Interview Coding Questions

Question:-  Given three integers x,y and z you need to find the all the possible combination of numbers having 4 at most x times, having 5 at most y times and having 6 at most z times as a digit.
 Example:-
input x=1,y=1,z=1
output

4
5
6
45
54
56
65
46
64
456
465
546
564
645
654

Solution :-  Please note code is in c.

 Github link for c file can be found at:- https://github.com/nikhilmhala/C_code/blob/master/intergerrepetation.c

/******************************************************************************

 Given three integers x,y and z you need to find the all the possible combination of numbers having 4 at most x times, having 5 at most y times and having 6 at most z times as a digit.

 Example:-

input x=1,y=1,z=1

output

4

5

6

45

54

56

65

46

64

456

465

546

564

645

654

*******************************************************************************/

#include

#include

#define BASE 10

int main()

{

    int x,y,z,j,t;

    long long num, n;

    int i, lastDigit;

    int freq[BASE];

    printf("Enter frequncy of x y and  z");

    scanf("%d %d %d", &x,&y,&z);

    t=x+y+z;

    for(j=1;j<(6*pow(10,t));j++)

    {

    /* Input number from user */

   // printf("Enter any number: ");

    //scanf("%lld", &num);

    

    /* Initialize frequency array with 0 */

    for(i=0; i

    {

        freq[i] = -10;

       // printf("%d",freq[i]);

    }

    /* Copy the value of 'num' to 'n' */

    num = j;

    n = num; 

    /* Run till 'n' is not equal to zero */

    while(n != 0)

    {

        /* Get last digit */

        lastDigit = n % 10;

        /* Remove last digit */

        n /= 10;

        /* Increment frequency array */

        if(lastDigit==4 || lastDigit==5 || lastDigit==6)

        {

                if(freq[lastDigit]==-10)

                    freq[lastDigit] = -1;

                freq[lastDigit]++;

        }

        else

        {

                num=-1;

                break;

        }

    }

    if(freq[4] ==(x-1) && freq[5]==-10 && freq[6] == -10 && num>0)

        printf(" %d\n",num);

    if(freq[5] ==(y-1) && freq[4]==-10 && freq[6] == -10 && num>0)

        printf(" %d\n",num);

    if(freq[6] ==(z-1) && freq[5]==-10 && freq[4] == -10 && num>0)

        printf(" %d\n",num);

    if(freq[4] ==(x-1) && freq[5]==(y-1) && freq[6]==-10 && num>0)

        printf(" %d\n",num);

    if(freq[5] ==(y-1) && freq[6]==(z-1) && freq[4]==-10 && num>0)

        printf(" %d\n",num);

    if(freq[6] ==(z-1) && freq[4]==(x-1) && freq[5]==-10 && num>0)

        printf(" %d\n",num);

    if(freq[4]==(x-1) && freq[5]==(y-1) && freq[6]==(z-1) && num>0)

        printf(" %d\n",num);

    }

        

    

    return 0;

}

Problem Statement

The Head Librarian at a library wants you to make a program that calculates the fine for returning the book after the return date. You are given the actual and the expected return dates. Calculate the fine as follows:

1. If the book is returned on or before the expected return date, no fine will be charged, in other words fine is 0.
2. If the book is returned in the same month as the expected return date, Fine = 15 Hackos × Number of late days
3. If the book is not returned in the same month but in the same year as the expected return date, Fine = 500 Hackos × Number of late months
4. If the book is not returned in the same year, the fine is fixed at 10000 Hackos.

Input Format

You are given the actual and the expected return dates in D M Y format respectively. There are two lines of input. The first line contains the D M Y values for the actual return date and the next line contains the D M Y values for the expected return date.

Constraints 
1≤D≤31 
1≤M≤12 
1≤Y≤3000

Output Format

Output a single value equal to the fine.

Sample Input

9 6 2015
6 6 2015

Sample Output

45

Explanation

Since the actual date is 3 days later than expected, fine is calculated as 15×3=45 Hackos.

Solution:-
 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
 int AD,ED,AM,EM,AY,EY;
    int D,M,Y,Fine=0;
    scanf("%d %d %d",&AD,&AM,&AY);
    scanf("%d %d %d",&ED,&EM,&EY);
    if((1<=AD<=31 && 1<=AM<=12 && 1<=AY<=3000)||(1<=ED<=31 && 1<=EM<=12 && 1<=EY<=3000))
    {

    Y=AY-EY;
    M=AM-EM;
    D=AD-ED;
    if(Y<0)
    {

        printf("0");
        return 0;
    }
      if(Y==0 && M<0)
    {

        printf("0");
        return 0;
    }
      if(Y==0 && M<=0 && D<0)
    {

        printf("0");
        return 0;
    }
    if(Y>0)
    {
        printf("10000");
        return 0;
    }
    if(M==0)
    {
        printf("%d",(D*15));
        return 0;
    }
    if(Y==0 && M>0)
    {
        printf("%d",(M*500));
    }
    }

    return 0;
}

Problem Statement

You are given time in AM/PM format. Convert this into a 24 hour format.

Note Midnight is 12:00:00AM or 00:00:00 and 12 Noon is 12:00:00PM.

Input Format

Input consists of time in the AM/PM format i.e. hh:mm:ssAM or hh:mm:ssPM
where
01≤hh≤12
00≤mm≤59
00≤ss≤59

Output Format

You need to print the time in 24 hour format i.e. hh:mm:ss
where
00≤hh≤23
00≤mm≤59
00≤ss≤59

Sample Input

07:05:45PM

Sample Output

19:05:45

Solution:-

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    char str[100];
    const char s[2]=":";
    char ans[10],ans1[10],ans3[10];
    char ss[10];
    char read;
    char *h;
    int arr[2],i=0;
    scanf("%s",str);
    // printf("%s",str);

    h=strtok(str,&s);
    //printf("%s",h);
    while(h!=NULL)
    {
        if(i==2)
        {
            strcpy(ss,h);
            //printf("%s",ss);
            read=ss[2];
            ss[2]='\0';
            strcpy(ans3,ss);
            arr[i]=atoi(ss);
         //   printf("%d\n",arr[i]);

        }
        else{

       // printf("%d\n",arr[i]);
      if(i==0)
      {
           strcpy(ans,h);

      }
      if(i==1)
      {
           strcpy(ans1,h);
       //printf("%s",ans);
      }
       arr[i]=atoi(h);
     //  ans++;
        }
        i++;
        h=strtok(NULL,&s);
    }
    if(read=='A'&&arr[0]==12)
    {
        arr[0]=00;
        strcpy(ans,"00");
    }
    if(read=='P')
    {
        if(arr[0]!=12)
        arr[0]=arr[0]+12;
    }
    if((1<=arr[0]<=12)&&(0<=arr[1]<=59)&&(0<=arr[2]<=59))
    {
        if(read=='P')
        {
            if(arr[0]==12)
                   printf("%s:",ans);
            else
                   printf("%d:",arr[0]);
    printf("%s:",ans1);
    printf("%s\n",ans3);
        }
    else
    {
        printf("%s:",ans);
            printf("%s:",ans1);
    printf("%s",ans3);
    }
    }
   // int hh=atoi()
    return 0;
}