Can we make the sum more efficient? Generally no. But what if we know that the array starts with 1, is sorted, and has no gaps? Then we can apply the formula:

where n is the last element of the array.

const sumContiguousArray = function (arr) {
  //get the last item
  const lastItem = arr[arr.length - 1];
  //Gauss's trick
  return (lastItem * (lastItem + 1)) / 2;
};
const nums = [1, 2, 3, 4, 5];
const sumOfArray = sumContiguousArray(nums);

O(N^2)

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