841 lines
22 KiB
TeX
841 lines
22 KiB
TeX
\chapter{Amortized analysis}
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\index{amortized analysis}
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The time complexity of an algorithm
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is often easy to analyze
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just by examining the structure
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of the algorithm:
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what loops does the algorithm contain
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and how many times the loops are performed.
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However, sometimes a straightforward analysis
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does not give a true picture of the efficiency of the algorithm.
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\key{Amortized analysis} can be used for analyzing
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algorithms that contain operations whose
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time complexity varies.
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The idea is to estimate the total time used for
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all such operations during the
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execution of the algorithm, instead of focusing
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on individual operations.
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\section{Two pointers method}
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\index{two pointers method}
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In the \key{two pointers method},
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two pointers are used for
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iterating through the elements in an array.
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Both pointers can move during the algorithm,
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but each pointer can move to one direction only.
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This restriction ensures that the algorithm works efficiently.
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We will next discuss two problems that can be solved
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using the two pointers method.
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\subsubsection{Subarray sum}
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As the first example,
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we consider a problem where we are
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given an array of $n$ positive numbers
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and a target sum $x$,
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and we should find a subarray whose sum is $x$
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or report that there is no such subarray.
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For example, the array
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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contains a subarray whose sum is 8:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (2,0) rectangle (5,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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It turns out that the problem can be solved in
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$O(n)$ time by using the two pointers method.
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The idea is to use
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left and right pointers that indicate the
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first and last element of an subarray.
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On each turn, the left pointer moves one step
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forward, and the right pointer moves forward
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as long as the subarray sum is at most $x$.
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If the sum becomes exactly $x$,
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a solution has been found.
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As an example, consider the following array
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with target sum $x=8$:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Initially, the subarray contains the elements
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1, 3 and 2, and the sum of the subarray is 6.
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The subarray cannot be larger, because
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the next element 5 would make the sum larger than $x$.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (0,0) rectangle (3,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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\draw[thick,->] (0.5,-0.7) -- (0.5,-0.1);
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\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Then, the left pointer moves one step forward.
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The right pointer does not move, because otherwise
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the sum would become too large.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (1,0) rectangle (3,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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\draw[thick,->] (1.5,-0.7) -- (1.5,-0.1);
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\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Again, the left pointer moves one step forward,
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and this time the right pointer moves three
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steps forward.
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The sum is $2+5+1=8$, so we have found a subarray
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where the sum of the elements is $x$.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (2,0) rectangle (5,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$2$};
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\node at (3.5,0.5) {$5$};
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\node at (4.5,0.5) {$1$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$2$};
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\node at (7.5,0.5) {$3$};
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\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
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\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The time complexity of the algorithm depends on
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the number of steps the right pointer moves.
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There is no useful upper bound how many steps the
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pointer can move on a single turn.
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However, the pointer moves \emph{a total of}
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$O(n)$ steps during the algorithm,
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because it only moves forward.
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Since both the left and right pointer
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move $O(n)$ steps during the algorithm,
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the time complexity is $O(n)$.
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\subsubsection{2SUM problem}
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\index{2SUM problem}
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Another problem that can be solved using
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the two pointers method is the following problem,
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also known as the \key{2SUM problem}:
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we are given an array of $n$ numbers and
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a target sum $x$, and our task is to find two numbers
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in the array such that their sum is $x$,
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or report that no such numbers exist.
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To solve the problem, we first
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sort the numbers in the array in increasing order.
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After that, we iterate through the array using
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two pointers.
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The left pointer starts at the first element
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and moves one step forward on each turn.
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The right pointer begins at the last element
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and always moves backward until the sum of the
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elements is at most $x$.
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If the sum of the elements is exactly $x$,
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a solution has been found.
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For example, consider the following array
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with target sum $x=12$:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$6$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$9$};
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\node at (6.5,0.5) {$9$};
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\node at (7.5,0.5) {$10$};
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The initial positions of the pointers
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are as follows.
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The sum of the numbers is $1+10=11$
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that is smaller than $x$.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (0,0) rectangle (1,1);
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\fill[color=lightgray] (7,0) rectangle (8,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$6$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$9$};
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\node at (6.5,0.5) {$9$};
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\node at (7.5,0.5) {$10$};
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\draw[thick,->] (0.5,-0.7) -- (0.5,-0.1);
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\draw[thick,->] (7.5,-0.7) -- (7.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Then the left pointer moves one step forward.
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The right pointer moves three steps backward,
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and the sum becomes $4+7=11$.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (1,0) rectangle (2,1);
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\fill[color=lightgray] (4,0) rectangle (5,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$6$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$9$};
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\node at (6.5,0.5) {$9$};
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\node at (7.5,0.5) {$10$};
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\draw[thick,->] (1.5,-0.7) -- (1.5,-0.1);
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\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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After this, the left pointer moves one step forward again.
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The right pointer does not move, and a solution
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$5+7=12$ has been found.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (2,0) rectangle (3,1);
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\fill[color=lightgray] (4,0) rectangle (5,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$6$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$9$};
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\node at (6.5,0.5) {$9$};
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\node at (7.5,0.5) {$10$};
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\draw[thick,->] (2.5,-0.7) -- (2.5,-0.1);
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\draw[thick,->] (4.5,-0.7) -- (4.5,-0.1);
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%
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% \footnotesize
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% \node at (0.5,1.4) {$1$};
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% \node at (1.5,1.4) {$2$};
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% \node at (2.5,1.4) {$3$};
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% \node at (3.5,1.4) {$4$};
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% \node at (4.5,1.4) {$5$};
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% \node at (5.5,1.4) {$6$};
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% \node at (6.5,1.4) {$7$};
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% \node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The algorithm consists of two phases:
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First, sorting the array takes $O(n \log n)$ time.
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After this, the left pointer moves $O(n)$ steps
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forward, and the right pointer moves $O(n)$ steps
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backward. Thus, the total time complexity
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of the algorithm is $O(n \log n)$.
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Note that it is possible to solve the problem
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in another way in $O(n \log n)$ time using binary search.
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In such a solution, we iterate through the array
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and for each number, we try to find another
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number such that the sum is $x$.
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This can be done by performing $n$ binary searches,
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and each search takes $O(\log n)$ time.
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\index{3SUM problem}
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A more difficult problem is
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the \key{3SUM problem} that asks to
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find \emph{three} numbers in the array
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such that their sum is $x$.
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Using the idea of the above algorithm,
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this problem can be solved in $O(n^2)$ time\footnote{For a long time,
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it was thought that solving
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the 3SUM problem more efficiently than in $O(n^2)$ time
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would not be possible.
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However, in 2014, it turned out \cite{gro14}
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that this is not the case.}.
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Can you see how?
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\section{Nearest smaller elements}
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\index{nearest smaller elements}
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Amortized analysis is often used for
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estimating the number of operations
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performed on a data structure.
|
|
The operations may be distributed unevenly so
|
|
that most operations occur during a
|
|
certain phase of the algorithm, but the total
|
|
number of the operations is limited.
|
|
|
|
As an example, consider the problem
|
|
of finding for each element
|
|
of an array the
|
|
\key{nearest smaller element}, i.e.,
|
|
the first smaller element that precedes
|
|
the element in the array.
|
|
It is possible that no such element exists,
|
|
in which case the algorithm should report this.
|
|
It turns out that the problem can be solved
|
|
in $O(n)$ time using an appropriate data structure.
|
|
|
|
An efficient solution to the problem is to
|
|
iterate through the array from left to right
|
|
and maintain a chain of elements where the
|
|
first element is the current element
|
|
and each following element is the nearest smaller
|
|
element of the previous element.
|
|
If the chain only contains one element,
|
|
the current element does not have a nearest smaller element.
|
|
At each step, elements are removed from the chain
|
|
until the first element is smaller
|
|
than the current element, or the chain is empty.
|
|
After this, the current element is added to 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, the numbers 1, 3 and 4 are added to the chain,
|
|
because each number is larger than the previous number.
|
|
Thus, the nearest smaller element of 4 is 3,
|
|
and the nearest smaller element of 3 is 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 the two first numbers in the chain.
|
|
Thus, the numbers 4 and 3 are removed from the chain,
|
|
and then the number 2
|
|
is added to the chain.
|
|
Its nearest smaller element is 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, the number 5 is larger than the number 2,
|
|
so it will be added to the chain, and
|
|
its nearest smaller element is 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}
|
|
|
|
The algorithm continues in the same way
|
|
and finds 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 of the chain, it is directly added
|
|
to 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 from the chain.
|
|
Thus, each element causes $O(1)$ chain operations,
|
|
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 a constant-size subarray
|
|
that moves through the array.
|
|
At each position of the window,
|
|
we want to calculate some information
|
|
about the elements inside the window.
|
|
An interesting problem is to maintain
|
|
the \key{sliding window minimum},
|
|
which means that at each position of the window,
|
|
we should report the smallest element inside the window.
|
|
|
|
The sliding window minimum can be calculated
|
|
using the same idea that we used for calculating
|
|
the nearest smaller elements.
|
|
We maintain a chain whose
|
|
first element is always the last element in the window,
|
|
and each element in the chain 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 from the chain
|
|
all elements that are larger than the new element.
|
|
After this, we add the new element to the chain.
|
|
Finally, if the last element in the chain
|
|
does not belong to the window anymore,
|
|
it is removed from 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) {$2$};
|
|
\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}
|
|
|
|
Suppose that the size of the sliding window is 4.
|
|
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 from the chain
|
|
and the number 3 is added to the chain.
|
|
The smallest element is still 1.
|
|
\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
|
|
does not 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 $O(n)$.
|
|
|
|
|
|
|