[코딜리티] codility lesson 7 Stacks and Queues - Fish100%

문제.

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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.

Write a function:

int solution(int A[], int B[], int N);

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).

Elements of input arrays can be modified.

Copyright 2009–2017 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

풀이.

1234567891011121314151617181920212223242526272829

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

import java.util.*;
class Solution {
    public int solution(int[] A, int[] B) {
        // write your code in Java SE 8
  int fishs = 0;
  Stack<Integer> fishNet = new Stack<Integer>();
  
  for( int i=0; i < A.length; i++){
   if(B[i]==1){ 
    fishNet.push(A[i]);
   }else{
    
    while(!fishNet.isEmpty()){
     if(fishNet.peek() > A[i]) break; //new fish is killed.
     if(fishNet.peek() < A[i]) fishNet.pop(); //old fish is killed.
    }
    if(fishNet.isEmpty() && B[i]==0) fishs++; //alive
   }
  }
  
  return fishNet.size()+fishs;
    }
}

설명.

직진만 하는 물고기들이 A배열의 순서대로 놓여져 있다. B배열은 물고기가 향하는 방향이 적혀있다.
이 직진물고기들은 앞에 있는 물고기를 잡아 먹는데 자기 덩치보다 큰놈에게 먹힌다.
물고기들 덩치는 A배열에 있는 값이다.

살아남은 물고기는 몇마리?

물고기를 0 의 방향으로 보낸다. 중간에 아무것도 없으면 살아남은 물고기 다.
역방향 물고기를 만나면 스택에 쌓아둔다.
물고기가 스택에 쌓인 물고기를 만나서 서로 겨룬다.
이제 살아남은 물고기 수와 스택에 남아 있는 수를 합친다.


아니 문제는 풀었지만은 이게 O(N) 이 되는 이유가 뭘까?
이게 100%가 되는 이유가 뭘까?
나의 엄청난 고민과 삽질이 다 무의미했다니....

여튼 하도 말이 안되는거 같아서 했더만 이게 답이라네, 운 좋게 빈틈을 통과 한건가?