Input array with unknown variable column size

Question

I have to solve the following problem:

Raghav is an online shopper. He wants to order some products. But before he orders the products he wants to compare it with the other merchants. Each row specifies a product but the limitation of comparing the prices are only with maximum of three merchants. Find the minimum in each row and add the price and display.

Assume the limit to buy a product is 10. Assume the number of merchant for each product may vary from 1 to 3.

Test cases

Input 1

Enter the number of products:
3
Enter the price per product:
110 102 100
250 200 312
412 450 475

Output 1

The minimum cost of each product is as follows
100
200
412
Total amount you need to pay Rs.712

Input 2

Enter the number of products:
13

Output 2

You are allowed to buy maximum of 10 products

Input 3

Enter the number of products:
3
Enter the price per product
100 250
500 300
450 350

Output 3

The minimum cost of each product is as follows
100
300
350
Total amount you need to pay Rs.750

This is my code:

#include<stdio.h>
#include<stdlib.h>
#define NN 31

int main()
{
int n;
printf("Enter the number of products:\n");
scanf("%d",&n);

if (n>10)
{ 
 printf("You are allowed to buy maximum of 10 products"); return 0;
}

printf("Enter the price per product:\n");

//Storing  the input in a 2D string first

char **str=malloc((n)*sizeof(char *));
for(int i=0;i<n;i++)
{
    str[i]=malloc(NN);
}

scanf("\n");
for(int i=0;i<n;i++)
{
    fgets(str[i],NN,stdin);
}

int **arr=(int **)calloc(n,sizeof(int *));
for (int i=0;i<n;i++)
    arr[i]=(int *)calloc(3,sizeof(int));

//Conversion of 2D string to 2D integer array

int max;
for(int i=0;i<n;i++)
{
    int num, j = 0, len;
    max=j;
while ( sscanf( str[i], "%d%n", &num, &len) == 1 ) 
{
    arr[i][j]=num;
    str[i] += len;    
    j++;
    if(j>max) max=j;        //find the column size of 2D array   
}
}

//Calculation of minimum possible cost
int *min,amount=0;
 min=(int *)calloc(n,sizeof(min));

printf("The minimum cost of each product is as follows:\n");

for (int i=0;i<n;i++)
    {
    min[i]=arr[i][0];
    for(int j=1;j<max;j++)
    {
        if (arr[i][j]<min[i])
        {
            min[i]=arr[i][j];
        }
    }
    printf("%d\n",min[i]);
    amount=amount+min[i];
    }
 printf("Total amount you need to pay Rs.%d",amount);
 free(min);

for(int i=0;i<n;i++)
{
    free(str[i]);
    free(arr[i]);
}
free(str);
free(arr);
return 0;
}

This code works fine if the input is provided as shown above (I read each row separately and terminate the row when \n is encountered).

But how to handle the cost inputs if they are provided as follows:

Input 4

Enter the number of products:
3
Enter the price per product
100
250
500
300
450
350

Again, the column size can vary between 1 to 3 and the user does not specify the column size.

I need help regarding how to stop the array input as and when the user stops. He/She may enter 3 or 6 or 9 prices for 3 products(n=3).

Minor: allocation size and fgets() size can be the same #define NN 31 ... str[i]=malloc(NN); ... fgets(str[i],NN,stdin); also no need for all that casting like (char *) cast. — chux
– chux, Commented Jun 16, 2017 at 2:59
The last case is confusing. number of products is 3, so why 6 lines of prices? Should not the 4th line onward be considered errors? — chux
– chux, Commented Jun 16, 2017 at 3:07
your are loosing str[i] here ` str[i] += len;` the free cannot work — Ôrel
– Ôrel, Commented Jun 16, 2017 at 3:33
IMO, dis-allow case 4. After the 3rd line is entered, give the result just like in case 3 — chux
– chux, Commented Jun 16, 2017 at 3:33

Ben · Accepted Answer · 2017-06-16 03:49:19Z

1

To stop the user input you can read all lines the user enters (up to 9 for the last case) and if the user has entered more then n lines, check for each line after, if the line is empty. If line is empty the user has completed their input.

edited Jun 16, 2017 at 3:49

answered Jun 16, 2017 at 3:29

Ben

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3 Comments

chux Over a year ago

So in cases 2,3, user now needs to do something more to get the results.

Ben Over a year ago

@chux updated my answer, if >n lines check for empty line.

Aritra Dasgupta Over a year ago

Thanks. This worked: int i=0; while(i<n1&&scanf("%d",&a[i])==1) i++;

Ôrel · Accepted Answer · 2017-06-16 03:36:04Z

0

here a solution

Add a loop to enter several value, break the loop when the user set number of product to 0.

Fix free of str[i]

#include<stdio.h>
#include<stdlib.h>
#define NN 31

int main() 
{       
    int n;  
    while (42) {
        printf("Enter the number of products, 0 to quit:\n");
        scanf("%d",&n);

        if (n > 10)
        {
            printf("You are allowed to buy maximum of 10 products");
            continue;
        } else if (n == 0) {
            printf("Bye");
            return 0;
        }

        printf("Enter the price per product:\n");

        //Storing  the input in a 2D string first

        char **str=malloc((n)*sizeof(char *));
        for(int i=0;i<n;i++)
        {   
            str[i] = malloc(NN);
        }

        scanf("\n"); 
        for(int i=0;i<n;i++)
        {
            fgets(str[i], NN, stdin);
        }

        int **arr= (int **)calloc(n,sizeof(int *));
        for (int i = 0; i < n; i++)
            arr[i] = (int *)calloc(3,sizeof(int));

        //Conversion of 2D string to 2D integer array

        int max;
        for(int i=0;i<n;i++)
        {       
            int num, j = 0, len;
            char *p = str[i];
            max=j;
            while ( sscanf(p, "%d%n", &num, &len) == 1 )
            {
                arr[i][j]=num;
                p += len;
                j++;
                if(j>max) max=j;        //find the column size of 2D array
            }
        }

        //Calculation of minimum possible cost
        int *min,amount=0;
        min=(int *)calloc(n,sizeof(min));

        printf("The minimum cost of each product is as follows:\n");
        for (int i=0;i<n;i++)
        {       
            min[i]=arr[i][0];
            for(int j=1;j<max;j++)
            {
                if (arr[i][j]<min[i])
                {
                    min[i]=arr[i][j];
                }
            }
            printf("%d\n",min[i]);
            amount=amount+min[i];
        }
        printf("Total amount you need to pay Rs.%d\n",amount);
        free(min);

        for(int i=0;i<n;i++)
        {
            free(str[i]);
            free(arr[i]);
        }
        free(str);
        free(arr);
    }
    return 0;
}

answered Jun 16, 2017 at 3:36

Ôrel

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6 Comments

Aritra Dasgupta Over a year ago

Thanks, but can't change the input from test case, as it will be evaluated as incorrect.

Ôrel Over a year ago

What to you expect as output for your input 4 ?

Aritra Dasgupta Over a year ago

The input of 6 prices values is interpreted as a 3 x 2 matrix with 2 prices for each of the 3 products. The output is sum of the minimum in each row.

Ôrel Over a year ago

How you detect the user has finished to give input ? perhaps he will give 9 prices, 3 prices for 3 products ?

Aritra Dasgupta Over a year ago

I used this: int i=0; n1=3*n; while(i<n1&&scanf("%d",&a[i])==1) i++;

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