## My solution runtime is 0ms ans O(n) space

```
bool dp[1010];
bool isInterleave(char* s1, char* s2, char* s3) {
int aLen = strlen(s1), bLen = strlen(s2), cLen = strlen(s3);
if(aLen + bLen != cLen) return false;
dp[0] = true;
for(int j = 1; j <= bLen; j++)
dp[j] = dp[j - 1] && (s2[j - 1] == s3[j - 1]);
for(int i = 1; i <= aLen; i++){
char cur = s1[i - 1];
dp[0] = dp[0] && (s1[i - 1] == s3[i - 1]);
for(int j = 1; j <= bLen; j++)
dp[j] = (dp[j] && cur == s3[i + j - 1]) || (dp[j - 1] && s2[j - 1] == s3[i + j - 1]);
}
return dp[bLen];
}
```