탐욕법 04 - 구명보트, 프로그래머스, 알고리즘, Programmers, Stack, Queue, Hash, 코딩테스트, Algorithm, 풀이과정, Leetcode, 릿코드, 코딩테스트, Tech interview, search, greedy

 

// Process

// 1. Input

// 2. Sort people in ascending way

// 3. Iterate till front and rear index are met

//  3.1. Check if front and rear index value can take the same boat

//   3.1.1. If so -> move both indices

//   3.1.2. If not -> move rear index

//  3.2. ++boatCount

// 4. Return boatCount

 

import java.util.*;



class Solution {

    public int solution(int[] people, int limit) {

        int boatCount = 0;

        int frontIndex = 0;

        int rearIndex = people.length-1;

            

        // 2.

        // 내부라이브러리 (오름차순만 됨)

        Arrays.sort(people);

        

        // Selection sort (내림차순이 라이브러리 안돼서 그냥 만들었었음)

        // for (int i = 0; i < people.length-1; ++i) {

        //     int maxIndex = i;

        //     int max = people[i];

        //     for (int j = i+1; j < people.length; ++j) {

        //         if (max < people[j]) {

        //             max = people[j];

        //             maxIndex = j;

        //         }

        //     }

        //     int temp = people[i];

        //     people[i] = max;

        //     people[maxIndex] = temp;

        // }

        

        // 3.

        while (frontIndex <= rearIndex) {

            if (people[frontIndex] + people[rearIndex] <= limit) {

                ++frontIndex;

                --rearIndex;

            } else {

                --rearIndex;

            }

            ++boatCount;

        }

        

        return boatCount;

    }

}

 

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