hahaha I feel you!

This is my noob code. It is not glamorous but it passed the tests

def simpleArraySum(n, ar):
    m = 1
    a = ar[0]
    while m < n:
        a = a + ar[m]
        m = m + 1
    return a

and I was saying to myself that if this is for beginners, I am gonna be Royally Screwed!!!

This worked for me :

def simpleArraySum(n, ar):
    sum = 0 
    i = 0 
    while i < n :
        sum += ar[i]
        i += 1
    
    return sum

feelsbadman

No, before you do algorithm challenges, you should know the basics of a programming language. You need to know about functions and arrays for this challenge, even though it's just adding two numbers together.

By the way, you don't need to understand all that code they put in there. You only need to put your code where they tell you. But they ought to explain that, because it's very confusing to see all that.

Right just like i thought. I glad Im not alone lmao smh

I think it's for beginners but there's just a lot in there. Like the "n" parameter for some reason is there but doesn't even have too be used. There's literally no point of it.

same bro

This is simple and gets the job done.

int sum=0,i,size,array[1000];
    scanf("%d",&size);
    for(i=size;i>=0;i--)
    {
    scanf("%d",&array[i]);
    }
    for(i=size;i>=0;i--)
    {
        sum+=array[i];
    }
    printf("%d",sum);
    
    return 0;

Here a sample from java static int simpleArraySum(int n, int[] ar) { int sum = 0; for (int i=0; i } return sum; }

same for me!!!

agree

Don't let this frustrate you. Over time, your coding skills will improve, to the point that, this will become trivial to you :)

In an attempt to help new coders doing this challenge, I will add a few ways this problem could have been resolved; in addition, some comparisons among them.

*****   Spoiler Alert

Different approaches

Note that I removed the n parameter. Dont see any use for it here, and len(arr) will provide the same value in constant time, O(1).

# Python comes with a Built-in function, sum, which
# takes an iterable and returns the sum of its values.
def array_sum_v1(arr):
    return sum(arr)

def array_sum_v2(arr):
    total = 0
    for number in arr:
        total += number
    return total

def array_sum_v3(arr):
    total = 0
    idx = len(arr) - 1
    while idx >= 0:
        total += arr[idx]
        idx -= 1
    return total

# If you need to preserve `arr`, create an auxiliar
# variable ``arr_cp`` and pop directly from that.
def array_sum_v4a(arr):
    total = 0
    # This is called slicing (can be used to copy the array)
    arr_cp = arr[:]
    # It will run until there are no more elements on arr_cp
    while arr_cp:
    	# `pop` removes (and returns) the last element
        # We could use it directly, without the
        # variable assignment
        number = arr_cp.pop()  
        total += number
    return total


# We will remove `number = arr_cp.pop()`
# and use it directly on the next line
def array_sum_v4b(arr):
    total = 0
    while arr:
        total += arr.pop()  # Avoids one line
    return total


# Overkill...
# But just wanted to show that there are
# much more ways to do this simple task; same as
# for every non-trivial problem.
# What is inside, its called a generator.
def array_sum_v5(arr):
    return sum(x for x in arr)

Time comparison

Below I created an array of 10000 random integers in the range [0, 10^9]. This will be used to compare among the functions.

import random

random.seed(0)
arr = [random.randint(0, 10**9) for _ in range(10**4)]

%timeit array_sum_v1(arr[:])
1000 loops, best of 3: 334 µs per loop

%timeit array_sum_v2(arr[:])
1000 loops, best of 3: 543 µs per loop

%timeit array_sum_v3(arr[:])
1000 loops, best of 3: 1.38 ms per loop

%timeit array_sum_v4a(arr)
1000 loops, best of 3: 1.55 ms per loop

%timeit array_sum_v4b(arr[:])
1000 loops, best of 3: 1.47 ms per loop

%timeit array_sum_v5(arr[:])
1000 loops, best of 3: 798 µs per loop

As expected, using the Built-in sum function is faster than looping over every item, loops in Python are kinda inefficient, but again, that is a topic for another day.

Hope this helps :)

def simpleArraySum(ar):
   return sum(ar)

That's it's, just the lines return sum(ar) don't overthink it some people get overwhelemd by the wealth of data in front of them. :)

i feel the same. i thought this would help to build some basics. It's way to beyond for starters like me.

yess,true.....

what is the meaning of stray error 240 and 302 also what is the use of all the lengthy stuff given in the program originally..

:-)

you are right indeed

import java.io.; import java.math.; import java.text.; import java.util.; import java.util.regex.*;

public class Solution {

public static void main(String[] args) throws IOException 
{
    Scanner sc = new Scanner(System.in);
    int l = sc.nextInt();
    int[] arr = new int[l];
    int sum =0;
    for(int i=0;i<l;i++)
    {
        arr[i]= sc.nextInt();
    } 
    for(int i=0;i<l;i++)
    {
        sum = sum+arr[i];
    }
    System.out.println(sum);
}

}

totally relatable right now!!

damn! I know right the wording sucks it make no sense for beginners

int main() { int a[1000],n=1,i=0,result=0; scanf("%d",&n); for(i=0;i

/* Enter your code here. Read input from STDIN. Print output to STDOUT */    
return 0;

}

I'm having the same trouble. I've tested this locally and it works, and I've passed test case #0 and #1, but mysteriously, not #2 (apologies for the Pascal, I'm a Delphi developer and it's my most comfortable language):

(* Enter your code here. Read input from STDIN. Print output to STDOUT *)

program ArraySum;
var 
    arraySize, i, sum, valErr: integer;
    intArray: array of integer;
    inputIntString, subIntString: string;
begin
    readln(arraySize);
    SetLength(intArray, arraySize);
    readln(inputIntString);
    i := 0;
    while (i < arraySize) and (inputIntString <> '') do
    begin
      subIntString := Copy(inputIntString, 0, Pos(' ', inputIntString) -1);
      if subIntString = '' then
      begin
        subIntString := inputIntString;
      end;
      Val(subIntString, intArray[i], valErr);
      inputIntString := Copy(inputIntString, Pos(' ', inputIntString) + 1, Length(inputIntString));
      i := i + 1;
    end;
    i := 0;
    sum := 0;
    while (i < arraySize) do
    begin
        sum := sum + intArray[i];
        i := i+1;
    end;
    writeln(sum);
end.

You can download the testcases and verify yourself.

I feel you, there should be a feedback mechanism for beginners. My code solve the Test case and then it fails two other cases and nothing else. How do I know what was wrong? I ask for help here and I dont get none

Memory savings!

nice one

thanks

Or you can just :

static int simpleArraySum(int[] ar) 
{
    return ar.Sum();
 }

This absolutely helped me solve this challenge! thanks a lot

Hi! How can you do this without initializing the array?

I tried your way but it keeps not working. So I tried

static int simpleArraySum(int[] ar) {

int n = 1;
int[] arrayFdp = new int [n];

int soma = 0;

for (int i = 0; i < arrayFdp.Length; i++)
    {
    soma = soma + arrayFdp[i];
     }

return soma; }

    and it does't pass all the sample testcases!
    What am I doing wrong?
    Thanks

return IntStream.of(ar).sum();

How to check this if we are passing the arguments in different editor?

So elegant! I always forget about it and end up using reduce. It is fast too. Granted, it's by a little margin and only ~8 / 10 times, but still, it's a winner:

# Benchmark.bmbm on an array of 100000 * 7 digits integers
Rehearsal -------------------------------------------
sum    0.000157   0.000002   0.000159 (  0.000157)
reduce 0.000183   0.000003   0.000186 (  0.000187)
each   0.006423   0.000001   0.006424 (  0.006429)
---------------------------------- total: 0.006769sec

              user     system      total        real
sum    0.000140   0.000002   0.000142 (  0.000139)
reduce 0.000170   0.000002   0.000172 (  0.000171)
each   0.006728   0.000034   0.006762 (  0.006841)

You need to save "i" as a variable like below

function simpleArraySum(ar) { var ans = 0; for (var i = 0; i < ar.length; i++) { ans += ar[i]; } return ans; }