CSE 1002 PPS1 new Winter ’16 (Id=2228)
13-Jan-2017 00:00 to 29-Jan-2017 00:00
Total time : 500 mins
Challenges : 5
Question 1 (Second Smallest Number)
Given a set of elements, design an Algorithm and write the subsequent C program to determine the second smallest number in that set.
Input Format
Number of elements in ‘n’
element-1
element-2
…
element-n
Output Format
Second smallest element in the set
Solution
C CODE
#include< stdio.h >
void main()
{
int n,i,num,small=999,sec=999;
scanf("%d",&n);
for(i=0;i < n;i++)
{
scanf("%d",&num);
if(num < =small)
{
sec=small;
small=num;
}
else if(num < =sec)
sec=num;
}
printf("%d",sec);
}
c++ code(updated later)
#include <iostream>
using namespace std;
int main() {
int n;
cin>>n;
int num,min,min2;
for(int i=0;i<n;i++)
{
cin>>num;
if(num<=min)
{
min2=min;
min=num;
}
else if(num<=min2)
min2=num;
}
cout<<min2;
return 0;
}
Input
INPUT: n - number of elements
Output
OUTPUT:
printf("%d",sec);
Processing
for(i=0;i < n;i++)
{
scanf("%d",&num);
if(num < =small)
{
sec=small;
small=num;
}
else if(num < =sec)
sec=num;
}
Pseudocode
BEGIN
Read n
for(i=0;i < n;i++)
{
scanf("%d",&num);
if(num < =small)
{
sec=small;
small=num;
}
else if(num < =sec)
sec=num;
}
printf("%d",sec);
END
Question 2 (Recursive Fibonacci)
Given the value of ‘n’, write a recursive routine in C to print the first ‘n’ elements of the Fibonacci series.
Input Format
Value of ‘n’
Output Format
Fibonacci series of ‘n’ terms, each term separated by a space
Solution
#include< stdio.h >
void main()
{
int i,n,f=0,sum=0,num=1;
scanf("%d",&n);
printf("0 1");
for(i=0;i < n-2;i++)
{
sum=f+num;
f=num;
num=sum;
printf(" %d",sum);
}
}
Input
INPUT: n - number of terms
Output
OUTPUT:
printf("0 1");
printf(" %d",sum);
Processing
for(i=0;i < n-2;i++)
{
sum=f+num;
f=num;
num=sum;
}
Pseudocode
BEGIN
Read n
printf("0 1");
for(i=0;i < n-2;i++)
{
sum=f+num;
f=num;
num=sum;
printf(" %d",sum);
}
END
Question 3 (Cyclic Right shift of Elements)
Given a set of elements stored in an array and a number ‘m’, design an Algorithm and write the subsequent C program to perform cyclic right shift of the array by ‘m’ places. For example, if the elements are 12, 13, 16, 7, 10 and m =2 then the resultant set will be 7, 10, 12, 13, 16.
Input Format
Number of elements in the set: ‘n’
element-1
element-2
…
element-n
value of ‘m’
Output Format
Elements in the set after right shift by ‘m’ places
Solution
#include< stdio.h >
void main()
{
int a[20],n,i,m,t,j;
scanf("%d",&n);
for(i=0;i < n;i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(j=0;j < m;j++)
{
t=a[n-1];
for(i=n-1;i > 0;i--)
a[i]=a[i-1];
a[0]=t;
}
for(i=0;i < n;i++)
printf("%d\n",a[i]);
}
Input
INPUT:
n - number of terms
for(i=0;i < n;i++)
scanf("%d",&a[i]);
m - number of shifts
Output
OUTPUT:
for(i=0;i < n;i++)
printf("%d\n",a[i]);
Processing
for(j=0;j < m;j++)
{ t=a[n-1];
for(i=n-1;i > 0;i--)
a[i]=a[i-1];
a[0]=t;
}
Pseudocode
BEGIN
Read n, array and m
for(j=0;j < m;j++)
{ t=a[n-1];
for(i=n-1;i > 0;i--)
a[i]=a[i-1];
a[0]=t;
}
for(i=0;i < n;i++)
printf("%d\n",a[i]);
END
Qestion 4 (Leaders of Elements)
Given a set of ‘n’ elements in an order, identify all the leaders and print them. An element is said to be a leader if all the elements to its right are smaller than it. For example, if the elements are 12, 13, 16, 7, 10 then there is only one leader element 16. If there are no leaders in the given set of elements then print ‘No leaders’.
Input Format
Number of elements in the given set: ‘n’
element-1
element-2
…
element-n
Output Format
Elements that are leaders. Else, print ‘No leaders’ when there is no leader.
Solution
#include< stdio.h >
void main()
{
int a[20],f=1,flag=0,i,j,n;
scanf("%d",&n);
for(i=0;i < n;i++)
scanf("%d",&a[i]);
for(i=0;i < n-1;i++)
{
for(j=i+1;j < n;j++)
if(a[i] > a[j])
f=1;
else
{
f=0;
break;
}
if(f==1)
{
printf("%d\n",a[i]);
flag=1;
}
f=0;
}
if(flag==0)
printf("No leaders");
}
Input
INPUT:
n - number of elements
for(i=0;i < n;i++)
scanf("%d",&a[i]);
Output
OUTPUT:
if(f==1)
printf("%d\n",a[i]);
if(flag==0)
printf("No leaders");
Processing
for(i=0;i < n-1;i++)
{
for(j=i+1;j < n;j++)
if(a[i] > a[j])
f=1;
else
{
f=0;
break;
}
if(f==1)
flag=1;
f=0;
}
Pseudocode
BEGIN
Read n
for(i=0;i < n-1;i++)
{
for(j=i+1;j < n;j++)
if(a[i] > a[j])
f=1;
else
{
f=0;
break;
}
if(f==1)
{
printf("%d\n",a[i]);
flag=1;
}
f=0;
}
END
Question 5 (Recursive reverse)
Given a string, write a recursive routine to reverse it. For example, given the string ‘and, the reversal of the string is ‘dna’.
Input Format
A string
Output Format
Reverse of the string
Solution
#include< stdio.h >
#include< string.h >
void main()
{
char s[20],ch;
int i,len;
scanf("%s",s);
len=strlen(s);
for(i=0;i < (len/2);i++)
{
ch=s[i];
s[i]=s[len-1-i];
s[len-1-i]=ch;
}
puts(s);
}
Input
INPUT: s - string
Output
OUTPUT: puts(s);
Processing
for(i=0;i < (len/2);i++)
{
ch=s[i];
s[i]=s[len-1-i];
s[len-1-i]=ch;
}
Pseudocode
BEGIN
Read s
for(i=0;i < (len/2);i++)
{
ch=s[i];
s[i]=s[len-1-i];
s[len-1-i]=ch;
}
puts(s);
END
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