Firstly, I tried fixing a dish (say $$$i$$$) for a particular person so that person $$$i$$$ became happy! thereafter I got the logic to count the no of happy persons who become happy if I rotate the array by $$$k$$$.

I saw it by considering when it’s advantageous to swap adjacent elements and arrived at the above comparator. I recommend this resource for practicing exchange argument style problems: https://codeforces.com/blog/entry/63533

I proved this with exchange arguments.

Consider two strings that are adjacent in s, let them be $$$s_1$$$ and $$$s_2$$$.

Let the number of X in $$$s_1$$$ be $$$d_1$$$, sum of digits in $$$s_1$$$ be $$$x_1$$$. Let the number of X in $$$s_2$$$ be $$$d_2$$$, sum of digits in $$$s_2$$$ be $$$x_2$$$.

The change in the score when you swap $$$s_1$$$ and $$$s_2$$$ in s will be $$$d_2 x_1 - d_1 x_2$$$.

If this change is positive, we swap $$$s_1$$$ and $$$s_2$$$.

So sort in order of decreasing $$$x/d$$$.

Let there be two strings with index $$$i,j$$$.So lets say placing $$$i$$$ before $$$j$$$ gives us a better answer than placing $$$j$$$ before $$$i$$$ so it means that $$$xi{*}dj$$$ $$$>=$$$ $$$xj{*}di$$$ $$$=>$$$ $$$xi{/}di$$$ $$$>=$$$ $$$xj{/}dj$$$ as $$$x$$$ and $$$d$$$ are non-negative. Therefore placing $$$i$$$ before $$$j$$$ gives a better answer iff $$$xi{/}di$$$ $$$>=$$$ $$$xj{/}dj$$$. We can extend this idea and just sort the strings in order of decreasing $$$x/d$$$ and compute the answer.