**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))
```

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