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[CHAP. 5
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The second and third runs test boundary values; i.e., the extreme situations. One extreme is where the new item is larger than all the elements in the array. This is tested in the second run by inserting 400. The other extreme is where the new item is smaller than all the elements in the array. This is tested in the third run by inserting 200.
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5.10 Write and test the function
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int frequency(float
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a[], int n, int x)
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This function counts the number of times the item x appears among the first n elements of the array a and returns that count as the frequency of x in a.
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Here we initialize the array a
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with 40 randomly arranged integers to test the function
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int frequency(float [I, int, int); main0 { float a[]
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= (561, 508, 329, 375, 455, 508, 561, 455, int n = 40, x; tout CC "Item: 1'; tin >> x; tout << 'The frequency CC frequency(a, n,
400, 301, 329, 599, 455, 400, 346, 346, 561, 390, 399, 400, 401, 561, 405, 405, 473, 329, 561, 505, 329, 455, 561, 599, 346, 301, 455, 561, 399, 599, 508, 508);
of item ' CC x CC ' is I' x) CC endl;
int frequency(float
a[], int n, int x)
int count = 0; for (int i = 0; i c n; i++) if (a[i] == x) ++count; return count;
The function uses a counter count . It simply compares each element of the array with the item x and increments the counter each time a match is found.
CHAP. 51
ARRAYS
Implement the Insertion Sort. In this algorithm, the main loop runs from 1 to n- 1. On the ith iteration, the element a [ i ] is inserted into its correct position among the sub-array from a [ o ] t o a [ i ] . This is done by shifting all the elements in the sub-array that are greater than a [ i] one position to the right. Then a [ i ] is copied into the gap between the elements less than a [ i ] and those greater. (See Problem 5.9.)
Out test driver initializes the array a with 8 numbers in random order:
void print(float [I, const int); void sort(float [I, const int); main0 -i float a[81 = (88.8, print(a, 8); sort(a, 8); print(a, 8);
44.4,
77.7,
11.1,
33.3,
99.9,
66.6, 22.2);
void print(float a[], const int n) -t for (int i = 0; i < n-l; i++) { tout CC a[i] CC 'I, 1'; if ((i+1)%16 == 0) tout CC endl; > tout CC a[n-l] CC endl;
// Insertion Sort: void sort(float a[], const
in,t
float temp; for (int i = 1; i c n; i++) { // sort {a[O],...,a[i]}: temp = a[i]; for (int j = i; j > 0 SC& a[j-11 > temp; j--) a[j] = a[j-11; a[j] = temp;
On the ith iteration of the main loop of the Insertion Sort inserts, element a [ i ] is inserted so that the sub-array {a [ 0 ] , . . ., a [ i ] } will be in increasing order. This is done by storing a [ i ] tem= porarily in temp and then using the inner loop to shift the larger elements to the right with a [ j ] a[j-l].Thentempcanbecopiedintotheelement a[j].Notethat a[k] < a[j] forall k 5 j, and a[j] < a[k] for j < k 5 i.Thisensuresthatthesub-array{a[O],...,a[i]}issorted. When the last iteration of the main loop is finished, i = = n - 1, so {a [ 0 ] , . . ., a [ n- 1 ] } is sorted.
ARRAYS
[CHAP. 5
5.12 Rewrite and test the Bubble Sort function presented in Example 5.12, as an indirect sort. Instead
of moving the actual elements of the array, sort an index array instead.
The test driver requires a test array a initialized with some random numbers and an index array. initialized with index [ i ] = = i. This ensures that a [ index [ i ] ] will be the same as a [ i ] initially: void print(const float a[], const int n); void sort(float a[], int index[], int n); void print(const float a[], int index[], const int n); main0 1 float a[8] = (55, 22, 99, 66, 44, 88, 33, 77); int index[8] = (0, 1, 2, 3, 4, 5, 6, 7); print(a, 8); sort(a, index, 8); print(a, index, 8); print(a, 8);
void swap(int&,
int&);
// Indirect Bubble Sort: void sort(float a[], int index[], int n) 1 for (int i = 1; i c n; i++) for (int j = 0; j c n-i; j++> if (a[index[j]] > a[index[j+l]]) swap(index[j],index[j+l]);
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