ARRAYS

1
ARRAYS

• Lecture 14
• 11.2.2002.

2

• void reverse (int x, int size)
• int i
• for (i0 ilt (size/2) i)
• temp xsize-i-1
• xsize-1-1 xi
• xi temp

int findmax (int x, int size) int i,
max max x0 for (i1 ilt size
i) if (xi gt max) max xi
return max
3
int findmax (int x, int size) if (size
1) return x0 maxl findmax (x, size-1)
if (x0 gt max) max x0 return
max
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Strings

• Strings are 1-dimensional arrays of type char.
• By convention, a string in C is terminated by the
  end-of-string sentinel \0, or null character.
• String constant abc is a character array of
  size 4, with the last element being the null
  chaaracter \0.
• char s abc

a
b
c
\0
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Searching an ArrayLinear and Binary Search
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Searching

• Check if a given element (key) occurs in the
  array.
• If the array is unsorted
• start at the beginning of the array
• inspect every element to see if it matches the key

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Linear Search

• / If key appears in a0..size-1, return its
  location, pos, s.t. apos key. If key is not
  found, return -1 /
• int search (int a, int size, int key)
• int pos 0
• while (pos lt size apos ! key)
• pos
• if (posltn)
• return pos
• return -1

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Linear Search

• int x 12,-3, 78,67,6,50,19,10
• Trace the following calls
• search (x, 8,6)
• search (x,8,5)

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Searching a sorted array

• Binary search works if the array is sorted
• Look for the target in the middle
• If you dont find it, you can ignore half of the
  array, and repeat the process with the other half.

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Binary Search Strategy

• What we want Find split betwen values larger
  and smaller than x

0
L
R
x
n
ltkey
gtkey

• Situation while searching

0
n
L
R
x
ltkey
?
gtkey

• Step Look at (LR)/2. Move L or R to the
  middle depending on test.

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Binary Search

• / If key appears in x0..size-1, return its
  location, pos s.t. xposkey. If not found,
  return -1 /
• int binsearch (int x, int size, int key)
• int L, R, mid
• ______________________
• while (_________________)
• ____________________

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Binary Search

• / If key appears in x0..size-1, return its
  location, pos s.t. xposkey. If not found,
  return -1 /
• int binsearch (int x, int size, int key)
• int L, R, mid
• ______________________
• while (_________________)
• mid (LR)/2
• if (xmid lt key)
• L mid
• else R mid
• ____________________

mid (LR)/2 if (xmid lt key) L
mid else R mid
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Binary Search loop termination

• / If key appears in x0..size-1, return its
  location, pos s.t. xposkey. If not found,
  return -1 /
• int binsearch (int x, int size, int key)
• int L, R, mid
• ______________________
• while (_________________)
• mid (LR)/2
• if (xmid lt key)
• L mid
• else R mid
• ____________________

L1 ! R
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Binary Search Return result

• / If key appears in x0..size-1, return its
  location, pos s.t. xposkey. If not found,
  return -1 /
• int binsearch (int x, int size, int key)
• int L, R, mid
• ______________________
• while ( L1 ! R)
• mid (LR)/2
• if (xmid lt key)
• L mid
• else R mid

if (Lgt0 xLkey) return L else return -1
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Binary Search Initialization

• / If key appears in x0..size-1, return its
  location, pos s.t. xposkey. If not found,
  return -1 /
• int binsearch (int x, int size, int key)
• int L, R, mid
• ______________________
• while ( L1 ! R)
• mid (LR)/2
• if (xmid lt key)
• L mid
• else R mid
• if (Lgt0 xLkey) return L
• else return -1

L-1 Rsize
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Binary Search Examples
-17 -5 3 6 12 21 45 63 50
Trace binsearch (x, 9, 3) binsearch (x, 9,
145) binsearch (x, 9, 45)
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Is it worth the trouble ?

• Suppose you had 1000 elements
• Ordinary search (if key is a member of x) would
  require 500 comparisons on average.
• Binary search
• after 1st compare, left with 500 elements
• after 2nd compare, left with 250 elements
• After at most 10 steps, you are done.
• What if you had 1 million elements ?

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Sorting

• Given an array x0, x1, ... , xsize-1
• reorder entries so that
• x0ltx1lt . . . ltxsize-1

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Sorting Problem

• What we want Data sorted in order
• sorted x0ltx1lt . . . ltxsize-1

size
0
x

• Initial conditions

size
0
unsorted
x
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Selection Sort

• General situation

k
0
size
x
remainder, unsorted
smallest elements, sorted

• Step
• Find smallest element, mval, in xk..size-1
• Swap smallest element with xk, then increase
  k.

mval
0
size
k
x
smallest elements, sorted
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Subproblem

• / Yield location of smallest element intx in
  x0 .. size-1/
• int min_loc (int x, int , int size)
• int j, pos / xpos is the smallest element
  found so far /
• pos k
• for (jk1 jltsize j)
• if (xi lt xpos)
• pos j
• return pos

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Selection Sort

• / Sort x0..size-1 in non-decreasing order /
• int selsort (int x, int size)
• int k, m
• for (k0 kltsize-1 k)
• m min_loc(x, k, size)
• temp ak
• ak am
• am temp

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Example
x
x
3
12
-5
6
142
21
-17
45
-17
-5
3
6
12
21
142
45
x
-17
12
-5
6
142
21
3
45
x
-17
-5
3
6
12
21
45
142
x
-17
-5
12
6
142
21
3
45
x
-17
-5
3
6
142
21
12
45
x
-17
-5
3
6
12
21
142
45
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Analysis

• How many steps are needed to sort n things ?
• Total number of steps proportional to n2

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Insertion Sort

• define MAXN 100
• void InsertSort (int listMAXN, int size)
• main ()
• int index, size
• int numbersMAXN
• / Get Input /
• size readarray (numbers)
• printarray (numbers, size)
• InsertSort (numbers, size)
• printarray (numbers, size)

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• void InsertSort (int list, int size)
• for (i1 iltsize i)
• item listi
• for (ji-1 (jgt0) (listj gt i) j--)
• listj1 listj
• listj1 item

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Common pitfalls with arrays in C

• Exceeding the array bounds
• int array10for (i0 ilt10 i) arrayi
  0
• C does not support array declaratiions with
  variable expressions.
• void fun (int array, int size) int
  tempsize . . .