# 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;
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;
int p = 1, s;
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.IOException;

public class Solution
{
public static void main(String[] args) throws IOException {

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))```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## 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

## 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