whats wrong with the code I wrote, its passing 14/16 tests but theres two its failing due to timeout. at first I added them all to an array and reduced the array, I realized I can just add them in place and not have to do that, yet it still fails. I dont have access to the tests that it is failing because Im using the free trial on this platform, heres the code I wrote:

Is there a problem with converting the string into a number?

function solution(a) {
let sum = 0;
for (let i = 0; i < a.length; i++) {
const num = array[i];
for (let j = 0; j < a.length; j++) {
const secondNum = a[j];
sum += Number(`${num}${secondNum}`);
}
}
}

I dont know its from their daily timed free practice tests. the solution only got a 250/300. I cant see all the tests unless im a paying member. It passed the tests I was able to see.

I didnt understand can you please explain a bit moreâŚ thanks

I have arr[i] that I can access in the inner loop and I thought im using it effectively, I cant access arr[j] in outer loop.

If you mean to say that I should calculate that in the outer loop than in the inner loop I will still have to do a check to see if i===j and if so then I will skip over that, so how does that save me any work as I still have to do a check in the inner loop? or maybe you mean for me to use some sort of memoization.

I was wrong before when I said I didnt know, I think they will have no more than 105 items

Input/Output
[execution time limit] 4 seconds (js)
[input] array.integer a
A non-empty array of positive integers.
Guaranteed constraints:
1 â¤ a.length â¤ 105,
1 â¤ a[i] â¤ 106.
The sum of all a[i] â a[j]s. It's guaranteed that the answer is less than 253.

Sorry, this is a bit cleaner, you can likely still optimize here and there. I think this is what the other user suggested.

function solution(a) {
let sum = 0;
for (let i = 0; i < a.length; i++) {
const num = a[i];
sum+=Number(`${num}${num}`);
for (let j = i + 1; j < a.length; j++) {
const secondNum = a[j];
sum += Number(`${num}${secondNum}`) +
Number(`${secondNum}${num}`)
}
}
return sum
}
solution([10,2])

The ideas are 2:

compute the âsame-element in the outer loopâ and then compute the [i][j] and [j][i] at once.

I was dealing with some problem where I needed to get all possible pair sums of array items, and I was able(with some help from this forum) to figure out solution with single loop, not with nested loops.

Wrong idea but keeping it for anyone learning, good to see what does not workâŚ

function solution(a) {
let sum = 0;
let i = 0;
for (i; i < a.length-1; i++) {
const num = a[i];
sum+=Number(`${num}${num}`);
const secondNum = a[i+1];
sum += Number(`${num}${secondNum}`) +
Number(`${secondNum}${num}`)
};
const lastItem = a[i];
return sum + Number(`${lastItem}${lastItem}`)
}
solution([10,2])

Your solution gives correct result for [10, 2] thatâs for sure.

I came up with some while loop, also getting 1344 for [10,2], code needs some clean up also.

function solution(a) {
let [first, second, len, sum] = [0, 0, a.length, 0];
let pairSum;
while (first < len && second < len) {
if (first === second) {
pairSum = Number(String(a[first]) + String(a[second]));
sum += pairSum;
}
else {
pairSum = Number(String(a[first]) + String(a[second]));
sum += pairSum;
pairSum = Number(String(a[second]) + String(a[first]));
sum += pairSum;
}
if (first === len - 1 && second === len - 1) {
return sum;
}
else if (second === len - 1) {
first++;
second = first;
}
else {
second++;
}
}
}

And by the way, if we would need to find just sum of all possible pair sums(meaning we would need just to deal with numbers, not strings), it can be done with some maths just like that:

I donât think math.pow is needed for that algorithm described on stackoverflow? It should require a log10. In and case, adding a pow or a log is a fair trade to go from n^2 to n.

Edit: oh, 10^j is where the pow comes in. That shouldnât be expensive though.

Not for the first one suggested. The first one gives 1386. The second one, yes, for the reason I stated above. I had a brain-fart and compared all Xs with the next one.