BCSL-045 Algorithm Design Lab BCA (IV)/045/Assignment/14-15
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BCSL-045
Algorithm Design Lab
BCA (IV)/045/Assignment/14-15
Q.1
(i)
A Palindrome is a string that reads the same forward and backward. Write an
algorithm for determining whether a string of n characters is palindrome.
Solution:C program to check whether is a palindrome and also find the length of string
# include <stdio.h>
#include <string.h>
Void main ()
{
Char string [25] , reverse_string [25] = { ‘\0’ };
Int I, length = 0, flag = 0;
Printf(“Enter a string \n”);
Gets(string);
For( i = 0; string [i] ! = ‘\0’ ; i++)
{
Length++;
}
Printf (“the length of the string ‘%s’ = %d\n” , string, length);
for ( I = Length – 1; I >=0 ; I - - )
{
Reverse _ string [ length – i – I ] = string [i];
}
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For (flag = 1, i= 0; i< length; i++)
{
If (reverse _ string [i] != string [i] )
Flag = 0;
}
If (flag==1)
Printf (“%s is a palindrome \n”, string) ;
else
Printf (“%s is a not palindrome \n”, string) ;
}
Output:
Enter a string
How are you
The length of the string ‘how are you’ = 12
How are you is not a palindrome
-------------------------------------------------------------------------------------------Enter a string
Madam
The length of string ‘madam’ = 5
Madam is a palindrome
Algorithm:Step 1 : Start
Step 2 : Declare variables char_string [25], reverse_string [25]
Int i , length = 0 , flag = 0 ;
Step 3 : read values string.
Step 4 : for (I = 0 ; string [i] != ‘\0’ ; i++)
length++
Step 5 : display ‘the length of string =’ , string , length
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Step 6 : for (I = length – 1; i >=0 ; I -- )
do reverse _ string [length – I - 1] = string [i]
Step 7 : for (flag = 1, i= 0 ; i< length ; i++)
if reverse _ string [i] ! = string [i]
do flag = 0
If flag ==1
display string is palindrome.
else
display string is not palindrome.
Step 8 : Stop.
(ii)
Find the largest number in an array and count a number of comparison
operations as well as running time complexity of each statement and total
complexity of the problem.
Solution:#include <stdio.h>
int main() {
Int I, n, count = 0;
Float arr [100] ;
Printf(“Enter total number of elements (1 to 100) : ”) ;
Scanf (“%d”, &n);
Printf ( i=0 ; i<n ; ++i )
{
Printf (“Enter Number %d:” , i+1);
scanf (“%f”, &arr [i] ) ;
}
For ( i=1; i<n;++i)
{
If (arr [0] < arr [i] )
{
Count++;
}
Arr [0] = arr [i] ;
}
Printf(“Largest element = %. 2f \n” , arr [0]);
printf (“the number of comparison %d”, count);
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Return 0 ;
Getch();
}
Q 1(iii) Write a programme which interchanges the values the variable using only
assignment operations. What is the minimum number of assignments operations
required?
Ans.
#include <stdio.h>
int main()
{
int x, y, temp;
printf("Enter the value of x and y\n");
scanf("%d%d", &x, &y);
printf("Before Swapping\nx = %d\ny = %d\n",x,y);
temp = x;
x
= y;
y
= temp;
printf("After Swapping\nx = %d\ny = %d\n",x,y);
return 0;
}
Minimum number of assignments are three in this case of swapping the values.
Q. (iv) Write a programme to do bubblesort to sort an array X={10,5,7,8,3,6,4,14}
showing the list obtained at each step. Also count no. of comparison operations
in the programme.
Ans.
#include <stdio.h>
int main()
{
int array[100], n, c, d, swap;
printf("Enter number of elements\n");
scanf("%d", &n);
printf("Enter %d integers\n", n);
for (c = 0; c < n; c++)
scanf("%d", &array[c]);
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for (c = 0 ; c <
{
for (d = 0 ; d
{
if (array[d]
{
swap
array[d]
array[d+1]
}
}
}
( n - 1 ); c++)
< n - c - 1; d++)
> array[d+1]) /* For decreasing order use < */
= array[d];
= array[d+1];
= swap;
printf("Sorted list in ascending order:\n");
for ( c = 0 ; c < n ; c++ )
printf("%d\n", array[c]);
return 0;
}
Q 1(v) Write a programme to find multiplication of matrices of order 4*4 and find the
total number of comparison, total no. of assignments, total number of
multiplications and total number of addition operations required in the
programme.
Ans.
#include <stdio.h>
int main()
{
int m, n, p, q, c, d, k, sum = 0;
int first[10][10], second[10][10], multiply[10][10];
printf("Enter the number of rows and columns of first matrix\n");
scanf("%d%d", &m, &n);
printf("Enter the elements of first matrix\n");
for ( c = 0 ; c < m ; c++ )
for ( d = 0 ; d < n ; d++ )
scanf("%d", &first[c][d]);
printf("Enter the number of rows and columns of second matrix\n");
scanf("%d%d", &p, &q);
if ( n != p )
printf("Matrices with entered orders can't be multiplied with each other.\n");
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else
{
printf("Enter the elements of second matrix\n");
for ( c = 0 ; c < p ; c++ )
for ( d = 0 ; d < q ; d++ )
scanf("%d", &second[c][d]);
for ( c =
{
for ( d
{
for (
{
sum
}
0 ; c < m ; c++ )
= 0 ; d < q ; d++ )
k = 0 ; k < p ; k++ )
= sum + first[c][k]*second[k][d];
multiply[c][d] = sum;
sum = 0;
}
}
printf("Product of entered matrices:-\n");
for ( c = 0 ; c < m ; c++ )
{
for ( d = 0 ; d < q ; d++ )
printf("%d\t", multiply[c][d]);
printf("\n");
}
}
return 0;
}
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