HISTOGRA - Largest Rectangle in a Histogram
A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:
Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.
The input contains several test cases.
Each test case describes a histogram and starts with an integer
n, denoting the number of rectangles it is composed of.
You may assume that
1 <= n <= 100000.
h1, ..., hn, where
0 <= hi <= 1000000000.
These numbers denote the heights of the rectangles of the histogram in left-to-right order.
The width of each rectangle is
A zero follows the input for the last test case.
For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.
7 2 1 4 5 1 3 3 4 1000 1000 1000 1000 0
Instead of using c++ STL version of stack, create your own version of stack using arrays and also use scanf, printf instead of cin,cout to avoid TLE.Last edit: 2017-03-13 08:58:59
AC in one go! Easy if you know this problem: http://www.geeksforgeeks.org/find-the-nearest-smaller-numbers-on-left-side-in-an-array/
Awesome problem! Looks easy but gets on your nerves after a while and you check for solutions.
Take all variables of long long type
it seems easy first, but then you are screwed XD
O(n) without a stack ^_^
Accepted, But In g++ 5.1, I have to use long long for both n and heights....I thought long should suffice ( < 10^9).
the best problem on stack....O(n) solution
Wow, great problem. I looked at times after debating divide and conquer and O(n^2) approaches, and figured there must be an O(n) solution. Not sure if I got lucky thinking of a solution. Subconsciously at least, the experience of ARRAYSUB and HOTELS helped me.Last edit: 2016-10-16 23:23:18