m[i][j] = 0;
continue;
}
if (S[j-1] == T[i-1]){
m[i][j] = m[i][j-1] + m[i-1][j-1];
} else {
m[i][j] = m[i][j-1] ;
}
}
}
return m[T.size()][S.size()];
}
//=====================
// Dynamic Programming
//=====================
//
// The idea here is an optimization of above idea
// (It might be difficult to understand if you don't understand the above idea)
//
// For example:
//
// S = "abbbc" T="abb"
// posMap = { [a]={0}, [b]={2,1} } <- the map of T's every char.
// numOfSubSeq = {1, 0, 0, 0 }
//
// S[0] is 'a', pos is 0, numOfSubSeq = {1, 0+1, 0, 0};
//
// S[1] is 'b', pos is 2, numOfSubSeq = {1, 1, 0, 0+0};
// pos is 1, numOfSubSeq = {1, 1, 0+1, 0};
//
// S[2] is 'b', pos is 2, numOfSubSeq = {1, 1, 1, 0+1};
// pos is 1, numOfSubSeq = {1, 1, 1+1, 1};
//
// S[3] is 'b', pos is 2, numOfSubSeq = {1, 1, 2, 2+1};
// pos is 1, numOfSubSeq = {1, 1, 1+2, 3};
//
// S[4] is 'c', not found, numOfSubSeq = {1, 1, 3, 3};
//
//
int numDistinct2(string S, string T) {
map< char, vector<int> > pos_map;
int len = T.size();
vector<int> numOfSubSeq(len+1);
numOfSubSeq[0] = 1;
for (int i=len-1; i>=0; i--){
pos_map[T[i]].push_back(i);
}
for (int i=0; i<S.size(); i++){
char ch = S[i];
if ( pos_map.find(ch) != pos_map.end() ) {
for (int j=0; j<pos_map[ch].size(); j++) {
int pos = pos_map[ch][j];
numOfSubSeq[pos+1] += numOfSubSeq[pos];
}
}
}
return numOfSubSeq[len];
}
//random invoker
int numDistinct(string S, string T) {
srand(time(0));
if (rand()%2){
cout << "-----Dynamic Programming Method One-----" << endl;
return numDistinct1(S,T);
}
cout << "-----Dynamic Programming Method Two-----" << endl;
return numDistinct2(S,T);
}
int main(int argc, char** argv)
{
string s = "rabbbit";
string t = "rabbit";
if (argc>2){
s = argv[1];
t = argv[2];
}
cout << "S=\"" << s << "\" T=\"" << t << "\"" << endl;
cout << "numDistinct = " << numDistinct(s, t) << endl;
return 0;
}