febup here is my solution.
i implemented the function solution(...) to return an int array of p's. plz check it out!

@Crotonite

Very nice of you for doing this test. What you can do now is to add a timer at the start and end of your code to see how long it will take to return all the equilibrium indices of P if N = 100,000. Then rewrite your code to make it run faster using Big 0 Notation if needed.

check this one out, it is quite fast (50 iterations of N = 100000 takes under 3secs i.e including the overhead of printing to stdout).

@Crotonite

Please try your code on their page at this link: https://codility.com/demo/take-sample-test
I don't think I can grade your code as the test belongs to the website owner at the above link.

k

crotonite:
k

I went through your code, what I found was that you have prefixed the values of the array N as: int[] ar = {-1, 3, -4, 5, 1, -6, 2, 1} which was just an example they used in the test. You were meant to generate random integers within the range [−2,147,483,648..2,147,483,647].

You also have this loop: for (int j = 0; j < 50; j++) in which you are calling this function int[] result = solution(arr2, arr2.Length);

This means your code is not doing what is required.

@febup i am still newbie you know. tanx, i will try to fix that asap.

@crotonite

Nice try bro keep it up

larisoft:



This runs in O(n).

public static int solution2(int[] nums, int n){

int sum = sum(nums);

int i = 0;
int left_sum = 0;

while(i < nums.length){

int right_sum = sum - left_sum - nums[i];

if(right_sum == left_sum){
return i;
}

left_sum+= nums[i];
i++;

}

return -1;

}

private static int sum(int[] arr){

int a = 0;
for(int i: arr){
a+=i;
}
return a;
}

Can someone explain how this runs in O(n)
cc Febup

jacob05:

Can someone explain how this runs in O(n)
cc Febup

@larisoft
You need to come and explain your code, I have converted it to C# below, when I run it, it does nothing.

using System;

namespace larisoftBigO
{
class Program
{
static void Main(string[] args)
{
int[] array = new int[8];

solution2(array, ;
}

public static int solution2(int[] nums, int n)
{

int sum = Sum(nums);

int i = 0;
int left_sum = 0;

while (i < nums.Length)
{

int right_sum = sum - left_sum - nums[i];

if (right_sum == left_sum)
{
Console.WriteLine(i.ToString());
Console.Read();
return i;
}

left_sum += nums[i];
i++;
Console.Read();
}

Console.Read();
return -1;

}

private static int Sum(int[] arr)
{

int a = 0;
foreach(int i in arr)
{
a += i;
}
return a;
}
}
}

@Crotonite
Your code is the closest to the solution, you have now dethroned larisoft unless he can come up with a better solution than yours.

^^^ah, i don't like what i am hearing at all. Larisoft, oya dethrone crotonite back, who the heck does he think he is?
He should be thankful that i have retired from programming to hunting for monitor lizards (so that i can make frontpage)

dhtml18:
^^^ah, i don't like what i am hearing at all. Larisoft, oya dethrone crotonite back, who the heck does he think he is?
He should be thankful that i have retired from programming to hunting for monitor lizards (so that i can make frontpage)

boss, make i sit for ur right hand o jare!

the prototype of my recent solution:

 import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		//int[] data = {-1, 3, -4, 5, 1, -6, 2, 1};

		int[] data = new int[100000];
		Random rand = new Random();
		
		 for(int i = 0; i  < data.length; i++){
			data[i] = rand.nextInt();
			//System.out.println(data[i]);
		}
		
		System.out.println("please wait...";
		double start = System.currentTimeMillis();
		int equill = solution(data, data.length);
		
		System.out.println();
		
		System.out.println("equillibrum at : " + equill);

			
		double stop = System.currentTimeMillis();

		System.out.println("it took " + ((stop - start)/1000) + "seconds to run "  );
	}





	private static int solution(int[] data, int n)
	{
		long lSum = 0, rSum = 0; //left and right summ of array
		int p = 0; //index of suspected equillibrum of array

		int ret = -1;

		//sum array 
		for (int i : data)
			rSum += i;
			
			//System.out.println("sum.." + rSum);
	

		//if first element is an aquillibrum, add to array
		if (rSum - data[p] == 0)
		{
			
		return p; }

		lSum = data[0];	 
		for (int i = 0; i < n - 1; i++)
		{
			p++;
			
			//System.out.println("rsum: " + rSum + " lsum: " + lSum);
			
			rSum -= (data[p] + lSum);

			if (lSum == rSum)
			{

				return p;

			}//endif

			//increment leftsum
			lSum += data[p];
		}
		
		return ret;
	}
} 

Chai, is that language above JAVA? before me i no fit read am, it is too concentrated for me. Abi na objective Java Language?

codility verified my algorithm....

Santa maria! i don see new one again o!! Wia is larisoft make him come debug this your codibility nonsense because adonbilivit!!!

dhtml18:
Chai, is that language above JAVA? before me i no fit read am, it is too concentrated for me. Abi na objective Java Language?

it is java ooo...

Hmn, this your JAVA is so concentrated that e no be like that JAVA wey i sabi again.

crotonite:
codility verified my algorithm....

Now let me verify it too from my end.

Febup:


Now let me verify it too from my end.

ooops, me bad!

i did not test for cases like: array == null, array.lenght >100000 || <= 0, etc. i will hv to enhance my algorithm when chanced. thanks...

dhtml18:
Hmn, this your JAVA is so concentrated that e no be like that JAVA wey i sabi again.

code obfuscation at work sire.

^^^Interesting. . .

crotonite:
the prototype of my recent solution:

 import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		//int[] data = {-1, 3, -4, 5, 1, -6, 2, 1};

		int[] data = new int[100000];
		Random rand = new Random();
		
		 for(int i = 0; i  < data.length; i++){
			data[i] = rand.nextInt();
			//System.out.println(data[i]);
		}
		
		System.out.println("please wait...";
		double start = System.currentTimeMillis();
		int equill = solution(data, data.length);
		
		System.out.println();
		
		System.out.println("equillibrum at : " + equill);

			
		double stop = System.currentTimeMillis();

		System.out.println("it took " + ((stop - start)/1000) + "seconds to run "  );
	}





	private static int solution(int[] data, int n)
	{
		long lSum = 0, rSum = 0; //left and right summ of array
		int p = 0; //index of suspected equillibrum of array

		int ret = -1;
                
		//sum array 
		for (int i : data)
			rSum += i;
                
			
			//System.out.println("sum.." + rSum);
	

		//if first element is an aquillibrum, add to array
		if (rSum - data[p] == 0)
		{
			
		return p; }

		lSum = data[0];	
                 
		for (int i = 0; i < n - 1; i++)
		{
			p++;
			
			//System.out.println("rsum: " + rSum + " lsum: " + lSum);
			
			rSum -= (data[p] + lSum);

			if (lSum == rSum)
			{

				return p;

			}//endif

			//increment leftsum
			lSum += data[p];
		}
                
		
		return ret;
	}
} 

Is the function solution not running at O(2n) cc dhtml18 Febup.

I guess they grow "linearly" the same way.. so the solution might be valid.

I agree with the poster above me. . . .

jacob05:
Is the function solution not running at O(2n) cc dhtml18 Febup.

It is Linear 0(N)

crotonite:
febup here is my solution.
i implemented the function solution(...) to return an int array of p's. plz check it out!

In Python

def solution(data):
  lSum = 0; ret = []; rSum = sum(data )
  if (rSum - data[0] == 0 ) : ret.append(1 )
  lSum = data[0]
  for p in xrange(1, len( data ) ) :
    if (lSum == (rSum - (data[p] + lSum)) ) : ret.append(p)
    lSum += data[p]
  return ret
  
ar = [-1, 3, -4, 5, 1, -6, 2, 1 ]
result = solution(ar )
print result

^^^
I don't have PHP setup on my machine so I tried your code on their site but it gave the error message below:

PHP Parse error: syntax error, unexpected '=' in func.php on line 6
Parse error: syntax error, unexpected '=' in func.php on line 6
RUNTIME ERROR (tested program terminated unexpectedly)

Detected some errors.

/* Your code*/
function solution($A) {
// write your code in PHP5.5
lSum = 0;
ret = [];
rSum = sum(A) )
if (rSum - A[0] == 0 ) : ret.append(1 )
lSum = A[0]
for p in xrange(1, len( A ) ) :
if (lSum == (rSum - (A[p] + lSum)) ) : ret.append(p)
lSum += A[p]
return ret
}

jacob05:

In Python

def solution(data):
lSum = 0; ret = []; rSum = sum(data )
if (rSum - data[0] == 0 ) : ret.append(1 )
lSum = data[0]
for p in xrange(1, len( data ) ) :
if (lSum == (rSum - (data[p] + lSum)) ) : ret.append(p)
lSum += data[p]
return ret
ar = [-1, 3, -4, 5, 1, -6, 2, 1 ]
result = solution(ar )
print result

Febup:
^^^
I don't have PHP setup on my machine so I tried your code on their site but it gave the error message below:

PHP Parse error: syntax error, unexpected '=' in func.php on line 6
Parse error: syntax error, unexpected '=' in func.php on line 6
RUNTIME ERROR (tested program terminated unexpectedly)

Detected some errors.

/* Your code*/
function solution($A) {
// write your code in PHP5.5
lSum = 0;
ret = [];
rSum = sum(A) )
if (rSum - A[0] == 0 ) : ret.append(1 )
lSum = A[0]
for p in xrange(1, len( A ) ) :
if (lSum == (rSum - (A[p] + lSum)) ) : ret.append(p)
lSum += A[p]
return ret
}

Attention to details ... It's in Python 2

jacob05:
Attention to details ... It's in Python 2

Edited

Ok I will use use .. Repl.it - Python 2 as you have suggested

Febup:


I need someone to run your code for us. Over to you dhtml18 if you have some time to spare.

use .. Repl.it - Python 2