\chapter{Amortized analysis}

\index{amortized analysis}

Often the time complexity of an algorithm
is easy to analyze by looking at the structure
of the algorithm:
what loops there are and how many times
they are performed.
However, sometimes a straightforward analysis
doesn't give a true picture of the efficiency of the algorithm.

\key{Amortized analysis} can be used for analyzing
an algorithm that contains an operation whose
time complexity varies.
The idea is to consider all such operations during the
execution of the algorithm instead of a single operation,
and estimate the total time complexity of the operations.

\section{Two pointers method}

\index{two pointers method}

In the \key{two pointers method},
two pointers iterate through the elements in an array.
Both pointers can move during the algorithm,
but the restriction is that each pointer can move
to only one direction.
This ensures that the algorithm works efficiently.

We will next discuss two problems that can be solved
using the two pointers method.

\subsubsection{Subarray sum}

Given an array that contains $n$ positive integers,
our task is to find out if there is a subarray
where the sum of the elements is $x$.
For example, the array
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
contains a subarray with sum 8:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

It turns out that the problem can be solved in
$O(n)$ time using the two pointers method.
The idea is to iterate through the array
using two pointers that define a range in the array.
On each turn, the left pointer moves one step
forward, and the right pointer moves forward
as long as the sum is at most $x$.
If the sum of the range becomes exactly $x$,
we have found a solution.

As an example, we consider the following array
with target sum $x=8$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

First, the pointers define a range with sum $1+3+2=6$.
The range can't be larger
because the next number 5 would make the sum
larger than $x$.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (0,0) rectangle (3,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\draw[thick,->] (0.5,-0.7) -- (0.5,-0.1);
\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

After this, the left pointer moves one step forward.
The right pointer doesn't move because otherwise
the sum would become too large.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (1,0) rectangle (3,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\draw[thick,->] (1.5,-0.7) -- (1.5,-0.1);
\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

Again, the left pointer moves one step forward,
and this time the right pointer moves three
steps forward.
The sum is $2+5+1=8$, so we have found a subarray
where the sum of the elements is $x$.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$2$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$1$};
\node at (5.5,0.5) {$1$};
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$3$};

\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

The time complexity of the algorithm depends on
the number of steps the right pointer moves.
There is no upper bound how many steps the
pointer can move on a single turn.
However, the pointer moves \emph{a total of}
$O(n)$ steps during the algorithm
because it only moves forward.

Since both the left and the right pointer
move $O(n)$ steps during the algorithm,
the time complexity is $O(n)$.

\subsubsection{Sum of two numbers}

\index{2SUM problem}

Given an array of $n$ integers and an integer $x$,
our task is to find two numbers in array
whose sum is $x$ or report that there are no such numbers.
This problem is known as the \key{2SUM} problem,
and it can be solved efficiently using the
two pointers method.

First, we sort the numbers in the array in
increasing order.
After this, we iterate through the array using
two pointers that begin at both ends of the array.
The left pointer begins from the first element
and moves one step forward on each turn.
The right pointer begins from the last element
and always moves backward until the sum of the range
defined by the pointers is at most $x$.
If the sum is exactly $x$, we have found a solution.

For example, consider the following array when
our task is to find two elements whose sum is $x=12$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$4$};
\node at (2.5,0.5) {$5$};
\node at (3.5,0.5) {$6$};
\node at (4.5,0.5) {$7$};
\node at (5.5,0.5) {$9$};
\node at (6.5,0.5) {$9$};
\node at (7.5,0.5) {$10$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

The initial positions of the pointers
are as follows.
The sum of the numbers is $1+10=11$
that is smaller than $x$.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (0,0) rectangle (1,1);
\fill[color=lightgray] (7,0) rectangle (8,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$4$};
\node at (2.5,0.5) {$5$};
\node at (3.5,0.5) {$6$};
\node at (4.5,0.5) {$7$};
\node at (5.5,0.5) {$9$};
\node at (6.5,0.5) {$9$};
\node at (7.5,0.5) {$10$};

\draw[thick,->] (0.5,-0.7) -- (0.5,-0.1);
\draw[thick,->] (7.5,-0.7) -- (7.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

Then the left pointer moves one step forward.
The right pointer moves three steps backward,
and the sum becomes $4+7=11$.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (1,0) rectangle (2,1);
\fill[color=lightgray] (4,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$4$};
\node at (2.5,0.5) {$5$};
\node at (3.5,0.5) {$6$};
\node at (4.5,0.5) {$7$};
\node at (5.5,0.5) {$9$};
\node at (6.5,0.5) {$9$};
\node at (7.5,0.5) {$10$};

\draw[thick,->] (1.5,-0.7) -- (1.5,-0.1);
\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

After this, the left pointer moves one step forward again.
The right pointer doesn't move, and the solution
$5+7=12$ has been found.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (3,1);
\fill[color=lightgray] (4,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$4$};
\node at (2.5,0.5) {$5$};
\node at (3.5,0.5) {$6$};
\node at (4.5,0.5) {$7$};
\node at (5.5,0.5) {$9$};
\node at (6.5,0.5) {$9$};
\node at (7.5,0.5) {$10$};

\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

At the beginning of the algorithm,
the sorting takes $O(n \log n)$ time.
After this, the left pointer moves $O(n)$ steps
forward, and the right pointer moves $O(n)$ steps
backward. Thus, the total time complexity
of the algorithm is $O(n \log n)$.

Note that it is possible to solve
in another way in $O(n \log n)$ time using binary search.
In this solution, we iterate through the array
and for each number, we try to find another
number such that the sum is $x$.
This can be done by performing $n$ binary searches,
and each search takes $O(\log n)$ time.

\index{3SUM-problem}
A somewhat more difficult problem is 
the \key{3SUM} problem where our task is
to find \emph{three} numbers whose sum is $x$.
This problem can be solved in $O(n^2)$ time.
Can you see how it is possible?

\section{Nearest smaller elements}

\index{nearest smaller elements}

Amortized analysis is often used for
estimating the number of operations
performed for a data structure.
The operations may be distributed unevenly so
that the most operations appear during a
certain phase in the algorithm, but the total
number of the operations is limited.

As an example, let us consider a problem
where our task is to find for each element
in an array the
\key{nearest smaller element}, i.e.,
the nearest smaller element that precedes
the element in the array.
It is possible that no such element exists,
and the algorithm should notice this.
It turns out that the problem can be efficiently solved
in $O(n)$ time using a suitable data structure.

An efficient solution for the problem is to
iterate through the array from the left to the right,
and maintain a chain of elements where the
first element is the active element in the array
and each following element is the nearest smaller
element of the previous element.
If the chain only contains one element,
the active element doesn't have a nearest smaller element.
At each step, we remove elements from the chain
until the first element is smaller
than the active element, or the chain is empty.
After this, the active element becomes the first element
in the chain.

As an example, consider the following array:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$5$};
\node at (5.5,0.5) {$3$};
\node at (6.5,0.5) {$4$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

First, numbers 1, 3 and 4 are added to the chain
because each element is larger than the previous element.
This means that the nearest smaller element of
number 4 is number 3 whose nearest smaller element
is number 1.
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (3,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$5$};
\node at (5.5,0.5) {$3$};
\node at (6.5,0.5) {$4$};
\node at (7.5,0.5) {$2$};

\draw[thick,->] (2.5,-0.25) .. controls (2.25,-1.00) and (1.75,-1.00) .. (1.6,-0.25);
\draw[thick,->] (1.4,-0.25) .. controls (1.25,-1.00) and (0.75,-1.00) .. (0.5,-0.25);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

The next number 2 is smaller than two first numbers in the chain.
Thus, numbers 4 and 3 are removed, and then number 2 becomes
the first element in the chain.
Its nearest smaller element is number 1:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (3,0) rectangle (4,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$5$};
\node at (5.5,0.5) {$3$};
\node at (6.5,0.5) {$4$};
\node at (7.5,0.5) {$2$};

\draw[thick,->] (3.5,-0.25) .. controls (3.00,-1.00) and (1.00,-1.00) .. (0.5,-0.25);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

After this, number 5 is larger than number 2,
so it will be added to the chain and
its nearest smaller element is number 2:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (4,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$1$};
\node at (1.5,0.5) {$3$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$2$};
\node at (4.5,0.5) {$5$};
\node at (5.5,0.5) {$3$};
\node at (6.5,0.5) {$4$};
\node at (7.5,0.5) {$2$};

\draw[thick,->] (3.4,-0.25) .. controls (3.00,-1.00) and (1.00,-1.00) .. (0.5,-0.25);
\draw[thick,->] (4.5,-0.25) .. controls (4.25,-1.00) and (3.75,-1.00) .. (3.6,-0.25);

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

Algorithm continues in a similar way
and finds out the nearest smaller element
for each number in the array.
But how efficient is the algorithm?

The efficiency of the algorithm depends on
the total time used for manipulating the chain.
If an element is larger than the first
element in the chain, it will only be inserted
to the beginning of the chain which is efficient.
However, sometimes the chain can contain several
larger elements and it takes time to remove them.
Still, each element is added exactly once to the chain
and removed at most once.
Thus, each element causes $O(1)$ operations
to the chain, and the total time complexity
of the algorithm is $O(n)$.

\section{Sliding window minimum}

\index{sliding window}
\index{sliding window minimum}

A \key{sliding window} is an active subarray
that moves through the array whose size is constant.
At each position of the window,
we typically want to calculate some information
about the elements inside the window.
An interesting problem is to maintain
the \key{sliding window minimum}.
This means that at each position of the window,
we should report the smallest element inside the window.

The sliding window minima can be calculated
using the same idea that we used for calculating
the nearest smaller elements.
The idea is to maintain a chain whose
first element is the last element in the window,
and each element is smaller than the previous element.
The last element in the chain is always the
smallest element inside the window.
When the sliding window moves forward and
a new element appears, we remove all elements
from the chain that are larger than the new element.
After this, we add the new number to the chain.
In addition, if the last element in the chain
doesn't belong to the window anymore, it is removed from the chain.

As an example, consider the following array
when the window size is $k=4$:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}

The sliding window begins from the left border of the array.
At the first window position, the smallest element is 1:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (0,0) rectangle (4,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};

\draw[thick,->] (3.5,-0.25) .. controls (3.25,-1.00) and (2.75,-1.00) .. (2.6,-0.25);
\draw[thick,->] (2.4,-0.25) .. controls (2.25,-1.00) and (1.75,-1.00) .. (1.5,-0.25);

\end{tikzpicture}
\end{center}

Then the window moves one step forward.
The new number 3 is smaller than the numbers
5 and 4 in the chain, so the numbers 5 and 4
are removed and the number 3 is added to the chain.
The smallest element is 1 as before.
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (1,0) rectangle (5,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};

\draw[thick,->] (4.5,-0.25) .. controls (4.25,-1.00) and (1.75,-1.00) .. (1.5,-0.25);

\end{tikzpicture}
\end{center}

After this, the window moves again and the smallest element 1
doesn't belong to the window anymore.
Thus, it is removed from the chain and the smallest
element is now 3. In addition, the new number 4
is added to the chain.
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (2,0) rectangle (6,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};

\draw[thick,->] (5.5,-0.25) .. controls (5.25,-1.00) and (4.75,-1.00) .. (4.5,-0.25);
\end{tikzpicture}
\end{center}

The next new element 1 is smaller than all elements
in the chain.
Thus, all elements are removed from the chain
and it will only contain the element 1:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (3,0) rectangle (7,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};

\fill[color=black] (6.5,-0.25) circle (0.1);

%\draw[thick,->] (5.5,-0.25) .. controls (5.25,-1.00) and (4.75,-1.00) .. (4.5,-0.25);
\end{tikzpicture}
\end{center}

Finally the window reaches its last position.
The number 2 is added to the chain,
but the smallest element inside the window
is still 1.
\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (4,0) rectangle (8,1);
\draw (0,0) grid (8,1);

\node at (0.5,0.5) {$2$};
\node at (1.5,0.5) {$1$};
\node at (2.5,0.5) {$4$};
\node at (3.5,0.5) {$5$};
\node at (4.5,0.5) {$3$};
\node at (5.5,0.5) {$4$};
\node at (6.5,0.5) {$1$};
\node at (7.5,0.5) {$2$};

\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};

\draw[thick,->] (7.5,-0.25) .. controls (7.25,-1.00) and (6.75,-1.00) .. (6.5,-0.25);
\end{tikzpicture}
\end{center}

Also in this algorithm, each element in the array
is added to the chain exactly once and
removed from the chain at most once.
Thus, the total time complexity of the algorithm is $O(n)$.



