Every time you use memset with the third parameter $$$sizeof(a)$$$, it fills the whole array (or matrix) with corresponding values. But because the number of test cases in this problem can be up to $$$10^4$$$, you can fill up $$$10^4 \cdot 2 \cdot 10^5 \cdot 26 = 52 \cdot 10^9$$$ elements in the worst case. Despite the fact that memset is a very fast function, $$$52$$$ billion is a huge number of operations. But when you fill up all values iteratively, you can be sure that you will do this at most $$$\sum n \cdot 26$$$ times (i.e. $$$2 \cdot 10^5 \cdot 26 = 5.2 \cdot 10^6$$$ times.