CS1010E Programming Methodology
Tutorial 7
Arrays and Matrices
Question 1 (a)
 Reverse a portion of a list
 void reverse(int list[],int size,int start,int n);
 Algorithm:
1. find start and end
2. exchange list[start] and list[end]
3. move start to the right and list to the left
4. stop when start >= end
List:
1
2
Reverse(list, 8, 2, 4)
3
4
5
6
7
8
Question 1 (a)
 Reverse a portion of a list
 void reverse(int list[],int size,int start,int n);
 Algorithm:
1. find start and end
2. exchange list[start] and list[end]
3. move start to the right and list to the left
4. stop when start >= end
List:
1
2
3
4
Start =2
Reverse(list, 8, 2, 4)
5
6
7
8
end = start + n -1 = 5
Question 1 (a)
 Reverse a portion of a list
 void reverse(int list[],int size,int start,int n);
 Algorithm:
1. find start and end
2. exchange list[start] and list[end]
3. move start to the right and list to the left
4. stop when start >= end
List:
1
2
Reverse(list, 8, 2, 4)
6
4
start
5
3
end
7
8
Question 1 (a)
 Reverse a portion of a list
 void reverse(int list[],int size,int start,int n);
 Algorithm:
1. find start and end
2. exchange list[start] and list[end]
3. move start to the right and list to the left
4. stop when start >= end
List:
1
2
Reverse(list, 8, 2, 4)
6
5
start
4
3
end
7
8
Start >= end, stop!
Question 1 (a)
void reverse(int list[], int size, int start, int n) {
int end = start + n - 1;
int tmp;
while (start <= end) {
tmp = list[start];
list[start] = list[end];
list[end] = tmp;
start++;
end--;
}
}
Code for question 1 (a)
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm I:
 for each element list2[i]
 insert it into list at position index + i
 insertion can be done by rightshift algorithm from last
tutorial
list
list2
1
2
3
4
5
Merge
6
7
8
1
2
3
4
5
6
7
9
rightshift this range
insert(list,5,list2,4,2)
8
9
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm I:
 for each element list2[i]
 insert it into list at position index + i
 insertion can be done by rightshift algorithm from last
tutorial
list
list2
1
2
3
4
5
Merge
6
7
8
1
2
6
3
4
5
7
8
9
rightshift this range
insert(list,5,list2,4,2)
9
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm I:
 for each element list2[i]
 insert it into list at position index + i
 insertion can be done by rightshift algorithm from last
tutorial
list
list2
1
2
3
4
5
Merge
6
7
8
1
2
6
7
3
4
5
8
9
9
rightshift this range
insert(list,5,list2,4,2)
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm I:
 for each element list2[i]
 insert it into list at position index + i
 insertion can be done by rightshift algorithm from last
tutorial
list
list2
1
2
3
4
5
Merge
6
7
8
1
2
6
7
8
3
4
5
9
9
rightshift this range
insert(list,5,list2,4,2)
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm I:
 for each element list2[i]
 insert it into list at position index + i
 insertion can be done by rightshift algorithm from last
tutorial
list
list2
1
2
3
4
5
Merge
6
7
8
1
2
6
7
8
9
9
done
insert(list,5,list2,4,2)
3
4
5
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm II:
 move every element in list[index, size]
 list[i+size2] = list[i];
 insert it into list at position index + I
Copy list2 into position
 use listcopy from last tutorial
list
1 2 3 4 5 0 0 0 0
list2
6
7
8
9
insert(list,5,list2,4,2)
1 2 3 4 5 0 3 4 5
Move elements
Question 1 (b)
 int insert(int list[],int size,int list2[],int size2,int index);
 Algorithm II:
 move every element in list[index, size]
 list[i+size2] = list[i];
 insert it into list at position index + I
Copy list2 into position
 use listcopy from last tutorial
list
1 2 3 4 5 0 0 0 0
1 2 3 4 5 0 3 4 5
Move elements
list2
6
7
8
9
insert(list,5,list2,4,2)
1 2 6 7 8 9 3 4 5
listcopy(list2, list + 2, 4)
Question 1 (c)
 Delete a rang of element within a list
 int delete(int list[],int size,int index,int n);
 Algorithm:
 shift element from list[index+n, size-1] to left
 list[i-n] = list[i];
list
1
2
3
4
to be deleted
5
6
7
8
9
1
5
3
[4, 8] to be shift left
i-3 =1 i=4
delete(1,3)
4
5
6
7
8
9
Question 1 (c)
 Delete a rang of element within a list
 int delete(int list[],int size,int index,int n);
 Algorithm:
 shift element from list[index+n, size-1] to left
 list[i-n] = list[i];
list
1
2
3
4
to be deleted
5
6
7
8
9
1
5
6
[4, 8] to be shift left
i-3 =2 i=5
delete(1,3)
4
5
6
7
8
9
Question 1 (c)
 Delete a rang of element within a list
 int delete(int list[],int size,int index,int n);
 Algorithm:
 shift element from list[index+n, size-1] to left
 list[i-n] = list[i];
list
1
2
3
4
to be deleted
5
6
7
8
9
1
5
6
[4, 8] to be shift left
i-3 =3 i=6
delete(1,3)
7
5
6
7
8
9
Question 1 (c)
 Delete a rang of element within a list
 int delete(int list[],int size,int index,int n);
 Algorithm:
 shift element from list[index+n, size-1] to left
 list[i-n] = list[i];
list
1
2
3
4
to be deleted
5
6
7
8
9
1
5
6
[4, 8] to be shift left
i-3 =4 i=7
delete(1,3)
7
8
6
7
8
9
Question 1 (c)
 Delete a rang of element within a list
 int delete(int list[],int size,int index,int n);
 Algorithm:
 shift element from list[index+n, size-1] to left
 list[i-n] = list[i];
list
1
2
3
4
to be deleted
5
6
7
8
9
1
5
6
8
9
7
[4, 8] to be shift left
i-3 =5 i=8
delete(1,3)
7
Done!
8
9
Question 2
ID
 Score is represented by a 2-D array (Matrix)
 let’s call the matrix as student[][3]
student[3]
 student[i]
 is a 1-D array of length 3
 Two problems:
1. Reading in matrix
2. Sorting matrix based on Mark
student[9]
196
179
158
198
146
165
134
118
159
136
171
137
154
175
157
188
106
135
151
164
112
186
173
189
Mark
46
92
56
32
31
73
67
54
55
59
34
98
76
57
76
57
37
70
47
33
59
80
73
9
Grade
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
Question 2
 Reading in matrix
int students[NUM_STUDENTS][3];
int matric, mark;
for(i = 0; i < NUM_STUDENTS; i++){
//Scans for data, saves the
into 2 columns.
scanf("%d %d", &matric,
&mark);
students[i][0] = matric;
students[i][1] = mark;
}
for ( c = 0 ; c < ( n - 1 ) ; c++ ) { position = c; for ( d = c + 1 ; d < n ; d++ ) { if ( array[position] > array[d] ) position = d; } if ( position != c ) { swap = array[c]; array[c] =
Question 2
 Sorting Matrix by Mark (student[i][1])
 Similar to 1-D sorting
for (i = 0 ; i < ( n - 1 ); i++) {
for (j = 0 ; j < n - i - 1; j++) {
if (array[j] > array[j+1]) {
temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
}
}
}
Bubble sort for 1-D
The difference
between 1-D and 2-D
is in:
Compare
SWAP
Question 2
 Sorting Matrix by Mark (student[i][1])
 Compare two students’ mark
 student[i][1] < student[j][1]
 Swap the records of two students
student i
198
32
?
student j
136
59
?
temp
int tmp[2];
tmp[0][0]=student[j][0];
tmp[0][1]=student[j][1]
Question 2
 Sorting Matrix by Mark (student[i][1])
 Compare two students’ mark
 student[i][1] < student[j][1]
 Swap the records of two students
student i
198
32
?
student j
136
59
?
136
59
temp
student[j][0] = student[i][0];
student[j][1] = student[i][1];
Question 2
 Sorting Matrix by Mark (student[i][1])
 Compare two students’ mark
 student[i][1] < student[j][1]
 Swap the records of two students
student i
136
59
?
student j
198
32
?
temp
136
59
student[i][0] = tmp[0];
student[i][1] = tmp[1];
Question 2
 Sorting the matrix:
for (i=0;i<(n-1); i++) {
for (j=0; j<n-i-1;j++) {
if (array[j]>array[j+1]) {
temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
}
}
}
1-D Bubble Sort
for (i=0;i<(n-1); i++) {
for (j=0; j<n-i-1;j++) {
if (student[j][1]>student[j+1][1]) {
temp[0] = array[j][0];
temp[1] = array[j][1];
student[j][0] = student[j+1][0];
student[j][1] = student[j+1][1];
array[j+1][0] = temp[0];
array[j+1][1] = temp[1];
}
}
}
2-D Bubble Sort
Question 3
 Summing up surroundings:
 Problem Solving:
 Step 1: What is input and what is output?
 Step 2: How to store the input and output?
 Store row and column into two variables: width, height
 Store M*N digits into one 2D array - board[][]
 Store M*N digits into another 2D array - result[][]
 Step 3: How to derive the output based on the input?
 Each cell in result[][] equals the sum of the eight adjacent
cells from board[][]
Question 3
 Algorithm:
 For each mine[i][j], summing up its surroundings:
 mine[i-1][j] + mine[i+1][j] + mine[i][j-1] + mine[i][j+1]
 Write output to a new matrix
 result[i][j] = mine[i-1][j] + mine[i+1][j] + mine[i][j-1] +
mine[i][j+1];