Sorting program query - c

Question

I came across a sorting code in quora and I was very confused with some of the lines of code, if someone explains it to me then it will be a big help.
The code is as follows:
#include <stdio.h>
int array[1000001] = {0};
int main(){
int i,j;
scanf("%d",&j);
int x;
for (i=0;i<j;i++){
scanf("%d",&x);
array[x]++;
}
for (i=0;i<1000001;i++){
while(array[i]>0){
printf("%dn",i);
array[i]--;
}
}
return 0;
}
Can someone explain what is happening in
for(i=0;i<j;i++)
{
scanf("%d",&x);
array[x]++;
}
and here
for (i=0;i<1000001;i++)
{
while(array[i]>0)
{
printf("%dn",i);
array[i]--;
}
}
and what does this dn stands for?

Answer 1

This is by far the best reason not to copy anything from quora or other sources without understanding what it is doing.
No matter what, it doesn't ensure if any of the inputs is right or not.
for(i=0;i<j;i++)
{
scanf("%d",&x);
// Are you sure that 'x' is always a valid int value and is less than your array size?
// No, if a monkey enters 200000 here, kabooom!
array[x]++;
}
In this for loop, which will iterate for j times, an input of type int is taken, and value at that index will be incremented, to know what value was entered by marking that index as 1.
for (i=0;i<1000001;i++)
{
while(array[i] > 0)
{
printf("%dn",i);
array[i]--;
}
}
Its bad by design, because if I have entered only 3 numbers to sort, it still iterates for 1000001 times.
But, it works for an expected/deliberately bounded input values.
For example, lets say user entered 6 values: 4, 2, 7, 0, 6, 8
So, j = 6, and contents of array would be all zeroes except for 4th, 2nd, 7th , 0th, 6th and 8th index.
Now this while loop does a job: it checks if ith index is set, i.e. if that index was entered by user not not. So the code inside while loop will only be executed for 4th, 2nd, 7th , 0th, 6th and 8th indexes.
To explain how the value is printed, lets dry run the code, Initially i is 0.
// Is ith index 1?
while (array[i] > 0) // i is 0
{
// Yes it is,
printf("%d ", i);
// See above, Prints 0 - not 1!
// Printing 'i' is the trick, instead of printing array[i]..
array[i]--; // Note, makes it zero!
}
for next iteration, i is 1
// Is ith index 1? - No, so move on to next index
while (array[i] > 0) // i is 1
{
//
}
lets say iteration has reached 7th index,
// Is ith index 1?
while (array[i] > 0) // i is 7
{
// Yes it is,
printf("%d ", i);
// See above, Printing 'i' is the trick,
// instead of printing array[i]..
array[i]--; // Note, makes it zero!
}
This will sort the numbers, without a doubt, but with some predefined constaints.
Additionally, you can/should add some implementation to get the sorted values in an array, based on if your are required to do it or not.
Finally, "%dn" is not a format specifier. It was probably meant to be "%d\n".
If it is kept unchanged the way it is, it will append n as a character after every number it prints, pointless.

Answer 2

This attempts to read in a number and ingores that the reading might fail.
scanf("%d",&j);
Afterwards j is therefor potentially uninitialised.
This makes a loop, which will do j steps, i.e. potentially an uninitialised number of loops.
for(i=0;i<j;i++)
This attempts to read a number once per loop, without caring about potential failures.
scanf("%d",&x);
It then uses the potentially uninitialised or non-updated number x to access an array.
array[x]++;
In case x is inconsitently for any reason, the access may well be beyond the array, which introduces potential undefined behaviour.
Should it however succeed, the array entry is increased by one.
Most of the array entries are implicitly initialised by the way, not because of the {0}.
So the array entry at the index which matches the read number is counted up.
In total, the number of occurrences of each number in the hopefully correctly formatted input is counted.
Nothing is sorted.
Yet.
This loops over all entries in the array, starting at the lower indexes, i.e. at the lower values of occurred values in the input.
for (i=0;i<1000001;i++)
If an entries in the array is not 0, the index/value is printed and the count is decreased until the entry is zero.
{
printf("%dn",i);
array[i]--;
}
I.e. the current index/value is printed as often as it has been seen in the input.
Implicitly, any value which did not occur in the input is not printed.
Concerning the "%dn" have a look here:
Is %dn a format string?
The most plausible explanation why the "n" does not occur in output, while the output does contain otherwise unexplained newlines, is that the "n" is actually an "\n" in the code which produced the shown output but got lost when posting the code.
(I agree with #WedaPashi that the code will have trouble with negative numbers in input on top of all the other vulnerabilities. Well spotted.)
I believe that the apparent speed of the solution for low values is sabotaged by looping over many empty entries for high values. It gives misleadingly fast output, while taking a long time doing nothing to reach the end of the array before actually being finished.