Unable to traverse array fully in C - arrays

Question

I had this question to solve:
Find the length of increasing subarrays in the given array and print length of each segment in a new line
Here's what I have done so far:
#include <stdio.h>
int main(){
int c = 0;
int a[] = {1, 3, 2, 4, 5, 7};
for (int i = 0; i < 6; i++) {
if (a[i] > a[i-1]) {
c++;
}
else {
printf("%d\n", c);
c=1;
}
}
return 0;
}
My expected answer is:
2
4
but I am getting only 2, on increasing i to 7, I am getting correct answer, am I doing something wrong in traversing the array because the length is 6 and i<6 should have given the correct answer
Apologies in advance if the question is too trivial, just getting started.

Answer 1

It's simple. Because you have included in this line:
if (a[i] > a[i-1])
i starts with a 0 but a[-1] doesn't exist.
To fix your code:
#include <stdio.h>
int main(void){
int c = 0;
int a[] = {1, 3, 2, 4, 5, 7};
for (int i = 1; i < 6; i++) {
if (a[i] > a[i-1])
c++;
else {
printf("%d\n", c);
c=1;
}
}
return 0;
}

Answer 2

It is possible to do it with a single printf in the loop if you loop over all values from 0 to the array length, and treat the first and last iterations as special cases:
#include <stdio.h>
int main(void){
int c = 0;
int a[] = {1, 3, 2, 4, 5, 7};
const int len = sizeof a / sizeof a[0];
for (int i = 0; i <= len; i++) {
if (i < len && (i == 0 || a[i] > a[i-1])) {
c++;
} else {
printf("%d\n", c);
c = 1;
}
}
return 0;
}
The i < len makes sure that the code does not access a[len], which is out of bounds. When i equals len the code reaches the else part to print the length of the final segment.
The i == 0 makes sure that the code does not access a[-1], which is out of bounds, but we want it to increment c for the first entry in the array.
The code can be simplified a bit by initializing c = 1 and starting the loop at i = 1. By doing that, the first element of the array has already been accounted for:
#include <stdio.h>
int main(void){
int c = 1;
int a[] = {1, 3, 2, 4, 5, 7};
const int len = sizeof a / sizeof a[0];
for (int i = 1; i <= len; i++) {
if (i < len && a[i] > a[i-1]) {
c++;
} else {
printf("%d\n", c);
c = 1;
}
}
return 0;
}
The code will run slightly faster if the code inside the loop is simplified as much as possible, eliminating the i < len test inside the loop, and terminating the loop when i reaches len (i.e. continue the loop if i < len), but that will require the length of the final segment to be printed separately after the for loop has terminated:
#include <stdio.h>
int main(void){
int c = 1;
int a[] = {1, 3, 2, 4, 5, 7};
const int len = sizeof a / sizeof a[0];
for (int i = 1; i < len; i++) {
if (a[i] > a[i-1]) {
c++;
} else {
printf("%d\n", c);
c = 1;
}
}
printf("%d\n", c);
return 0;
}
If there is a requirement for the code to work when the overall sequence has zero length, then the final printf would print the wrong value 1. That could be dealt with either by setting c to 0 when len is 0, or by skipping the final printf if len is 0. (It depends whether you want to print anything at all when the overall sequence has zero length.)