Find the longest arithmetic progression from an array - c

Question

I have an array of numbers ex.[5,1,2,4,6,8,12], and I want to find the length of longest arithmetic progression within the sequence and to print it. Longest arithmetic progression means an increasing sequence with common difference, in this case [2,4,6,8].
#include <stdio.h>
void main()
{
int array[100], i, num,diff;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
printf("\n Numbers in a.p: ");
for (i = 0; i < num; i++) {
diff = array[i+1]-array[i];
if (array[i]-diff == array[i+1]-diff);
{
printf("%d, ", array[i]);
}
}
printf("\n Common difference:%d", diff);
}

Answer 1

like this
#include <stdio.h>
int main(void){
#if DEBUG
int array[] = {5,1,2,4,6,8,12};
int num = sizeof(array)/sizeof(*array);
int i, diff;
#else
int array[100], i, num,diff;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
#endif
int j, len, longest_len = 2, longest_i = 0;
for (i = 0; i < num - longest_len; i += len-1) {
diff = array[i+1]-array[i];
for(j = i+2; j < num && array[j-1]+diff == array[j]; ++j);
len = j - i;
if(longest_len < len){
longest_len = len;
longest_i = i;
}
}
printf("\n Numbers in a.p: ");
diff = array[longest_i+1] - array[longest_i];
printf("[ ");
for(i = 0; i < longest_len; ++i){
printf("%d", array[longest_i + i]);
if(i == longest_len-1)
printf(" ]\n");
else
printf(", ");
}
printf("\n Common difference:%d", diff);
}

Answer 2

Since this seems to be a homework or challenge I will only help by solving your immediate problems, caused by very strange code.
Here is code which at leat detects the progressions correctly.
You can yourself count the length, store the longest length and its index, then print that sequence after parsing all the array.
There are two assumptions here, which you might want to avoid for challenge/homework:
no identical numbers entered
no overlapping progressions,
i.e. no number is the last of one and the first of next progression
The representation of more than one sequence in the output is a little jittery (missing ")"), but that is not relevant for finding and exclusively printing the longest one.
I did not bother about your ratio output, no idea what that is supposed to be.
Code:
#include <stdio.h>
// avoid a warning
int main() {
int array[100], i, num, diff=0, lastdiff=0, first=1;
printf("Enter the size of an array \n");
// for good code, this should check the result
scanf("%d", &num);
// in case of scaf-failure, cleanup here for good code
// asking for the number of elements
// and then relying on that number to be entered
// is less elegant than checking for and end condition
// like EOF or negative input
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
// for good code, this should check the result
scanf("%d", &array[i]);
// in case of scaf-failure, cleanup here for good code
}
printf("\n Numbers in a.p: ");
for (i = 1; i < num; i++) {
lastdiff=diff;
diff = array[i]-array[i-1];
if (diff==lastdiff)
{ if(first==1)
{ first=0;
printf("(%d, %d",array[i-2], array[i-1]);
}
printf(", %d", array[i]);
} else
{ first=1;
}
}
printf(")\n Ratio:%d", diff);
// avoid a warning
return 0;
}

Answer 3

There are a number of ways to approach this challenge. You can either check each diff against each value in the array or work the other way around. Given you are not sorting the values, you may benefit by nesting the check of values within your loop over all possible diffs. Something similar to the following works:
#include <stdio.h>
int main (void) {
int a[] = {5, 1, 2, 4, 6, 8, 12},
n = sizeof a / sizeof *a,
startidx = 0,
maxlen = 0;
for (int d = 1; d < n; d++) { /* loop over diffs */
int idx = -1,
len = 1;
for (int i = 1; i < n; i++) /* loop over values */
if (a[i - 1] + d == a[i]) {
if (idx < 0) /* if index not set */
idx = i - 1; /* set to 1st index */
len++; /* increment length */
}
if (idx >= 0 && len > maxlen) { /* if progression found */
maxlen = len; /* save length */
startidx = idx; /* save start index */
}
}
printf ("longest progression: '%d' elements.\n", maxlen);
for (int i = 0; i < maxlen; i++)
printf (i ? ", %d" : "%d", a[startidx + i]);
putchar ('\n');
return 0;
}
Example Use/Output
$ ./bin/maxprogression
longest progression: '4' elements.
2, 4, 6, 8
Investigate several ways to approach it, and finally settle on the one that makes the most sense to you. You can work on optimizing later.
As far as the code you posted goes, always validate all user input by, at minimum, checking that the number of expected conversions to type took place, e.g.
if (scanf("%d", &num) != 1) {
fprintf (stderr, "error: input conversion failed for 'num'.\n");
return 1;
}
You would do the same in your values loop. Let me know if you have any questions. Good luck with your coding.

Answer 4

Following is the complete working code. You can see it working here:
#include <stdio.h>
int main()
{
int array[100], i, num,diff, resDiff, startIndx, resStartIndx, countCur, countPre;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
//Now code changes
startIndx =0, resStartIndx=0, countPre=0, resDiff=0;
for (i = 0; i < num; /*No increment required here*/) {
countCur =0;
startIndx=i;
countCur++;
if(++i < num)
{
diff = array[i] - array[startIndx];
countCur++;
i++;
while((i < num) && (diff == (array[i] - array[i-1])))
{
countCur++;
i++;
}
if(countCur > countPre)
{
resStartIndx = startIndx;
countPre = countCur;
resDiff = diff;
}
}
}
countPre += resStartIndx;
printf("\n Numbers in a.p: ");
for (i = resStartIndx; i < countPre; i++) {
printf("%d, ", array[i]);
}
printf("\n Common difference:%d", resDiff);
return 0;
}