# Optimised DP solution enclosed with a very tidy yet fastest solution too C++

• ``````class Solution {
public:
//AC - 20ms - using recorder;
bool isMatch(string s, string p)
{
int sIndex=0, pIndex=0;
int sAIndex=-1, pAIndex=-1;
int sLen=s.length(), pLen=p.length();
while(sIndex < sLen)
{
if(s[sIndex]==p[pIndex] || p[pIndex]=='?') { sIndex++, pIndex++; continue;  }
if(p[pIndex] == '*') { pAIndex = pIndex++; sAIndex = sIndex; continue;  }
if(pAIndex > -1) { pIndex = pAIndex+1; sIndex = ++sAIndex; continue;  }
return false;
}
while(p[pIndex] == '*') pIndex++;
return pIndex==pLen;
}

//AC - 28ms - typical DP solution;
bool isMatch(string s, string p)
{
int sLen = s.length(), pLen = p.length();
int count = 0;
for(int i = 0; i < pLen; i++) if(p[i] == '*') count++;
if((count==0 && pLen!=sLen) || (pLen-count>sLen)) return false;
int match[sLen+1]{0};
match[0] = 1;
for(int i = 0; i < pLen; ++i)
{
if(p[i] == '*')
{
for(int j = 1; j <= sLen; ++j)
match[j] = match[j-1] || match[j];
}
else
{
for(int i = sLen; j > 0; --j)
match[j] = match[j-1] && (p[i]=='?' || p[i]==s[j-1]);
/*int t0, t1 = match[0];*/
/*for(int j = 1; j <= sLen; ++j)*/
/*{*/
/*t0 = match[j];*/
/*match[j] = t1 && (p[i]=='?' || p[i]==s[j-1]);*/
/*t1 = t0;*/
/*}*/
match[0] = 0;
}
}
return match[sLen];
}
};``````

• @LHearen Awesome as always, but could you please explain why the first solution is quicker? Thank you!

• @Salkexy2 I personally think there are two major reasons:

• DP solution will always allocate extra space for recording;
• DP solution will fill up all the `cells` while the recursive won't, it just trys the possible path till the path is dead.

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