Turbo Sort encountered runtime error.

help
,
1

hey can anyone find the bug in this program problem link


program TSORT;

var 
   t,x,N,y,temp:longint;
   num:array[1..100000] of longint;
begin
	read(t);
	
	for x := 1 to t do
	begin
	    readln(N);
	    num[x] := N;
	end;
	
	for x := 1 to t do
	begin
	for y := 1 to t do
	begin
	    if num[y] > num[y+1] then
	    begin
	    temp := num[y];
	    num[y] := num[y+1];
	    num[y+1] := temp;
	    end;
	    
	end;
	end;
	
	for x := 2 to t+1 do
	begin
	    writeln(num[x]);
	end;
end.

i run it on codechef code,compile & run it ran there succefully with giving right results but when i submitted it encountered an runtime error.

2

@vinay1729
Array out of bound.

Althogh this aproach wont fetch you Correct answer.
You need a better approach.

1 Like
3

please tell on which line i have array out of bound?

4

This is not the right approach to sort 10^6 numbers.

5

Use counting sort.

6

use counting sort.

7

i am not quite familiar with pascal.
But by seeing…

First of all…the two nested loop are wrong.
it should be…

for x := 1 to t do

begin

for y := x+1 to t do

 begin
8

if num[x] > num[y] then
begin
temp := num[x];
num[x] := num[y];
num[y] := temp;
end;

9

is this only the bug?, if there is more please do tell.

10

This is correct code

program TSORT;

var 
 t,x,N,y,temp:longint;
 num:array[1..100000] of longint;
begin
read(t);

for x := 1 to t do
begin
    read(N);
    num[x] := N;
end;



for x := 1 to t do
begin
for y := x+1 to t do
begin
    if num[x] > num[y] then
    begin
    temp := num[x];
    num[x] := num[y];
    num[y] := temp;
    end;

end;
end;

for x := 1 to t do
begin
    writeln(num[x]);
end;

end.

11

I have corrected your code.
see below.

12

Hey but it not worked.

13

hey is the problem is caused because of read() so rather i should use readln() also i have made array of size 1^100000 only however problem says [t <= 10^6] & [0 <= N <= 10^6]?

14

but it wont fetch you ac…

15

yup…use read()…
and 1000000…

16

will it fetch me AC it not fetched me AC rather it says now TLE so i think solution is ok but algorithms is not efficiently fast right?

17

Yes…Its not efficient u can use counting sort.
here is a method.
Just count how many u input 1,2 3,…
and print.

take an array of size 10^6.

initialize it with 0.

now when you input a number x.
then you should do a[x]=a[x]+1.

suppose you input 1 1 2 2 2 4 5 5.
then a becomes(1-indexing) 2 3 0 1 2 0 0 0 0…

it means 1 is occuring 2 times 2 is occuring 3 times…

18

How to declare and array of 10^6 size,since the above code is not working on arr[1000000] array size