Thread: Array Comparison and Modification

Originally Posted by R.Stiltskin

Still working on this, I see. I was hoping that by the time you finished implementing the step-counting you would have realized how simply (and efficiently) you could complete the assignment with only a minor modification to your original program. Does this give you a hint?

Code:

#include <stdio.h>#include <stdlib.h>
#define MAXVALUE 10
int compact (int[], int);
int main()
{
	int arrSeries[MAXVALUE], i=0, newArrayLength;
	int arrLength = MAXVALUE;
	printf("Please enter %d numbers.\n", MAXVALUE);
	while(i != MAXVALUE)
	{
		scanf("%d", &arrSeries[i]);
		i++;
	}
    newArrayLength=compact(arrSeries, arrLength);
    printf("The new array length is %d.\n", newArrayLength);
    getchar();
    return 0;
}
int compact(int arrSeries[], int arrLength)
{
	int i=0, j= 0, counter=0, newArray[MAXVALUE];
    newArray[0]=arrSeries[0];
    for(i = 0; i != arrLength; i++)
        {
            if(newArray[j] != arrSeries[i])
            {
                newArray[++j] = arrSeries[i];
            }
        }
    j++;
    while(counter < j)
        {
            printf("%d  ", newArray[counter]);
            counter++;
        }
	printf("\nThe number of steps: %d.\n", j);
	return j;
}

Isn't this the solution you intended? Or is there an even more elegant method other than in place compacting ?
Thanks.

Originally Posted by laserlight

Code:

(i < max_size && scanf("%d", &numbers[i]) == 1)

Sorry for my so less experience in coding, but I have never seen a while loop number reading like this. Can you please explain the logic so that I can implement loops like this myself? The implementation seems very elegant to me. I know that a while(1) loop is an infinite loop, but I don't' get the ==1 testing part.
Thanks !.

Originally Posted by laserlight

Basically, the idea is to keep reading integers until the max_size is reached, or there are no more integers left to read. Read up on scanf to understand its return value.

Sorry for being late, here's the solution that I did, don't know how efficient it is...

Code:

#include <stdio.h>#include <stdlib.h>
size_t read_numbers(int numbers[], int max_size)
{
    size_t i = 0;
    printf("Please enter up to %d integers (trigger EOF to stop early).\n", max_size);
    while (i < max_size && scanf("%d", &numbers[i]) == 1)
    {
        ++i;
    }
    return i;
}


void print_numbers(const int numbers[], size_t size)
{
    size_t i;
    printf(">>> ");
    for (i = 0; i < size; ++i)
    {
        printf("%d ", numbers[i]);
    }
    printf("<<<\n");
}


size_t compact(int numbers[], size_t size)
{
    size_t i, j=0, k=0;
    if (size <= 1)
    {
        return size;
    }
    else
    {
        for(i=(size-1); i>0; i--)
        {
            if(numbers[i-1]==numbers[i])
            {
				k=i;
				while(k!=size)
				{
                numbers[k-1]=numbers[k];
				k++;
				}
				j++;
            }
        }
		size=size-j;
		return (size);
    }
}


int main(void)
{
    int numbers[10];
    size_t size = read_numbers(numbers, sizeof(numbers) / sizeof(numbers[0]));
    size = compact(numbers, size);
    print_numbers(numbers, size);
	system("pause");
    return 0;
}

In the worst case, it's still O(n²).

You would do better to go back to your compact function in post #61 and make it in-place by simplifying it, rather than by complicating it. Essentially all you have to do is eliminate counter and newArray.

Originally Posted by R.Stiltskin

In the worst case, it's still O(n²).

You would do better to go back to your compact function in post #61 and make it in-place by simplifying it, rather than by complicating it. Essentially all you have to do is eliminate counter and newArray.

Will try to do the same.