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Author:Vaibhav Tulsyan
Testers:Jakub Safin
Editorialist:Praveen Dhinwa

DIFFICULTY:

easy-medium

PREREQUISITES:

prefix-sums, data structures

PROBLEM:

You are given two arrays $A, B$ each containing $N$ integers. In a single operation, you can choose any index $i$ ($1 \leq i \leq N$) and increment $A_i$ by $1$.

Find out minimum number of operations required such that $A_i \bmod B_i$ is equal for all $i, 1 \leq i \leq N$.

SOLUTION

We want to make $A_i \bmod B_i$ equal for all the $1 \leq i \leq N$. Let $K$ be this value. Note that $K$ can't greater than $min(B_i) - 1$.

The obvious approach is to iterate over all possible values of $K$, and for each of its value find the minimum cost required, and simply output the minimum of these costs.

So, lets see how to calculate the minimum cost to set all of the modulos to some particular value $K$.

1. For all the positions having $A_i \bmod B_i < K$, you need to increment them until the value reaches $K$.
2. For all the positions with $A_i \bmod B_i > K$, you need to make the values of $A_i \bmod B_i$ reach zero first, and then increment them each $K$ times.

Let us first rearrange the elements of the arrays $A$ and $B$ in increasing order of values $A_i \bmod B_i$.

Now, let's see how to calculate the total number of operations required for first type. For each element $i$, we have to increment it $K - A_i \bmod B_i$ times. Let $L$ be the number of such elements of type 1. You can see that these elements will form a prefix of the rearranged array. Summing over all such elements, we get the total number of opeartions $= K \cdot L - \sum\limits_{i = 1}^{L} {A_i \bmod B_i}$. The quantity $\sum\limits_{i = 1}^{L} {A_i \bmod B_i}$ can be computed by maintaining prefix sums over the values $A_i \bmod B_i$.

Now, let's see how to calculate the second type of costs. All the positions that have $A_i \bmod B_i > K$ will be some suffix of the rearranged array. You need to make the values of $A_i \bmod B_i$ reach zero first. For each of the element $i$, such that $A_i \bmod B_i > K$, the cost required to make $A_i \bmod B_i = 0$ will be $B_i - (A_i \bmod B_i)$ (note that it's guaranteed that $A_i \bmod B_i \neq 0$ as $A_i \bmod B_i > K, K \geq 0$. The sum of the above quantity for each $i$ in the suffix can be found using the suffix sums. Afterwards, we increase each of these elements by $K$, which contribution will be $K \times$ number of such elements.

You can see #hemanth_1's solution for a detailed explanation of this idea along with some examples.

Alternate approaches.

Please check meooow's solution. It's an alternate view of seeing this solution.

Setters and Tester Solution Codes

Tester's solution