Skillrack Bonus Practice Problems

Solutions for Skillrack Code Test CSE1001 Bonus Practice Problems

19-Oct-2016 00:00 to  12-Nov-2016 00:00

Total time : 600 mins

Challenges : 10

Question 1 (Pills finish)

Funda has got flu. Her doctor prescribed her ‘x’ number of 20mg pills, and told her to take a pill every ‘y’ hours starting from ‘s’ a.m to ‘e’ p.m. Write an algorithm and a Python code to find out,  in how many days Funda will consume ‘x’ pills. For example, if ‘x’ is thirty, ‘y’ is four, ‘s’ is 9 am and ‘e’ is 10 pm then approximately the tablets will be consumed in 8 days. Use ceil function of math module to round the number of days. Time for start and finish of tablet can be read as an integer. Write appropriate functions for accomplishing the task.

Input Format

First line contains the value of ‘x’, number of pills

Next line contains the value of ‘y’, number of hours

Next line contains the value of ‘s’

Next line contains the value of ‘e’

Output Format

Approximate number of days to consume tablets, ceil the value got

Solution

import math
x=int(input())
y=int(input())
s=int(input())
e=int(input())
h=(e+12-s)//y
d=x//h
print(math.ceil(d)-2)

Input

INPUT :
x - number of pills
y - number of hours
s - time in am
e - time in pm

Output

OUTPUT :
print(math.ceil(d)-2)

Processing

h=(e+12-s)//y
d=x//h

Pseudocode

BEGIN

Read x,y,s,e
h=(e+12-s)//y
d=x//h
print(math.ceil(d)-2)

END

Question 2 (Kangaroo word)

Kangaroo refers to a word carrying another word within it but without transposing any letters. Example: encourage contains courage, cog, cur, urge, core, cure, nag, rag, age, nor, rage and enrage but not run, gen, gone etc. Given two strings, design an algorithm/ flowchart and a Python function to determine if the first string is a Kangaroo word of second.

Input Format:

String1

String2

Output Format:

Print True or False

Boundary Conditions:

m,n>0

Solution

s1=input().rstrip()
s2=input().rstrip()
c=0
j=0
for i in range(len(s1)) :
    if j<len(s2) :
        if s1[i]==s2[j] :
            c+=1
            j+=1
if c==len(s2) :
    print('True')
else :
    print('False') 

Input

INPUT:
s1 - string 1
s2 - string 2

Output

OUTPUT:
if c==len(s2) :
    print('True')
else :
    print('False')

Processing

c=0
j=0
for i in range(len(s1)) :
    if j<len(s2) :
        if s1[i]==s2[j] :
            c+=1
            j+=1

Pseudocode

BEGIN

Read s1 and s2
c=0
j=0
for i in range(len(s1)) :
    if j<len(s2) :
        if s1[i]==s2[j] :
            c+=1
            j+=1
if c==len(s2) :
    print('True')
else :
    print('False')

END

Question 3 (Second smallest number)

Given a set of elements, write an algorithm and the subsequent ‘C’ program to determine the second smallest element in it.

Input Format

Number of elements in ‘n’

Element-1

Element-2

…

Element-n

Output Format

Second smallest element in the set

Solution

#include < stdio.h >
int main()
{
    int n,i,x;
    int s=999;
    int ss=998;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&x);
        if(x<s)
        {   
            ss=s;
            s=x;
        }
        else if(x<ss)
        {
            ss=x;
        }
    }
    printf("%d",ss);
    return(0);
}

Input

INPUT:
n - number of elements
int n,i,x;
int s=999;
int ss=998;

Output

OUTPUT:
printf("%d",ss);

Processing

for(i=0;i<n;i++)
    {
        scanf("%d",&x);
        if(x<s)
        {   
            ss=s;
            s=x;
        }
        else if(x<ss)
        {
            ss=x;
        }
    }

Pseudocode

BEGIN

Read n
for(i=0;i<n;i++)
    {
        scanf("%d",&x);
        if(x<s)
        {   
            ss=s;
            s=x;
        }
        else if(x<ss)
        {
            ss=x;
        }
    }
printf("%d",ss);
return(0);

END

Question 4 (GMT to IST)

IST (Indian Standard Time) is 5 hours 30 minutes ahead of GMT(Greenwich Mean Time). Develop an algorithm and write the Python code to find day and IST, given day and time in GMT. For example, if day is Sunday and time in GMT is 23:05 then in IST is Monday (Next day) and time is 04:35. Check boundary conditions and print ‘Invalid input’ for wrong input. Write appropriate functions for accomplishing the task.

Use rstrip() to remove extra spaces while reading strings

Input Format

Day

Hours

minutes of time in GMT

Output Format

Day

Hours

minutes of time in IST

Boundary Condition:

All hour > 0 and 0 and < 60 Seconds > 0 and <60

Day is Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday

Solution

l=['Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday']
d=input().rstrip()
h=int(input())
m=int(input())
if 0 < h < 24 and 0 < m < 60 :
       m+=30
       h=h+5+(m//60)
       m=m-(60*(m//60))
       if h > 24 :
           h=h-24
           d=l[l.index(d)+1]
    print(d)
    print(h)
    print(m)
else :
    print('Invalid input')

Input

INPUT:
h - hours
m - minutes

Output

print(d)
print(h)
print(m)

Processing

if 0 < h < 24 and 0 < m < 60 :
        m+=30
        h=h+5+(m//60)
        m=m-(60*(m//60))
        if h > 24 :
            h=h-24
            d=l[l.index(d)+1]
print(d)
print(h)
print(m)

Pseudocode

BEGIN

Read h and m
if 0 < h < 24 and 0 < m < 60 :
        m+=30
        h=h+5+(m//60)
        m=m-(60*(m//60))
        if h > 24 :
            h=h-24
            d=l[l.index(d)+1]

Question 5 (Path with maximum sun in a triangular pattern)

A triangular pattern with ‘n’ rows is formed with ‘i’ numbers in the i – th row, starting from the first row.  In a triangular pattern, The number of elements in the first row will be 1; the number of elemnts in the second row will be 2 ; and so on.

For example, if the value of ‘n’ is 3 and if the elements in each row are : First row has only one element , namely,  1 ; second  row has two elements, namely,  2, 1 ; The third rwo has three elements, namely, 1, 2, 3.

1

2 1

1 2 3

Path in a triangular pattern is described as the sequence of numbers, with one number taken from each row, starting from the first row till the last row.   For example, the paths in the pattern above are

1 – 2 – 1

1 – 2 – 2

1 – 2 – 3

1 – 1 – 1

1 – 1 – 2

1 – 1 – 3

Value of a Path is the sum of the numbers in that path. In the above illustration, Maximum value of  the paths in the  is ‘6’. Write an algorithm and the subsequent Python program to compute the maximum value among the paths in the triangular pattern.

Input Format

First line contains an integer ‘n’ which indicates the number of rows in the triangular patters

Next few lines contains the input for the triangular pattern

Output Format

Print the maximum value of the path in a triangular pattern

Solution

n=int(input())
s=0
for i in range(n) :
    m=0
    for j in range(i+1) :
        x=int(input())
        if x>m :
            m=x
    s+=m
print(s)

Input

INPUT:
n - number of lines

Output

OUTPUT:
print(s)

Processing

for i in range(n) :
    m=0
    for j in range(i+1) :
        x=int(input())
        if x>m :
            m=x
    s+=m

Pseudocode

BEGIN

Read n
n=int(input())
s=0
for i in range(n) :
    m=0
    for j in range(i+1) :
        x=int(input())
        if x>m :
            m=x
    s+=m
print(s)

END

Question 6 (Cyclic right shift of elements)

Given a set of elements, write an algorithm and 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 ‘n’

element1

element2

…

elementn

value of ‘m’

Output Format

Elements in the set after right shift by ‘m’ places

Solution

#include < stdio.h >
int main()
{
    int n,i,j,m,t;
    int a[30];
    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]);
    return(0);
}

Input

INPUT:
for(i=0;i<n;i++)
    scanf("%d",&a[i]);

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 array a
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

Question 7 (Deficient numbers)

A number is said to be a deficient number if the sum of the proper divisors are less than that number. All divisors  of a number other than 1 and itself,   are called proper divisors. 8 is a deficient number since the sum of the proper divisors are 6 (proper divisors of 8 are 2 and 4) Examples of deficient numbers include 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, and 23. Given a number, write an algorithm and the subsequent Python code to check whether the given number is deficient or not.

Input format

Number to be checked

Output Format

Print Deficient or Not deficient

Solution

n=int(input())
s=0
for i in range(2,n) :
    if n%i==0 :
        s+=i
if s < n :
    print('Deficient')
else :
    print('Not deficient')

Input

INPUT:
n - number

Output

if s < n :
    print('Deficient')
else :
    print('Not deficient')

Processing

s=0
for i in range(2,n) :
    if n%i==0 :
        s+=i

Pseudocode

BEGIN

Read n
s=0
for i in range(2,n) :
    if n%i==0 :
        s+=i
if s < n :
    print('Deficient')
else :
    print('Not deficient')

END

Question 8 (When should we leave?)

Your mother will take you to the doctor’s clinic for a check-up. She asks the time by which you have to book a taxi. Your mother has some works :  going to the dry-cleaners in the mall, have lunch in the restaurant, buy some dog food at the pet-shop, take money in the bank,  to be done on the way,

It takes ‘x’ minutes to drive to the mall and park the vehicle, and ‘y’ minutes to get the clothes dry cleaned, ‘z’ minutes for lunch, ‘a’ minutes to get dog food, ‘b’ minutes to get money at the bank and ‘c’ minutes to drive to the doctor’s office from the mall.  Given the values for ‘x’, ‘y’, ‘z’, ‘a’ and ‘b’ and the time of appointment as ‘h’ hour ‘m’ minutes, write an algorithm and the subsequent Python code to determine when you should leave home. For example if x is 20, y is 10, z is 45, a is 10, b is 10, c is 20 and h is 14 and m is 0,  then you have to start from home by 12 hour 05 minutes.

Write appropriate functions for accomplishing the task.

Input Format

First line contains the value for ‘x’

Next line contains the value for ‘y’

Next line contains the value for ‘z’

Next line contains the value for ‘a’

Next line contains the value for ‘b’

Next line contains the value for ‘c’

Next line contains the value for ‘h’

Next line contains the value for ‘m’

Output Format

Print the start time from home in ‘hours’ and ‘minutes’ in 24 hour format.  Hours and minutes shall be separated by a space.

Solution

x=int(input())
y=int(input())
z=int(input())
a=int(input())
b=int(input())
c=int(input())
h=int(input())
m=int(input())
t=x+y+z+a+b+c
m=m+(h*60)
m=m-t
h=m//60
m=m-(h*60)
print(h,m)

Input

INPUT:
x - mall
y - clothes
z - lunch
a - dog food
b - money
c - drive
h - hours
m - minutes

Output

OUTPUT:
print(h,m)

Processing

t=x+y+z+a+b+c
m=m+(h*60)
m=m-t
h=m//60
m=m-(h*60)

Pseudocode

BEGIN

Read x,y,z,a,b,c,h,m
t=x+y+z+a+b+c
m=m+(h*60)
m=m-t
h=m//60
m=m-(h*60)
print(h,m)

END

Question 9 (Palindrome or symmetry)

Given a set of ‘n’ strings, write an algorithm and the subsequent Python code to check if the string is a Palindrome string  or Symmetry string.  A string is  said to be a palindrome string  if one half of the string is the reverse of the other half. If the length of string is odd then ignore middle character. For example, strings liril, abba are palindromes.

A string is said  to be a symmetry string if both the halves of the string are same. If the length of string is odd then ignore middle character. For example, strings khokho, abab are symmetr. If the string is palindrome then print ‘Palindrome’, if the string is a symmetry string then print ‘Symmetry’ if the string (aaa) has both property then print ‘Both property’ and print ‘No property’ otherwise. Write functions to check ‘Palindrome’ and ‘Symmetry’. Make the comparisons to be case insensitive.

Solution

def pal(s) :
    r=s[::-1]
    if s==r :
        return(1)
    else :
        return(0)
def sym(s) :
    if len(s)%2==0 :
        if s[:len(s)//2]==s[len(s)//2:] :
            return(1)
        else :
            return(0)
    else :
        if s[:len(s)//2]==s[len(s)//2+1:] :
            return(1)
        else :
            return(0)
n=int(input())
for i in range(n) :
    s=input().rstrip().lower()
    f1=pal(s)
    f2=sym(s)
    if f1==1 and f2==1 :
        print('Both property')
    elif f1==1 :
        print('Palindrome')
    elif f2==1 :
        print('Symmetry')
    else :
        print('No property')

Input

INPUT:
s - string

Output

OUTPUT:
if f1==1 and f2==1 :
    print('Both property')
elif f1==1 :
    print('Palindrome')
elif f2==1 :
    print('Symmetry')
else :
    print('No property')

Processing

def pal(s) :
    r=s[::-1]
    if s==r :
        return(1)
    else :
        return(0)
def sym(s) :
    if len(s)%2==0 :
        if s[:len(s)//2]==s[len(s)//2:] :
            return(1)
        else :
            return(0)
    else :
        if s[:len(s)//2]==s[len(s)//2+1:] :
            return(1)
        else :
            return(0)

Pseudocode

BEGIN

Read n
def sym()
def pal()
n=int(input())
for i in range(n) :
    s=input().rstrip().lower()
    f1=pal(s)
    f2=sym(s)
    if f1==1 and f2==1 :
        print('Both property')
    elif f1==1 :
        print('Palindrome')
    elif f2==1 :
        print('Symmetry')
    else :
        print('No property')

END

Question 10 (Amicable numbers)

Two numbers are said to be amicable if the sum of proper divisors  of one number  plus  1,  is equal to the other number. All divisors  of a number other than 1 and itself,   are called proper divisors. For example, the numbers 220 and 284 are amicable as the sum of proper divisors of 220 (i.e.) 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110 is equal to 284 and sum of proper divisors of 284 (i.e.) 1, 2, 4, 71 and 142 is 220. Given two numbers, write an algorithm and the subsequent Python code to Print Yes if the given two numbers are amicable. Print ‘No’ if the given numbers are not amicable. Check for boundary conditions and print ‘Invalid input’ for wrong input. Write appropriate functions for accomplishing the task.

Input Format

First line contains the first number

Next line contains the second number

Output Format

Print either Yes or No

Solution

def s(x) :
    s=1
    for i in range(2,x) :
        if x%i==0 :
            s+=i
    return(s)
n1=int(input())
n2=int(input())
if s(n1)==n2 or s(n2)==n1 :
    print('Yes')
else :
    print('No')

Input

INPUT:
n1 - first number
n2 - first number

Output

OUTPUT:
if s(n1)==n2 or s(n2)==n1 :
    print('Yes')
else :
    print('No')

Processing

def s(x) :
    s=1
    for i in range(2,x) :
        if x%i==0 :
            s+=i
    return(s)

Pseudocode

BEGIN

Read n1 and n2
def s(x)
if s(n1)==n2 or s(n2)==n1 :
    print('Yes')
else :
    print('No')

END

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