[Codility] Fish

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 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:

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

that, given two non-empty 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.

Write an efficient algorithm for the following assumptions:

• 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.
  

 

  하류로 가는 물고기의 크기를 스택(downFish)에 저장하고 상류로 가는 물고기의 위치(Index)에 도착했을 때 해당 물고기 크기와 스택에 저장된 크기를 비교한다. 크기 비교는 스택에 저장된값중 현재 위치의 물고기 크기보다 큰 값이 나올때까지다. 만약, 현재 물고기 크기가 스택의 모든 값보다 컸다면 모든 물고기를 잡아먹은 것이기 때문에 cnt++ 을 하고 다시 하류로 가는 물고기들을 스택에 넣고 비교를 반복한다.

 

import java.util.*;

class Solution {
    public int solution(int[] A, int[] B) {
        
        Stack<Integer> downFish = new Stack<>();
        
        int cnt = 0;

        for(int i = 0; i < A.length; i++){
            
            if(B[i] == 0 ){
                // 하류로 가는 물고기가 없고 상류로 가는 물고기
                if(downFish.size() == 0){
                    cnt++;
                }
                else{
                    //상류로 가는 물고기가 죽을때까지 비교
                    while(downFish.size() > 0){
                        if(downFish.peek() > A[i]){
                            break;
                        }
                        else{
                            downFish.pop();
                        }
                    }
                    // 하류로 가는 물고기 모두 잡아먹힌 경우 ++
                    if(downFish.empty()){
                        cnt++;
                    }
                }
            }
            else{
                downFish.push(A[i]);
            }
        }
        cnt += downFish.size();
        
        return cnt;
    }
}