For each i , the max # of perfect square is always i (when square = 1*1). So the boundary of each dp[i] = i.

```
class Solution {
public:
int numSquares(int n) {
if (n == 0) return 0;
vector<int> dp(n+1, 0);
for (int i=0; i<=n; ++i) {
dp[i] = i;
for (int j = 2; j<=sqrt(i); ++j) {
dp[i] = min(dp[i], 1 + dp[i - j*j]);
}
}
return dp[n];
}
};
```