Lesson7 - Fishes

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.

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내 답안이자 베스트답안

Stack에 쌓아있는 생선(left)을 꺼내서, 집어 넣으려는 생선(right)과 비교하여 left의 방향이 1이고 right가 0 일때, right의 size가 더 큰 경우에는 계속해서 꺼내고, 아닌 경우 if 처리를 해주었다. O(N)으로 100% 성공하였다.

Fish Class를 만들때 생성자에서 this.FishSize = FishSize 가 아닌 FishSize = FishSize 해서 원치않는 답이 계속 나왔다.

실무에서는 코틀린을 다루다보니 자바 클래스를 만드는일이 없어 당황하였다.

import java.util.Stack;
class Solution {
    public int solution(int[] A, int[] B) {
        Stack fishes = new Stack<Fish>();

        for (int i=0; i<A.length; i++) {
            int fishSize = A[i];            
            int streamDirection = B[i];
            Fish right = new Fish(fishSize, streamDirection);
            
            while(true) {
                if (fishes.isEmpty()) {
                    fishes.add(right);
                    break;
                }

                Fish left = (Fish) fishes.pop();                        
                
                if (left.getStreamDirection() == 1 && right.getStreamDirection() == 0) {
                    if (left.getFishSize() > right.getFishSize()) {
                        fishes.add(left);
                        break;
                    }
                } else {
                    fishes.add(left);
                    fishes.add(right);
                    break;
                }
            }
        
        }

        return fishes.size();
    }
}

class Fish {
    private int fishSize = 0;
    private int streamDirection = 0;

    public Fish(int fishSize, int streamDirection) {
        this.fishSize = fishSize;
        this.streamDirection = streamDirection;
    }

    public int getFishSize() {
        return fishSize;
    }

    public int getStreamDirection() {
        return streamDirection;
    }    
}