The Interview Question Today#92(26-5-19)

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The Interview Question Today#27(26-3-19)(Size of structure)

Question:

Write a c program to find the size of structure without using sizeof operator?

Solution:

C program:

#include<stdio.h>
struct  XYZ{
    int x;
    float y;
    char z;
};

int main(){
    int sz = (int) (((struct XYZ *)0) + 1);
    printf("Size of Struct= %d",sz);
    return 0;

}

Output:

Size of Struct= 12

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Integer Promotions in C

Integer Promotions in C

Some data types like char, short int take less number of bytes than int, these data types are automatically promoted to int or unsigned int when an operation is performed on them. This is called integer promotion. For example, no arithmetic calculation happens on smaller types like char, short and enum. They are first converted to int or unsigned int, and then arithmetic is done on them. If an int can represent all values of the original type, the value is converted to an int. Otherwise, it is converted to an unsigned int.

For example see the following program:

#include <stdio.h>
int main()
{
    char a = 30, b = 40, c = 10;
    char d = (a * b) / c;
    printf ("%d ", d);
    return 0;
}

Output:

120
At first look, the expression (a*b)/c seems to cause arithmetic overflow because signed characters can have values only from -128 to 127 (in most of the C compilers), and the value of subexpression ‘(a*b)’ is 1200 which is greater than 128. But integer promotion happens here in arithmetic done on char types and we get the appropriate result without any overflow.

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The Interview Question Today#54(18-4-19)

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Tricky Interview #1

Question:

W A C Program That Prints All The vowels Given In String Without Using IF and Break Statements In program? 

Submit Ur Answers (Follow The Link)

Link: https://docs.google.com/forms/d/e/1FAIpQLSeBXalNTB5cLX8dC8BlFmzGcItV52GBiUkqaaCM2q_bpaDnKg/viewform?usp=sf_link

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The Interview Question Today#36(4-4-19)

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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

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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

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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.

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;

}

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Print “Even” or “Odd” without using conditional statement

Question:

Print “Even” or “Odd” without using a conditional statement

C program:

#include<stdio.h>

int main()

{

    int no;

    printf("Enter a no: ");

    scanf("%d", &no);

    (no & 1 && printf("odd"))|| printf("even");

    return 0;

}

Output:

Enter a no: 45

odd

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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

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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

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To find sum of two numbers without using any operator

Write a program to find the sum of positive integers without using any operator.

C program:

#include<stdio.h>

int add(int x, int y)

{

    return printf("%*c%*c", x, ' ', y, ' ');

}

int main()

{

    printf("Sum = %d", add(3, 4));

    return 0;

}

Output:

Sum = 7

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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.

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

//  f[] -->  An array that contains
//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

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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

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The Interview Question Today#34(2-4-19)

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The Interview Question Today#33(1-4-19)

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The Interview Question Today#32(31-3-19)

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The Interview Question Today#31(30-3-19)

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The Interview Question Today#30(29-3-19)

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