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Inside the Array.prototype.upperBound function you have access to the "this" keyword. You can access array elements, values, and methods. For example "this[0]", "this[1]", "this.length", "this.map()", etc.
The most efficient way to solve this problem is with binary search.
Choose the middle element and check if it's less than or equal to the goal value. If so, you can rule out the left side of the array.
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Yes, variations of upper bound or binary search boundary problems frequently appear in technical interviews, including FAANG companies. They test a candidate's understanding of binary search edge cases and search space reduction.
The optimal approach is to use binary search on the sorted array. By repeatedly halving the search space, you can efficiently locate the first element that is strictly greater than the target value.
A sorted array is the key data structure for this problem. Its ordered nature allows binary search to be applied, enabling faster lookups compared to scanning every element.
Lower bound returns the first index where the value is greater than or equal to the target, while upper bound returns the first index where the value is strictly greater than the target. Both are commonly solved using modified binary search.
