Question:
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
A) 32
B) 48
C) 36
D) 60
Sunday 31 March 2019
Saturday 30 March 2019
Friday 29 March 2019
Thursday 28 March 2019
Print a long int in C using putchar() only
Print a long int in C using putchar() only
C program:
#include <stdio.h>
void print(long n)
{
/* If the number is smaller than 0, put a - sign and change number to positive */
if (n < 0) {
putchar('-');
n = -n;
}
if (n/10)
print(n/10);
// Print the last digit
putchar(n%10 + '0');
}
int main()
{
long int n = 12045;
print(n);
return 0;
}
Output:
12045
Print characters without using format specifiers
Question:
C program to print characters without using format specifiers
C Program:
// Prints characters without format specifiers
#include <stdio.h>
int main()
{
//ASCII Values
printf("\x41 \n");
printf("\x4d \n");
printf("\x49 \n");
printf("\x54");
return 0;
}
Output:
A
M
I
T
Round off
Write a one-line C function to round floating point numbers
Algorithm: roundNo(num)1. If num is positive then add 0.5.
2. Else subtract 0.5.
3. Typecast the result to int and return.
2. Else subtract 0.5.
3. Typecast the result to int and return.
Example:num = 1.67, (int) num + 0.5 = (int)2.17 = 2
C program:
/* Program for rounding floating point numbers */
# include<stdio.h>
int roundNo(float num)
{
return num < 0 ? num - 0.5 : num + 0.5;
}
int main()
{
printf("%d", roundNo(-1.777));
getchar();
return 0;
}
C to convert a number to a string
Question:
What is the best way in C to convert a number to a string?
Solution: Use sprintf() function.
C program:
#include<stdio.h>
int main()
{
char result[50];
float num = 23.34;
sprintf(result, "%f", num);
printf("\n The string for the num is %s",
result);
}
Output:
The string for the num is 23.340000
Program for Sum the digits of a given number
Program for Sum the digits of a given number
Examples :
Input : n = 687
Output : 21
C program:
# include<stdio.h>
/* Function to get sum of digits */
int getSum(int n)
{
int sum;
/* Single line that calculates sum */
for (sum = 0; n > 0; sum += n % 10, n /= 10);
return sum;
}
int main()
{
int n = 687;
printf(" %d ", getSum(n));
return 0;
}
Output:
21
Activity Selection Problem
Activity Selection Problem
1) Sort the activities according to their finishing time
2) Select the first activity from the sorted array and print it.
3) Do following for remaining activities in the sorted array.
a) If the start time of this activity is greater than or equal to the finish time of previously selected activity then select this activity and print it.
2) Select the first activity from the sorted array and print it.
3) Do following for remaining activities in the sorted array.
a) If the start time of this activity is greater than or equal to the finish time of previously selected activity then select this activity and print it.
Example:
Consider the following 3 activities sorted by finish time.
Input:
start[] = {10, 12, 20};
finish[] = {20, 25, 30};
Output:
A person can perform at most two activities. The
maximum set of activities that can be executed
is {0, 2}
C program:
#include<stdio.h>
/* Prints a maximum set of activities that can be done by a single person, one at a time.*/
// n --> Total number of activities
// s[] --> An array that contains
//start time of all activities
//start time of all activities
// f[] --> An array that contains
//finish time of all activities
//finish time of all activities
void printMaxActivities(int s[], int f[], int n)
{
int i, j;
printf ("Following activities are selected n ");
i = 0;
printf("%d ", i);
for (j = 1; j < n; j++)
{
// If this activity has start time greater than or
// equal to the finish time of previously selected
// activity, then select it
if (s[j] >= f[i])
{
printf ("%d ", j);
i = j;
}
}
}
int main()
{
int s[] = {1, 3, 0, 5, 8, 5};
int f[] = {2, 4, 6, 7, 9, 9};
int n = sizeof(s)/sizeof(s[0]);
printMaxActivities(s, f, n);
return 0;
}
Output:
Following activities are selected n 0 1 3 4
Minimum number of Coins
Find the Minimum number of Coins
1) Initialize result as empty.
2) find the largest denomination that is
smaller than V.
3) Add found denomination to result.
Subtract the value of found denomination from V.
4) If V becomes 0, then print result.
Else repeat steps 2 and 3
for the new value of V
Example:
Input: V = 70
Output: 2
We need a 50 Rs note and a 20 Rs note.
C program:
#include <stdio.h>
#define COINS 9
#define MAX 20
// All denominations of Indian Currency
int coins[COINS] = { 1, 2, 5, 10, 20, 50, 100, 200, 2000 };
void findMin(int cost)
{
int coinList[MAX] = { 0 };
int i, k = 0;
for (i = COINS - 1; i >= 0; i--) {
while (cost >= coins[i]) {
cost -= coins[i];
coinList[k++] = coins[i];
}
}
for (i = 0; i < k; i++) {
printf("%d ", coinList[i]); }
return;
}
int main(void)
{
int n;
printf("Enter the Rupees u want Change\n");
scanf("%d",&n);
printf("Following is minimal number of change for %d: ", n);
findMin(n);
return 0;
}
Output:
Enter the Rupees u want Change
1276
Following is minimal number of change for 1276: 200 200 200 200 200 200 50 20 5 1
Aptitude Hack#33(28-3-19)
Question:
5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?A) 20 days
B) 18 days
C) 16 days
D) 15 days
Breath First Search
ALGORITHM BFS(G)//Implements a breadth-first search traversal of a given graph
//Input: Graph G = V, E//Output: Graph G with its vertices marked with consecutive //integers in the order they are visited by the BFS traversal
mark each vertex in V with 0 as a mark of being “unvisited”count ← 0for each vertex v in V do
if v is marked with 0Bfs(v)
Algorithm Bfs(v)//visits all the unvisited vertices connected to vertex v//by a path and numbers them in the order they are visited
//via global variable count
count← count + 1;
mark v with count and initialize a queue with vwhile the queue is not empty do
for each vertex w in V adjacent to the front vertex do
if w is marked with 0count ← count + 1;
mark w with the countadd w to the queue
remove the front vertex from the queue
C program:
#include<stdio.h>
int count=0;
int n;
void bfs(int a[][n],int b[],int v);
main()
{
int i,j;
printf("\nEnter number of vertices:");
scanf("%d",&n);
int a[n][n];
int b[n];
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("\nEnter connection bt vertex %c & vertex %c",i+65,j+65);
scanf("%d",&a[i][j]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d\t",a[i][j]);
}
printf("\n");
}
for(i=0;i<n;i++)
b[i]=0;
bfs(a,b,0);
}
void bfs(int a[][n],int b[],int v)
{
int i,r=0,f=-1;
int q[n];
count++;
b[v]=count;
q[++f]=v;
do
{
v=q[r];
for(i=0;i<n;i++)
{
if(a[v][i]==1)
{
if(b[i]==0)
{
count++;
b[i]=count;
q[++f]=i;
}
}
}
printf("%c\n",q[r++]+65);
}while(f>=r);
}
Output:
Enter number of vertices:4
Enter connection bt vertex A & vertex A0
Enter connection bt vertex A & vertex B1
Enter connection bt vertex A & vertex C0
Enter connection bt vertex A & vertex D0
Enter connection bt vertex B & vertex A1
Enter connection bt vertex B & vertex B0
Enter connection bt vertex B & vertex C1
Enter connection bt vertex B & vertex D0
Enter connection bt vertex C & vertex A0
Enter connection bt vertex C & vertex B1
Enter connection bt vertex C & vertex C0
Enter connection bt vertex C & vertex D1
Enter connection bt vertex D & vertex A1
Enter connection bt vertex D & vertex B0
Enter connection bt vertex D & vertex C1
Enter connection bt vertex D & vertex D0
0 1 0 0
1 0 1 0
0 1 0 1
1 0 1 0
A
B
C
D
//Input: Graph G = V, E//Output: Graph G with its vertices marked with consecutive //integers in the order they are visited by the BFS traversal
mark each vertex in V with 0 as a mark of being “unvisited”count ← 0for each vertex v in V do
if v is marked with 0Bfs(v)
Algorithm Bfs(v)//visits all the unvisited vertices connected to vertex v//by a path and numbers them in the order they are visited
//via global variable count
count← count + 1;
mark v with count and initialize a queue with vwhile the queue is not empty do
for each vertex w in V adjacent to the front vertex do
if w is marked with 0count ← count + 1;
mark w with the countadd w to the queue
remove the front vertex from the queue
C program:
#include<stdio.h>
int count=0;
int n;
void bfs(int a[][n],int b[],int v);
main()
{
int i,j;
printf("\nEnter number of vertices:");
scanf("%d",&n);
int a[n][n];
int b[n];
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("\nEnter connection bt vertex %c & vertex %c",i+65,j+65);
scanf("%d",&a[i][j]);
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d\t",a[i][j]);
}
printf("\n");
}
for(i=0;i<n;i++)
b[i]=0;
bfs(a,b,0);
}
void bfs(int a[][n],int b[],int v)
{
int i,r=0,f=-1;
int q[n];
count++;
b[v]=count;
q[++f]=v;
do
{
v=q[r];
for(i=0;i<n;i++)
{
if(a[v][i]==1)
{
if(b[i]==0)
{
count++;
b[i]=count;
q[++f]=i;
}
}
}
printf("%c\n",q[r++]+65);
}while(f>=r);
}
Output:
Enter number of vertices:4
Enter connection bt vertex A & vertex A0
Enter connection bt vertex A & vertex B1
Enter connection bt vertex A & vertex C0
Enter connection bt vertex A & vertex D0
Enter connection bt vertex B & vertex A1
Enter connection bt vertex B & vertex B0
Enter connection bt vertex B & vertex C1
Enter connection bt vertex B & vertex D0
Enter connection bt vertex C & vertex A0
Enter connection bt vertex C & vertex B1
Enter connection bt vertex C & vertex C0
Enter connection bt vertex C & vertex D1
Enter connection bt vertex D & vertex A1
Enter connection bt vertex D & vertex B0
Enter connection bt vertex D & vertex C1
Enter connection bt vertex D & vertex D0
0 1 0 0
1 0 1 0
0 1 0 1
1 0 1 0
A
B
C
D
Wednesday 27 March 2019
Sieve of Eratosthenes Algorithm
ALGORITHM Sieve(n)
//Implements the sieve of Eratosthenes
//Input: A positive integer > 1
//Output: Array L of all prime numbers less than or equal to n
for p ← 2 to n do A[p] ← p
for p ← 2 to √n do
if A[p] = 0 //p hasn’t been eliminated on previous passes
j ← p ∗ p
while j ≤ n do
A[j] ← 0 //mark element as eliminated
j ← j + p
//copy the remaining elements of A to array L of the primes
i ← 0
for p ← 2 to n doifA[p] = 0
L[i] ← A[p]
i ← i + 1
return L
Method:
Algorithm to find all the prime numbers less than or equal to a given integer n by Eratosthenes’ method:
1. Create a list of consecutive integers from 2 to n: (2, 3, 4, …, n).
2. Initially, let p equal 2, the first prime number.
3. Starting from p2, count up in increments of p and mark each of these numbers greater than or equal to p2 itself in the list. These numbers will be p(p+1), p(p+2), p(p+3), etc..
4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this number (which is the next prime), and repeat from step 3.
C program:
// C program to print all primes smaller than or equal
//to n using Sieve of Eratosthenes
#include <stdio.h>
#include <stdlib.h>
void SieveOfEratosthenes(int n)
{
// Create a integer array "prime[0..n]" and initialize
// all entries it as 1(true). A value in prime[i] will
// finally be 0(false) if i is Not a prime, else true.
int prime[n+1];
int p,i;
for(i=0;i<=n;i++)
{
prime[i]=1;
}
for (p=2; p*p<=n; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] ==1)
{
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for ( i=p*p; i<=n; i+= p)
prime[i] = 0;
}
}
// Print all prime numbers
for (p=2; p<=n; p++)
if (prime[p]==1){
printf("%d ",p);
}
}
int main()
{
int n;
printf("Enter the N value\n");
scanf("%d",&n);
printf("Following are the prime numbers smaller than or
equal to %d\n",n);
SieveOfEratosthenes(n);
return 0;
}
Output:
Enter the N value
30
Following are the prime numbers smaller than or equal to 30
2 3 5 7 11 13 17 19 23 29
SortAnalysis Aglorithm
ALGORITHM SortAnalysis(A[0..n - 1])
//Input: An array A[0..n - 1] of n orderable elements//Output: The total number of key comparisons made
count ← 0
for i ← 1 to n - 1 do
v ← A[i]
j ← i - 1
while j ≥ 0 and A[j] > v do
count ← count + 1
A[j + 1] ← A[j]
j ← j - 1
A[j + 1] ← v
return count
C program:
#include<stdio.h>int main(){
int n,v,i,j,count=0,arr[100];
printf("Enter the No element u want to enter\n");
scanf("%d",&n);
printf("Enter the Elements\n");
for(i=0;i<n;i++){
scanf("%d",&arr[i]);
}
for(i=1;i<n;i++)
{
v=arr[i];
j=i-1;
while(j>=0 && arr[j]>v){
count++;
arr[j+1]=arr[j];
j--;
arr[j+1]=v;
}
}
for(i=0;i<n;i++)
{
printf("%d, ",arr[i]);
}
printf("\nNo of Steps:%d\n",count);
}
Output:
Enter the No element u want to enter5
Enter the Elements
1
32
54
6
4
1 ,4, 6, 32, 54
No of Steps:5
Factorial Algorithm
ALGORITHM F(n)
//Computes n! recursively//Input: A nonnegative integer n
//Output: The value of n!
if n = 0 return 1
else return F (n - 1) ∗ n
C program:
#include <stdio.h>
int factorial(int c){
if(c==1){
return 1;
}
else{
return c*factorial(c-1);
}
}
int main()
{
int c, n, fact = 1;
printf("Enter a number to calculate its factorial\n");
scanf("%d", &n);
fact=factorial(n);
// for (c = 1; c <= n; c++)
//fact = fact * c;
printf("Factorial of %d = %d\n", n, fact);
return 0;
}
Output:
Enter a number to calculate its factorial
5
Factorial of 5 = 120
Binary Algorithm
ALGORITHM Binary(n)
//Input: A positive decimal integer n
//Output: The number of binary digits in n’s binary representation
count ← 1
while n > 1 do
count ← count + 1
n ← n/2
return countAnother Way
ALGORITHM BinRec(n)
//Input: A positive decimal integer n
//Output: The number of binary digits in n’s binary representation
if n = 1 return 1
else return BinRec(n/2) + 1
C program:
// Iterative C program to count the number of
// digits in a number in binary form
#include <stdio.h>
int countDigit(long long n)
{
int count = 0;
while (n != 0) {
n = n /2;
++count;
}
return count;
}
int main(void)
{
long long n =1024;
printf("Number of digits : %d",countDigit(n));
return 0;
}
Output:
Number of digits : 11
