Given an array of integers numbers, compare the sum of elements on even positions against the sum of elements on odd positions (0-based). Return "even" if the sum of elements on even positions is greater, "odd" if the sum of elements on odd positions is greater, or "equal" if both sums are equal.
Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(numbers.length^2) will fit within the execution time limit.
Example
For numbers = [1, 2, 3, 4, 5], the output should be solution(numbers) = "even".
Explanation:
Sum of elements on even positions is 1 + 3 + 5 = 9.
Sum of elements on odd positions is 2 + 4 = 6.
9 > 6, so the expected output is "even".For numbers = [-1, 4, 3, -2], the output should be solution(numbers) = "equal".
Explanation:
Sum of elements on even positions is (-1) + 3 = 2.
Sum of elements on odd positions is 4 + (-2) = 2.
2 = 2, so the expected output is "equal".This problem asks you to split the array into even-indexed and odd-indexed elements using 0-based indexing, compute the two sums, and return "even", "odd", or "equal" accordingly. The simplest and best approach is a single pass through the array, accumulating values by index parity and comparing the two totals at the end. Since the required complexity is lenient, an O(n) solution is ideal for a fast OA implementation.
