CODILITY - LESSON 5 - Fish

THE PROBLEM:

You are given two non-empty zero-indexed 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 0s and/or 1s, 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.

THE SOLUTION (100% score, time complexity O(N))

(click here to see the full test results)

class Solution {

    public int solution(int[] A, int[] B) {

  Stack<int> aliveFishes = new Stack<int>();  

  for(int i = 0; i < A.Length; i++)

  {

       if(aliveFishes.Count==0)

       {

               aliveFishes.Push(i);

       }

       else

       {

               while(aliveFishes.Count!=0 
                        && B[i] - B[aliveFishes.Peek()] == -1 
                        && A[aliveFishes.Peek()] < A[i])  
               { aliveFishes.Pop(); }

               if (aliveFishes.Count !=0) 
               { if(B[i] - B[aliveFishes.Peek()] != -1) aliveFishes.Push(i); }

               else { aliveFishes.Push(i); }

        }

  }

  return aliveFishes.Count;

    }

}

This test took me a while, even though it is marked as 'easy'. I think because initially I tried without using a Stack.

Also because i din't read it throughly and missed the detail that while comparing two fishes, the first fish needs to have direction 1 and the second direction 0, and not 0-1.

The solution takes the first fish and pushes into the stack (as this is empty).

Then with the second and following fishes, it checks three possibilities:

• While the stack is not empty, if the direction of the fish is 1 and the direction of the last fish in the stack is 0 (aliveFishes.Peek()) and the size of the fish is bigger that the last fish in the stack, the we pop (delete) the last fish in the stack (as it has been eaten).
• If the stack is not empty, and the direction of the fish and the last fish of the stack is not the right one (gets to live) we push the fish into the stack.
• if above condition doesn't occur, so the stack is empty, we push the fish (it won't be eaten yet).

For more information about Stack follow the link:

msdn -> stack class and methods