탐욕법 01 - 체육복, 프로그래머스, 알고리즘, Programmers, Stack, Queue, Hash, 코딩테스트, Algorithm, 풀이과정, Leetcode, 릿코드, 코딩테스트, Tech interview, search, greedy

 

// Process

// 1. Input n, lost, reserve

// 2. Convert to list

// 3. Remove overlapped students

// 4. Iterate while lost and reserve are remained

//  4.1. Check around(+-1) value of reserve is in lost array

//   4.1.1. If so -> Remove both, and count student

// 5. Return student

 

import java.util.*;



class Solution {

    public int solution(int n, int[] lost, int[] reserve) {

        int student = 0;

        

        // 2. 

        List<Integer> list_lost = new ArrayList<Integer>();

        for (int number : lost)

            list_lost.add(number);

        List<Integer> list_reserve = new ArrayList<Integer>();

        for (int number : reserve)

            list_reserve.add(number);

        

        // 3. 

        for (int i = 0; i < list_lost.size(); ++i) {

            boolean isDone = false;

            for (int j = 0; j < list_reserve.size() && !isDone; ++j) {

                if (list_lost.get(i) == list_reserve.get(j)) {

                    list_lost.remove(i--);

                    list_reserve.remove(j);

                    isDone = true;

                }

            }

        }

        

        student = n - list_lost.size();

        

        // 4.

        int index = 0;

        while (index < list_lost.size()) {

            boolean isDone = false;

            for (int i = 0; i < list_reserve.size() && !isDone; ++i) {

                if (list_lost.get(index)-1 == list_reserve.get(i) || list_lost.get(index)+1 == list_reserve.get(i)) {

                    list_reserve.remove(i);

                    ++student;

                    isDone = true;

                }

            }

            ++index;

        }

        

        return student;

    }

}

 

 

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