Solution to Distinct Subsequences by LeetCode

7 Jul

July 7, 2014January 21, 2020 Sheng 1

Question: https://oj.leetcode.com/problems/distinct-subsequences/

Question Name: Distinct Subsequences

This is a dynamic programming challenge. Firstly I solved it with recursion. But that solution did not satisfy the time requirement. These two posts give more details about the right solution: LeetCode: Distinct Subsequences（不同子序列的个数）(in Chinese) and Distinct Subsequences DP explanation (in English).

class Solution:

# @return an integer

def numDistinct(self, S, T):

cache = [0] * (len(T)+1)

cache[0] = 1

for indexS in xrange(len(S)):

pre = cache[0]

for indexT in xrange(len(T)):

current = cache[indexT+1]

if S[indexS] == T[indexT]:

cache[indexT+1] += pre

pre = current

return cache[-1]

One Reply to “Solution to Distinct Subsequences by LeetCode”

Well, the algorithm running time can be reduced to half if you can recognize the fact that only half of the matrix needs to be filled, aka triangle matrix.

#include <vector>

#include <string>

#include <algorithm>

using namespace std;

class Solution {

public:

int numDistinct(string s, string t) {

size_t lent = t.length(), lens = s.length();

if(lens < lent)return 0;

else if(0 == lent)return 1;

vector<size_t> memo(lens, 0);

for(size_t j = 0; j < lens; ++j) {

if(t[0] == s[j])memo[j] = 1;

memo[j] += (0 == j) ? 0 : memo[j - 1];

}

size_t upperleft = 0, tmp = 0;

for(size_t i = 1; i < lent; ++i) {

upperleft = memo[i - 1], memo[i - 1] = 0;

for(size_t j = i; j < lens; ++j) {

tmp = memo[j];

if(s[j] == t[i])memo[j] = memo[j - 1] + upperleft;

else memo[j] = memo[j - 1];

upperleft = tmp;

}

return (int)memo[lens - 1];

}

};

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