Largest Rectangle
Problem Statement :
Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by h[ i ] where i [ 1, n ]. If you join k adjacent buildings, they will form a solid rectangle of area . For example, the heights array . A rectangle of height and length can be constructed within the boundaries. The area formed is h * k = 2 * 3 = 6. Function Description Complete the function largestRectangle int the editor below. It should return an integer representing the largest rectangle that can be formed within the bounds of consecutive buildings. largestRectangle has the following parameter(s): h: an array of integers representing building heights Input Format The first line contains n, the number of buildings. The second line contains n space-separated integers, each representing the height of a building. Constraints 1 <= n <= 10^5 1 <= h[ i ] <= 10^6 Output Format Print a long integer representing the maximum area of rectangle formed. .
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int i, j, k, n, h;
int ar[100000];
int area, max=0;
scanf("%d",&n);
for(i=0;i<=n;++i)
scanf("%d",&ar[i]);
for(i=0;i<=n;++i)
{
int c = 1;
j = i-1;
k = i+1;
while(j>=0 && ar[j]>ar[i]){
j--;
c++;
}
while(k<=n && ar[k]>ar[i]){
k++;
c++;
}
area = c * ar[i];
max = (area>max)? area : max;
}
printf("%d",max);
return 0;
}
Solution in C++ :
In C ++ :
#include <stack>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int N, h[100005];
int p = 1, s[100005];
int main() {
scanf("%d", &N);
for (int i = 1; i <= N; ++i) scanf("%d", &h[i]);
int ans = 0;
for (int i = 0; i < N + 2; ++i) {
while (h[i] < h[s[p - 1]]) {
int y = h[s[p - 1]];
p--;
ans = max(ans, (i - s[p - 1] - 1) * y);
}
s[p++] = i;
}
printf("%d\n", ans);
return 0;
}
Solution in Java :
In Java :
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Solution
{
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 64 * 1024);
final int N = Integer.parseInt(br.readLine().trim(), 10);
final String[] data = br.readLine().trim().split(" ");
final long[] hist = new long[N];
for (int i = 0; i < N; i++) {
final long v = Long.parseLong(data[i], 10);
hist[i] = v;
}
long res0 = 0L;
for (int i = 0; i < N; i++) {
int idx0 = i;
for (; idx0 >= 1; idx0--) {
if (hist[idx0 - 1] < hist[i]) {
break;
}
}
int idx1 = i;
for (; idx1 < hist.length - 1; idx1++) {
if (hist[idx1 + 1] < hist[i]) {
break;
}
}
final long area = hist[i] * (idx1 - idx0 + 1);
if (area > res0) {
res0 = area;
}
}
System.out.println(res0);
br.close();
br = null;
}
}
Solution in Python :
In Python3 :
def solve(H) :
s,i,m = [],0,0
while i < len(H) :
if not s or H[i] > H[s[-1]] :
s.append(i)
i += 1
else :
t = s.pop()
a = H[t] * ((i - s[-1] -1) if s else i)
if a > m :
m = a
while s :
t = s.pop()
a = H[t] * ((i - s[-1] -1) if s else i)
if a > m :
m = a
return m
N = int(input())
H = list(int(_) for _ in input().split())
print(solve(H))
View More Similar Problems
Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →