I found solving in Javascript

let sum = 0; for (let i = 0; i < ar.length; i++) { sum = sum + ar[i]; } return sum;

You can also use reduce() for the simple way const sum = ar.reduce((a,b) => a+b, 0); return sum;

TypeScript function signature is not correct. It shuld return bigint instead of number.

RUST:

fn aVeryBigSum(ar: &[i64]) -> i64 { let sum: i64 = {

    let mut temp_sum: i64 = 0;

    for i in ar {
        temp_sum += i;
    }

    temp_sum
};

sum

}

Java Short Solution

public static long aVeryBigSum(List ar) { // Write your code here long sum=0; for(long i:(ar)) sum+=i; return sum; }

I have spent so much time to transform between str and int and do the manual computation digit by digit to ensure the storage. And then it turns out we don't need to consider the storage at all? JUST SUM???