From addf9e4e697744de60f2f8af5554c733bc295271 Mon Sep 17 00:00:00 2001
From: sujunghwang <64738942+sujunghwang@users.noreply.github.com>
Date: Thu, 23 Feb 2023 09:45:08 +0900
Subject: [PATCH] =?UTF-8?q?Week05=20PRG=2042584=20=EC=A3=BC=EC=8B=9D?=
 =?UTF-8?q?=EA=B0=80=EA=B2=A9?=
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---
 sujunghwang/week05/PRG_42584.java | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)
 create mode 100644 sujunghwang/week05/PRG_42584.java

diff --git a/sujunghwang/week05/PRG_42584.java b/sujunghwang/week05/PRG_42584.java
new file mode 100644
index 0000000..6592743
--- /dev/null
+++ b/sujunghwang/week05/PRG_42584.java
@@ -0,0 +1,27 @@
+/*
+https://school.programmers.co.kr/learn/courses/30/lessons/42584?language=java
+
+주어진 배열을 탐색하면서 현재 값보다 작은 값이 언제 나오는지 구하면 되는 문제
+문제 카테고리는 스택/큐 였는데 모르겠어서 그냥 배열로 이중 for loop 돌림
+효율성 문제 생길 줄 알았는데 안생겨서 그냥 넘어가기로 함
+*/
+
+class Solution {
+    public int[] solution(int[] prices) {
+        int len = prices.length;
+        int[] answer = new int[len];
+        
+        for(int i = 0; i < len-1; i++){
+            answer[i] = 1;
+            for(int j = i+1; j < len-1; j++){
+                if(prices[j] >= prices[i]){
+                    answer[i] += 1;
+                } else {
+                    break;
+                }
+            }
+        }
+        
+        return answer;
+    }
+}