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Improve language

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Antti H S Laaksonen 2017-05-25 23:05:43 +03:00
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chapter08.tex
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@ -74,7 +74,7 @@ contains a subarray whose sum is 8:
This problem can be solved in This problem can be solved in
$O(n)$ time by using the two pointers method. $O(n)$ time by using the two pointers method.
The idea is maintain pointers that point to the The idea is to maintain pointers that point to the
first and last value of a subarray. first and last value of a subarray.
On each turn, the left pointer moves one step On each turn, the left pointer moves one step
to the right, and the right pointer moves to the right to the right, and the right pointer moves to the right
@ -171,9 +171,9 @@ whose sum is $x$ has been found.
The running time of the algorithm depends on The running time of the algorithm depends on
the number of steps the right pointer moves. the number of steps the right pointer moves.
There is no useful upper bound how many steps the While there is no useful upper bound on how many steps the
pointer can move on a single turn. pointer can move on a \emph{single} turn.
However, the pointer moves \emph{a total of} we know that the pointer moves \emph{a total of}
$O(n)$ steps during the algorithm, $O(n)$ steps during the algorithm,
because it only moves to the right. because it only moves to the right.
@ -188,8 +188,8 @@ the algorithm works in $O(n)$ time.
Another problem that can be solved using Another problem that can be solved using
the two pointers method is the following problem, the two pointers method is the following problem,
also known as the \key{2SUM problem}: also known as the \key{2SUM problem}:
we are given an array of $n$ numbers and given an array of $n$ numbers and
a target sum $x$, and our task is to find a target sum $x$, find
two array values such that their sum is $x$, two array values such that their sum is $x$,
or report that no such values exist. or report that no such values exist.
@ -335,8 +335,8 @@ certain phase of the algorithm, but the total
number of the operations is limited. number of the operations is limited.
As an example, consider the problem As an example, consider the problem
of finding the \key{nearest smaller element} of finding for each array element
of each array element, i.e., the \key{nearest smaller element}, i.e.,
the first smaller element that precedes the element the first smaller element that precedes the element
in the array. in the array.
It is possible that no such element exists, It is possible that no such element exists,
@ -558,11 +558,11 @@ always corresponds to the minimum element inside the window.
After each window move, After each window move,
we remove elements from the end of the queue we remove elements from the end of the queue
until the last queue element until the last queue element
is smaller than the last window element, is smaller than the new window element,
or the queue becomes empty. or the queue becomes empty.
We also remove the first queue element We also remove the first queue element
if it is not inside the window anymore. if it is not inside the window anymore.
Finally, we add the last window element Finally, we add the new window element
to the end of the queue. to the end of the queue.
As an example, consider the following array: As an example, consider the following array: