Consultation Hours
• Mubashir:
– Tuesday from 12:30 to 1:30 with ease of Students
.
• Zohaib
– Wednesday b/w 9:30 -10:30
• Location: TA Room (next to HOD Office)
Arrays
Arrays
• There are situations in which we would want to store
more than one value at a time in a single variable.
• For example, suppose we wish to arrange the
percentage marks obtained by 100 students in
ascending order.
– Construct 100 variables to store percentage marks
obtained by 100 different students, i.e. each variable
containing one student’s marks.
– Construct one variable (called array or subscripted
variable) capable of storing or holding all the hundred
values.
• it would be much easier to handle one
variable than handling 100 different variables.
Moreover, there are certain logics that cannot
be dealt with, without the use of an array.
Array Declaration and Initialization in C
type variable_name[lengthofarray];
double height[10];
float width[20];
int min[9];
char name[20];
int num[6]={2,4,12,5,45,5};
int n[]={2,4,12,5,45,5};
float p[]={1.5,2.5,3};
int k[3];
k[0]=1;
k[1]=2;
k[2]=3;
All elements of the array should be of the
same type
In C Language, arrays starts at position 0.
Array Declaration and Initialization
• The elements of the array occupy adjacent
locations in memory.
Arrays - No bounds checking
// An incorrect program. Do Not Execute!
int main()
{
int crash[10], i;
for(i=0; i<100; i++)
crash[i]=i;
return 1;
}
9
Linear/Sequential Search
int a[]={1,2,5,4,7};
int key=4,n=5,flag=0;
for(i=0; i<= n-1; i++){
if(a[i] == key){
flag=1;
break;
}
}
if(flag == 0)
printf("\nThe number is not in the list");
else
printf("\nThe number is found");
Finding Min and Max Number
// start min and max off as equal to the first number
min = nums[0];
max = nums[0];
// iterate through nums
int i;
for(i = 1; i < nums_length; ++i) {
// update max, if necessary
if( nums[i] > max ) {
max = nums[i];
}
// update min, if necessary
if(nums[i] < min) {
min = nums[i];
}
}
Bubble Sort
Third Iteration
Bubble Sort
for(i=0; i<n-1; i++){ // number of iterations
for(j=0; j<(n-i-1); j++){ // number of pair wise comparisons
if(array[j]>array[j+1]){
temp = array[j+1];
array[j+1] = array[j];
array[j] = temp;
}
}
}
Bubble Sort
1. for(i=0; i<n-1; i++){ // number of iterations
2. for(j=0; j<(n-i-1); j++){ // number of pair wise comparisons
3. if(array[j]>array[j+1]){
4. temp = array[j+1];
5. array[j+1] = array[j];
6. array[j] = temp;
}
}
}
1.
Step i
1 0
for(i=0; i<n-1; i++){
2 0
3 0
4,5,6
0
2. for(j=0; j<(n-i-1); j++){
2 0
3 0
3. if(array[j]>array[j+1]){ 4,5,6 0
2 0
3 0
4. temp = array[j+1]; 4,5,6 0
5. array[j+1] = array[j]; 2 0
3 0
6. array[j] = temp;
4,5,6
0
2 0
}
1 1
}
}
a[j]>
j a[j+1] a[0] a[1] a[2] a[3] a[4]
5
4
3
2 1
0
5
4
3
2 1
0T
5
4
3
2 1
0
4
5
3
2 1
1
4
5
3
2 1
1T
4
5
3
2 1
1
4
3
5
2 1
2
4
3
5
2 1
2T
4
3
5
2 1
2
4
3
2
5 1
3
4
3
2
5 1
3T
4
3
2
5 1
3
4
3
2
1 5
4
4
3
2
1 5
4
4
3
2
1 5
1. for(i=0; i<n-1; i++){
2. for(j=0; j<(n-i-1); j++){
3. if(array[j]>array[j+1]){
4. temp = array[j+1];
5. array[j+1] = array[j];
6. array[j] = temp;
}
}
}
Step i
1 1
2 1
3 1
4,5,6 1
2 1
3 1
4,5,6 1
2 1
3 1
4,5,6 1
2 1
1 2
a[j]>
j a[j+1] a[0] a[1] a[2] a[3] a[4]
4
4 3 2 1 5
0
4 3 2 1 5
0T
4 3 2 1 5
0
3 4 2 1 5
1
3 4 2 1 5
1T
3 4 2 1 5
1
3 2 4 1 5
2
3 2 4 1 5
2T
3 2 4 1 5
2
3 2 1 4 5
3
3 2 1 4 5
3
3 2 1 4 5
1. for(i=0; i<n-1; i++){
2. for(j=0; j<(n-i-1); j++){
3. if(array[j]>array[j+1]){
4. temp = array[j+1];
5. array[j+1] = array[j];
6. array[j] = temp;
}
}
}
Step i j
1 2 3
2 2 0
3 2 0T
4,5,6 2 0
2 2 1
3 2 1T
4,5,6 2 1
2 2 2
1 3 2
2 3 0
3 3 0T
4,5,6 3 0
2 3 1
1 4 1
a[j]>
a[j+1]
a[0] a[1] a[2] a[3] a[4]
3 2 1 4 5
3 2 1 4 5
3 2 1 4 5
2 3 1 4 5
2 3 1 4 5
2 3 1 4 5
2 1 3 4 5
2 1 3 4 5
2 1 3 4 5
2 1 3 4 5
2 1 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
Homework
• Trace Bubble Sort algorithm for the following
input
– Directly (show comparisons and swaps)
– Using tracing table (show variables also)
• Input: {2,6,12,5,9}
• Submission
– Handwritten
– Start of next Class
2-D Array
Initializing two dimensional array
Two Dimensional Array
Matrix Addition
int i, j;
for (i = 0; i < NRows; i++) {
for (j = 0; j < NCols; j++) {
C [i][j] = A [i][j] + B [i][j];
}
}
Appendix
Practice Questions
• Addition of Matrices
• Matrix Multiplication
• Average and Statistics
Matrix Multiplication
• It is defined between two matrices only if the number
of columns of the first matrix is the same as the
number of rows of the second matrix. If A is an m-by-n
matrix and B is an n-by-p matrix, then their product
A×B is an m-by-p matrix given by
Matrix Multiplication
int i = 0;
int j = 0;
int k = 0;
for(i = 0; i < 2; i++) // rows of matrix 1
for( j = 0; j < 4; j++) // columns of matrix 2
for( k = 0; k < 3; k++) // col of m1,row of m2
a3[i][j] += a1[i][k] * a2[k][j];
Computing Mean, Median and Mode
Using Arrays
• Mean
– Average (sum/number of elements)
• Median
– Number in middle of sorted list
– 1, 2, 3, 4, 5 (3 is median)
– If even number of elements, take average of middle
two
• Mode
– Number that occurs most often
– 1, 1, 1, 2, 3, 3, 4, 5 (1 is mode)
35
Bubble Sort
for(x=0; x<n; x++){ // number of iterations
for(y=x+1; y<n; y++){ // number of pair wise comparisons
if(array[y]>array[y+1]){
temp = array[y+1];
array[y+1] = array[y];
array[y] = temp;
}
}
}
Matrix Multiplication
int i, j, k;
int sum;
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
sum = 0;
for (k = 0; k < N; k++) {
sum += A[i][k] * B[k][j];
}
C[i][j] = sum;
}
}