Python 代码类(class) : stacks and queues fish bug

Question

我的代码似乎为 codility 上的一个测试用例返回了错误的答案。(其余测试用例都是正确的)

对于测试用例:大随机

大型随机测试，N = ~100,000我得到了一个

得到 868 预期 840

问题链接: https://app.codility.com/programmers/lessons/7-stacks_and_queues/fish/

<p></p>

<p>def solution(A, B):
    #declare stacks for fish traveling downstrea, upstream
    dstrm = []
    ustrm = []
    #set counter to zero for fish traveling upstream UNHINDERED
    #by fish traveling downstream, needed because iteration starts upstream
    counter = 0
    n = len(A)
    #iterate over input, starting from upstream
    for i in range(n):
        if B[i] == 0:
            ustrm.append(A[i])
        elif B[i] == 1:
            dstrm.append(A[i])</p>

# clear upstream stack of fish known to be UNHINDERED, increase counter
        if len(ustrm) > 0 and len(dstrm) == 0:
            counter += len(ustrm)
            ustrm.clear()

    #compare what's bigger and decrease stack of fish that is eaten

        while len(dstrm) > 0 and len(ustrm) > 0:
            if dstrm[-1] > ustrm[-1]:
                ustrm.pop()
            elif ustrm[-1] > dstrm[-1]:
                dstrm.pop()

    return len(dstrm) + len(ustrm) + counter

Answer 1

这是我的方法 - 也许有人能想到 - 我是如何解决它的Codility Python 中的可溶性 100 %

Code

def solution(A, B):
"""
Codility 100%
https://app.codility.com/demo/results/trainingF5HTCA-YFV/

Idea is to use stack concept
For each iteration if current fish is of type 1 add it to stack 1 fish
Create stack 1 fish - it always holds the downstream ie 1 kind of fish
 and keep killing the smaller fish from the list if it is greater
  else killed by current fish.
Now if stack 1 fish has no fish it means current fish can be counted as remaining fish.


0 represents a fish flowing upstream - 0 fish
1 represents a fish flowing downstream - 1 fish

A[0] = 4    B[0] = 0 alive fish
A[1] = 3    B[1] = 1 downstream
A[2] = 2    B[2] = 0 eaten by A[1]
A[3] = 1    B[3] = 0 eaten by A[1]
A[4] = 5    B[4] = 0 eat to A[1] and remain alive

"""

count = 0
# stores the 1 fish in stack
stack_1_fish = []
print(A)
print(B)
for index in range(len(A)):
    if B[index] == 0:
        # until stack has some 1 fish
        while stack_1_fish:
            # get last fish from stack and check if it can die or not
            # the larger fish eats the smaller one.
            # 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
            if stack_1_fish[-1] > A[index]:
                # stack 1 fish kill to current fish
                # exit from while loop and else block also execute next top for loop
                # check for other fish fight
                print("Killed by " + str(stack_1_fish[-1]) + " Die " + str(A[index]))
                break
            else:
                # stack 1 fish is killed by current fish
                p = stack_1_fish.pop()
                print("Killed by " + str(A[index]) + " Die " + str(p))

        # this is the case when stack becomes empty ie. no fish of kind 1
        # it would never meet previous fish but can fight with downstream fish
        else:
            print("Count fish as remaining......." + str(A[index]))
            count += 1
    else:
        # fish 1 found add to stack
        stack_1_fish.append(A[index])
        print("1 fish found, add to stack, it can kill or killed by someone..." + str(A[index]))
        print(stack_1_fish)

print("Count: " + str(count))
print("Stack 1 fish: " + str(len(stack_1_fish)))
return count + len(stack_1_fish)

Execution Output -

if __name__ == '__main__':
result = solution([4, 3, 2, 1, 5], [0, 1, 0, 0, 0])
# result = solution([4, 3, 2, 1, 5], [0, 0, 0, 0, 0])
# result = solution([4, 3, 2, 1, 5], [1, 1, 1, 1, 1])
print("")
print("Solution " + str(result))

[4, 3, 2, 1, 5]
[0, 1, 0, 0, 0]
Count fish as remaining.......4
1 fish found, add to stack, it can kill or killed by someone...3
[3]
Killed by 3 Die 2
Killed by 3 Die 1
Killed by 5 Die 3
Count fish as remaining.......5
Count: 2
Stack 1 fish: 0

Solution 2

[4, 3, 2, 1, 5]
[0, 0, 0, 0, 0]
Count fish as remaining.......4
Count fish as remaining.......3
Count fish as remaining.......2
Count fish as remaining.......1
Count fish as remaining.......5
Count: 5
Stack 1 fish: 0

Solution 5

[4, 3, 2, 1, 5]
[1, 1, 1, 1, 1]
1 fish found, add to stack, it can kill or killed by someone...4
[4]
1 fish found, add to stack, it can kill or killed by someone...3
[4, 3]
1 fish found, add to stack, it can kill or killed by someone...2
[4, 3, 2]
1 fish found, add to stack, it can kill or killed by someone...1
[4, 3, 2, 1]
1 fish found, add to stack, it can kill or killed by someone...5
[4, 3, 2, 1, 5]
Count: 0
Stack 1 fish: 5

Solution 5