Task:Fish

N voracious fish are moving along a river. Calculate how many fish are alive.

You are given two non$-$empty arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.

The fish are numbered from $0$ to N $1.$ If P and Q are two fish and P $<$ Q$,$ then fish P is initially upstream of fish Q$.$ Initially, each fish has a unique position.

Fish number P is represented by A$[$P$]$ and B$[$P$].$ Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only $0$s and$/$or $1$s$,$ where:

$0$ represents a fish flowing upstream,

$1$ represents a fish flowing downstream.

If two fish move in opposite directions and there are no other $($living$)$ fish between them, they will eventually meet each other. Then only one fish can stay alive $$ the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P $<$ Q$,$ B$[$P$]=1$ and B$[$Q$]=0,$ and there are no living fish between them. After they meet:

If A$[$P$]>$ A$[$Q$]$ then P eats Q$,$ and P will still be flowing downstream,

If A$[$Q$]>$ A$[$P$]$ then Q eats P$,$ and Q will still be flowing upstream.

We assume that all the fish are flowing at the same speed. That is$,$ fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.

For example, consider arrays A and B such that:

A$[0]=4$ B$[0]=0$

A$[1]=3$ B$[1]=1$

A$[2]=2$ B$[2]=0$

A$[3]=1$ B$[3]=0$

A$[4]=5$ B$[4]=0$

Initially all the fish are alive and all except fish number $1$ are moving upstream. Fish number $1$ meets fish number $2$ and eats it$,$ then it meets fish number $3$ and eats it too. Finally, it meets fish number $4$ and is eaten by it$.$ The remaining two fish, number $0$ and $4,$ never meet and therefore stay alive.