BOJ_20033_Square, Not Rectangle (Java)

[Gold IV] Square, Not Rectangle - 20033

문제 링크

성능 요약

메모리: 46272 KB, 시간: 396 ms

분류

이분 탐색, 자료 구조, 스택

제출 일자

2025년 2월 23일 01:02:13

문제 설명

A histogram is a polygon made by aligning $N$ adjacent rectangles that share a common base line. Each rectangle is called a bar. The $i$-th bar from the left has width 1 and height $H_i$.

Your goal is to find the area of the largest rectangle contained in the given histogram, such that one of the sides is parallel to the base line.

Figure 1. The histogram given in the example, with the largest rectangle shown on the right.

Actually, no, you have to find the largest square. Since the area of a square is determined by its side length, you are required to output the side length instead of the area.

Figure 2. The histogram given in the example, with the largest square shown on the right.

입력

On the first line, a single integer $N$ is given, where $1 \le N \leq 300\,000$.

On the next line, $N$ space-separated integers $H_1, H_2, \cdots, H_N$, are given. $H_i$ $(1 \le H_i \le 10^9)$ is the height of the $i$-th bar.

출력

Output the side length of the largest square in the histogram, such that one of the sides is parallel to the base line.

문제 풀이

최대 K x K 되는 정사각형 변 길이 K 결정문제다. 파라메트릭 서치 느낌으로 풀었다.

코드

package BOJ_20033_SquareNotRectangle;

/**
 * Author: nowalex322, Kim HyeonJae
 */

import java.io.*;
import java.util.*;

public class Main {
    static BufferedReader br;
    static BufferedWriter bw;
    static StringTokenizer st;
    static int N, H[];
    public static void main(String[] args) throws Exception {
        new Main().solution();
    }

    public void solution() throws Exception {
//        br = new BufferedReader(new InputStreamReader(System.in));
        br = new BufferedReader(new InputStreamReader(new FileInputStream("src/main/java/BOJ_20033_SquareNotRectangle/input.txt")));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));

        N = Integer.parseInt(br.readLine());
        H = new int[N];
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            H[i] = Integer.parseInt(st.nextToken());
        }

        int left = 1, right = N;
        int res = 0;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(isAble(mid)){
                res = mid;
                left = mid + 1;
            }
            else{
                right = mid - 1;
            }
        }
        System.out.println(res);
        bw.flush();
        bw.close();
        br.close();
    }

    private boolean isAble(int mid) {
        int cnt = 0;
        for (int i = 0; i < N; i++) {
            if(mid <= H[i]) {
                cnt++;
                if(cnt >= mid) return true;
            }
            else cnt = 0;
        }
        return false;
    }
}