Codility voracious fish are moving along a river I am trying to solve the codility problem:


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


Codility Problem

My code get only 50 percent of 100. Could you tell me what is wrong with my code?


  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 &lt;
  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 &lt; 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.
  
  Write a function:

class Solution { public int solution(int[] A, int[] B); }

  
  that, given two non-empty zero-indexed arrays A and B consisting of N
  integers, returns the number of fish that will stay alive.
  
  For example, given the arrays shown above, the function should return
  2, as explained above.
  
  Assume that:
  
  N is an integer within the range [1..100,000]; each element of array A
  is an integer within the range [0..1,000,000,000]; each element of
  array B is an integer that can have one of the following values: 0, 1;
  the elements of A are all distinct.
  
  Complexity:
  
  Expected worst-case time complexity is O(N); expected worst-case space
  complexity is O(N), beyond input storage (not counting the storage
  required for input arguments).


This is my algorithm:

public static int solution(int[] size, int[] direction)
{
    var stack = new Stack&lt;int&gt;();
    const int UPSTREAM = 0;
    const int DOWNSTREAM = 1;
    var updateStack = false;

    for (int i = 0; i &lt; size.Length; i++)
    {
        if (stack.Count &lt;= 0) //initial data
            stack.Push(i);
        else
        {
            var lastFish = stack.Peek();

            if (direction[i] == direction[lastFish] || direction[lastFish] == UPSTREAM &amp;&amp; direction[i] == DOWNSTREAM) 
            {
                stack.Push(i);
            }
            else
            {
                while (stack.Count &gt; 0 &amp;&amp; size[lastFish] &lt; size[i] &amp;&amp; direction[lastFish] == DOWNSTREAM &amp;&amp; direction[i] == UPSTREAM)
                {
                    updateStack = true;
                    stack.Pop();
                    if (stack.Count &gt; 0)
                        lastFish = stack.Peek();
                }
                if (updateStack)
                {
                    stack.Push(i);
                    updateStack = false;
                }
            }
        }

    }
    return stack.Count;
}
This is an example of a situation where good naming is your friend. Instead of parameters A and B, and magic numbers 0 and 1, and the rather long-winded expression stack.Peek(), I suggest you start by installing some useful names:

const int UPSTREAM = 0;
const int DOWNSTREAM = 1;

public static int solution(int[] size, int[] direction)
{
    var stack = new Stack&lt;int&gt;();
    var updateStack = false;
    for (int i = 0; i &lt; size.Length; i++)
    {
        if (stack.Count &lt;= 0) //initial data
            stack.Push(i);
        else
        {
            var lastFish = stack.Peek();

            if (direction[i] == direction[lastFish] 
             || direction[i] != direction[lastFish] 
                &amp;&amp; direction[lastFish] == UPSTREAM 
                &amp;&amp; direction[i] == DOWNSTREAM) 
            {
                stack.Push(i);
            }
            else
            {
                while (stack.Count &gt; 0 
                       &amp;&amp; size[lastFish] &lt; size[i] 
                       &amp;&amp; direction[lastFish] == DOWNSTREAM 
                       &amp;&amp; direction[i] == UPSTREAM)
                {
                    updateStack = true;
                    stack.Pop();
                    lastFish = stack.Peek();
                }
                if (updateStack)
                    stack.Push(i);
            }
        }

    }
    return stack.Count;
}


This shows me there is some confusion on your part. You are allowed to assume that a logical or (||) fails because it's false - there's no need to repeat the test in a false mode:

if (direction[i] == direction[lastFish] 
 || direction[i] != direction[lastFish] &amp;&amp; ...


Instead, just assume that failure implies failure, and go on:

if (direction[i] == direction[lastFish] 
  || direction[lastFish] == UPSTREAM 
     &amp;&amp; direction[i] == DOWNSTREAM) 


Furthermore, you can assume that since you know the directions are different, and since you know there are only two directions, testing one of them is enough:

if (direction[i] == direction[lastFish] || direction[i] == DOWNSTREAM)


That said, I notice that you initialize updateStack to false outside of your loop and never reset it to false anyplace else. This means that once you've set updateStack = true, it will stay true forever (inside you loop, anyway). That's probably the source of your error, because it should stop being true once you've updated the stack.