Implement a function findIndex(array, predicate, [fromIndex=0]) that takes an array of values, a function predicate, and an optional fromIndex number argument, and returns the index of the first element in the array that satisfies the provided testing function predicate.

findIndex(array, predicate, [(fromIndex = 0)]);

Arguments

1. array (Array): The array to inspect.
2. predicate (Function): The function invoked per iteration.
3. [fromIndex=0] (number): The index to search from.

Predicate signature: The predicate function is invoked with three arguments: (value, index, array).

• value: The current element being iterated.
• index: The index of the current element.
• array: The original input array.

Returns

(number): Returns the index of the found element, else -1.

Examples

const arr = [1, 2, 3, 4, 5];

// Search for the first value in the array that is greater than 3.
findIndex(arr, (num) => num > 3); // => 3

// Start searching from index 4 (inclusive).
findIndex(arr, (num) => num > 3, 4); // => 4

// No such element exists.
findIndex(arr, (num) => num > 10, 3); // => -1

Edge cases

Your function should handle negative and out of bound indices, as demonstrated in the examples below.

• Negative: negative integers count back from the last item in the array. 1 means the last element in the array, 2 means the second last element, and so on.
• Out of bound: if index < -(array.length), start searching from index 0. If index >= array.length, no search will take place and you should return -1.

const arr = [1, 2, 3, 4, 5];
// Start searching from index 2 (inclusive).
findIndex(arr, (num) => num > 2, -3); // => 2
findIndex(arr, (num) => num % 2 === 0, -3); // => 3
// Start from 0 if fromIndex < -(array.length)
findIndex(arr, (num) => num > 2, -10); // => 2
// Search rightwards from index that's already out of bounds,
// which will always result in -1.
findIndex(arr, (num) => num > 2, 10); // => -1