Problem statement
The ‘Largest Rectangle in Histogram’ problem is as follows:
Given an array of integers heights representing the histogram’s bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.
I love this problem as it provides an amazing introduction to problem solving, more specifically, formalizing proofs of why algorithms work. Let’s start with an example:
Let the array of heights be [1, 4, 7, 5, 7]
:
The answer is 16
:
If you have solved this problem before, or just looked at the answer, you may have found it hard to convince yourself why the algorithm works. Fret not, I am here to try formalizing the proof of the algorithm so that you can understand how exactly it works!
The general solution is to keep a monotonically increasing stack of indices
Invariants
We can think of the monotonically increasing property as an invariant that the stack must maintain.
As a refresher, an invariant is a property that is always true across iterations (of a loop) or for all elements in a data structure.
A monotonically increasing stack of indices maintains the following invariants:
 $S[0] < S[1] < \cdots < S[1]$
 $H[S[0]] \leq H[S[1]] \leq \cdots \leq H[S[1]]$
Notations
For each index $i \in [0, n)$ where $H$ is the array of heights and $n = H$. Let $S$ be the monotonically increasing stack.
Operations
Many monotonic stack problems involve thinking about two key operations:
 When to push?
 When to pop?
Push
When $S = 0$ or $H[i] \geq H[S[1]]$, then we push $i$ to the stack as it maintains the invariant.
Pop
When $S > 0$ and $H[i] < H[S[1]]$, then we pop from the stack, receiving index $j$, and try computing the largest rectangle $L$ using $H[j]$ as the shortest rectangle.
To calculate the $L$, let $S’ = S  [j]$. The height of $L$, $h = H[j]$.
To derive the width of $L$, $w$, we need to find the leftmost and rightmost rectangles that are still $\geq H[j]$. We can label the indices of these rectangles as $l$ and $r$.
$$ H[l] \geq H[j] \land H[r] \geq H[j] $$
Otherwise, $H[j]$ will no longer be the shortest rectangle. After computing $L$, we can push $i$ onto the stack as well.
Solving $L$
Finding the rightmost index
Notice that if $H[i] < H[j]$ and $S$ is monotonically increasing, then $H[i  1]$ must be the rightmost tallest rectangle. Formally, the claim can be written as
$$ \forall k \in [j, i  1], H[k] \geq H[j] $$
The above claim implies that $L$ can go from $j$ to $i  1$ without encountering a shorter rectangle in between.
Suppose that $\exists k’ \in [j, i  1] : H[k’] < H[j]$, then it must have caused $j$ to be popped off of $S$ prior to $i$. However, this causes a contradiction as $j \in S$. Therefore, the claim holds and as a result, $l = i  1$.
Finding the leftmost index
Case 1: $j  S’[1] = 1$
If $j  S’[1] = 1$, then $H[j] \geq H[S’[1]]$ by monotonic stack. Thus, the $l = S’[1] + 1$ as that implies that the leftmost rectangle that is $H[l] \geq H[j]$ is $j$ itself.
Case 2: $j  S’[1] > 1$
If $j  S’[1] > 1$, then what we want to show is that
$$ \forall k \in (S’[1], j), H[k] \geq H[S[1]] \land H[k] \geq H[j] $$
To prove this, notice that $\forall k \in (S’[1], j), \exists o \in (k, j) : H[o] < H[k]$ given that $k \not\in S’$. Also notice that $H[o] \geq H[S’[1]]$, otherwise, $S’[1]$ would not be in $S’$. This proves the first part of the claim where $H[k] \geq H[S[1]]$ as $H[o]$ can be interchanged with $H[k]$.
Given that $o \not\in S’$ as well (otherwise $S’[1] = o$), then $\forall o \in (k, j), H[o] \geq H[j]$, otherwise, $o \in S’$ which is a contradiction. That means $H[j]$ would have caused the final $o$ to be popped from the stack.
Therefore, the claim is proven and we know that $l = S’[1] + 1$ as the leftmost rectangle that is $H[l] \geq H[j]$ is $H[S’[1] + 1]$ given that $H[S’[1]] \leq H[j]$ by monotonic stack.
Case 3: $S’ = 0$
If $S’ = 0$. When this happens there is no leftmost rectangle, so we can set $l = 0$.
Calculating $w$
Therefore, we have found both the rightmost index where $H[r] \geq H[j]$, $r = i  1$ and the leftmost index where $H[l] \geq H[j]$, $l = S’[1] + 1$ or $l = 0$. The width of $L$ is therefore, $w = r  l + 1$.
If $S’ > 0$, then $w = (i  1)  (S’[1] + 1) + 1 = i  S’[1]  1$.
If $S’ = 0$, then $w = (i  1)  0 + 1 = i$.
Implementation
When we implement the algorithm, we see that we naturally derive the ideal stack solution:


Conclusion
There you have it! The largest rectangle in histogram algorithm has now been formally proven and the magic numbers have been answered! Of course, when working on these problems during an interview, it is not expected of you to derive these proofs from scratch, but having such an intuition about monotonic stacks help a lot in reasoning with the magic values that are used!