A16265. Grouped Carriages

Pak Chanek observes that the carriages of a train is always full on morning departure hours and afternoon departure hours. Therefore, the balance between carriages is needed so that it is not too crowded in only a few carriages.

A train contains $N$ carriages that are numbered from $1$ to $N$ from left to right. Carriage $i$ initially contains $A_i$ passengers. All carriages are connected by carriage doors, namely for each $i$ ( $1\leq i\leq N-1$ ), carriage $i$ and carriage $i+1$ are connected by a two-way door.

Each passenger can move between carriages, but train regulation regulates that for each $i$ , a passenger that starts from carriage $i$ cannot go through more than $D_i$ doors.

Define $Z$ as the most number of passengers in one same carriage after moving. Pak Chanek asks, what is the minimum possible value of $Z$ ?

The first line contains a single integer $N$ ( $1 \leq N \leq 2\cdot10^5$ ) — the number of carriages.

The second line contains $N$ integers $A_1, A_2, A_3, \ldots, A_N$ ( $0 \leq A_i \leq 10^9$ ) — the initial number of passengers in each carriage.

The third line contains $N$ integers $D_1, D_2, D_3, \ldots, D_N$ ( $0 \leq D_i \leq N-1$ ) — the maximum limit of the number of doors for each starting carriage.

An integer representing the minimum possible value of $Z$ .

One strategy that is optimal is as follows:

- $5$ people in carriage $1$ move to carriage $4$ (going through $3$ doors).
- $3$ people in carriage $5$ move to carriage $3$ (going through $2$ doors).
- $2$ people in carriage $6$ move to carriage $5$ (going through $1$ door).
- $1$ person in carriage $6$ moves to carriage $7$ (going through $1$ door).

The number of passengers in each carriage becomes $[2,4,5,5,4,5,4]$ .