My three C++ solutions (iterative (16ms) & DP (180ms) & modified recursion (88ms))

• The reason that the iterative solution is much faster for this case is we only need to save (and deal with) the positions (iStar for s, jStar for p) of the last "" we met. We only need to do traceback using iStar and jStar and all the previous "" can be ignored since the last "" will cover all the traceback cases for the previous "".
What we need to do are

• if the current p character is '' (i.e. p[j]==''), then we update iStar and jStar with the cureent i and j values. iStar/jStar will be used for traceback. Also we do --i to start the depth first search with the case that '*' represents a null string.
• if p[j]!='', then we check if mismatch occurs (i.e. p[j]!=s[i] and p[j]!='?'), if so we check if we met a '' before (iStar>=0), if not, then we return false since no match can achieve. Otherwise, we traceback to the positions at which the last '*' happens and do the next possible dfs search (i.e. i = iStar++; j = jStar; remember to update iStar too to save the i position to try in the next traceback).
• The loop will quit when we reach the end of s. At last, we need to skip all the '*' in p to see if we can reach the end of p. if so, match, otherwise mismatch

``````class Solution {
public:
bool isMatch(string s, string p) {
int  slen = s.size(), plen = p.size(), i, j, iStar=-1, jStar=-1;

for(i=0,j=0 ; i<slen; ++i, ++j)
{
if(p[j]=='*')
{ //meet a new '*', update traceback i/j info
iStar = i;
jStar = j;
--i;
}
else
{
if(p[j]!=s[i] && p[j]!='?')
{  // mismatch happens
if(iStar >=0)
{ // met a '*' before, then do traceback
i = iStar++;
j = jStar;
}
else return false; // otherwise fail
}
}
}
while(p[j]=='*') ++j;
return j==plen;
}
};
``````

A DP solution is also given here. It has O(N^2) time complexity and O(N) space

``````class Solution {
public:
bool isMatch(string s, string p) {
int pLen = p.size(), sLen = s.size(), i, j, k, cur, prev;
if(!pLen) return sLen == 0;
bool matched[2][sLen+1];
fill_n(&matched[0][0], 2*(sLen+1), false);

matched[0][0] = true;
for(i=1; i<=pLen; ++i)
{
cur = i%2, prev= 1-cur;
matched[cur][0]= matched[prev][0] && p[i-1]=='*';
if(p[i-1]=='*') for(j=1; j<=sLen; ++j) matched[cur][j] = matched[cur][j-1] || matched[prev][j];
else for(j=1; j<=sLen; ++j)            matched[cur][j] =  matched[prev][j-1] && (p[i-1]=='?' || p[i-1]==s[j-1]) ;
}
return matched[cur][sLen];
}
};
``````

A recursion version. A typical recursion version will give us TLE due to too many unnecessary recursive calls. As we explained, all the traceback recursive calls at the '' we met (except the last '') are unneccessary and should be avoided. In the below version, we use recLevel to track the recursion level (i.e the total '' we met) and we also use curLevel to save the order of '' we currently process. If it is not the last '' we met (i.e if(recLevel>curLevel+1) ), then we will return false directly ( if(recLevel>curLevel+1) return false;) to skip all unneccessary recursion call at the '' before the last '*'.

``````class Solution {
private:
bool helper(const string &s, const string &p, int si, int pi, int &recLevel)
{
int sSize = s.size(), pSize = p.size(), i, curLevel = recLevel;
bool first=true;
while(si<sSize && (p[pi]==s[si] || p[pi]=='?')) {++pi; ++si;} //match as many as possible
if(pi == pSize) return si == sSize; // if p reaches the end, return
if(p[pi]=='*')
{ // if a star is met
while(p[++pi]=='*'); //skip all the following stars
if(pi>=pSize) return true; // if the rest of p are all star, return true
for(i=si; i<sSize; ++i)
{   // then do recursion
if(p[pi]!= '?' && p[pi]!=s[i]) continue;
if(first) {++recLevel; first = false;}
if(helper(s, p, i, pi, recLevel)) return true;
if(recLevel>curLevel+1) return false; // if the currently processed star is not the last one, return
}
}
return false;
}
public:
bool isMatch(string s, string p) {
int recLevel = 0;
return helper(s, p, 0, 0, recLevel);
}
};``````

• This post is deleted!

• Doesn't the first solution read past the string for p if s is longer than p?

• For the recursive case, the line, if(recLevel>curLevel+1), is the key of the LTE problem.
When the result is true, it does not work; if false, LTE happens.

• I agree with you.

• Ur greedy algorithm is very neat and clear. Great work.

• the first solution will be wrong in this situation:s="aaa",p="a*a";

• the first solution will overflow eg: aaaa aa

• The last solution fails in this solution s="abab", p="ab". Should check if p[pi] != '*' first.

• I made the recursion code in java way. Finally solve my LTE. But could anyone explain how this recursion level help in this question?

``````public class Solution {
Integer level = 0;
public boolean isMatch(String sS, String pS) {
char[] s=sS.toCharArray();
char[] p=pS.toCharArray();
if(p.length==0&&s.length>0) return false;
return helper(s,p,0,0);
}
public boolean helper(char[] s,char[] p,int i, int j){
boolean first=true;
int curlevel= level;
if(j==p.length) return i==s.length||p[j-1]=='*';
if(i==s.length) {
while(p[j]=='*'&&j<p.length-1) j++;
return p[j]=='*';
}
if(p[j]=='?')
return helper(s,p,++i,++j);
if(p[j]!='*')
return p[j]==s[i]&&helper(s,p,++i,++j);
//p[j]=='*'
boolean res =false;
while(j<p.length&&p[j]=='*') j++;
for(int k=i;k<=s.length;k++){
if(first)
{++level;first=false;}
if(level>curlevel+1)
return false;
res|=helper(s,p,k,j);
if(res) return res;
}
return res;
}
}``````

• @omg_zozobra
The answer is yes. It is luck that it works.

• @dingess
Easy enough to fix though:

``````  bool IsWildcardMatch(string s, string p)
{
int slen = s.size(), plen = p.size(), i, j, iStar = -1, jStar = -1;

for (i = 0, j = 0; i < slen; i++, j++)
{
if (j < plen && p[j] == '*')
{ //meet a new '*', update traceback i/j info
iStar = i;
jStar = j;
--i;
}
else
{
if (j >= plen || (p[j] != s[i] && p[j] != '?'))
{  // mismatch happens
if (iStar >= 0)
{ // met a '*' before, then do traceback
i = iStar++;
j = jStar;
}
else return false; // otherwise fail
}
}

}
while (j<plen && p[j] == '*') ++j;
return j == plen;
}``````

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