Implementing On-Stack Replacement (OSR) in JavaScript

On-Stack Replacement (OSR) is a technique used in garbage collection to improve the efficiency of marking live objects. It's particularly useful in generational garbage collectors where young objects are frequently collected. This challenge asks you to implement a simplified version of OSR in JavaScript, focusing on the core logic of identifying and replacing pointers to objects that are about to be promoted to the old generation.

Problem Description

You are tasked with implementing an On-Stack Replacement (OSR) mechanism. Imagine a simplified scenario where you have a set of objects and a stack representing the call stack. When an object is about to be promoted from the young generation to the old generation, OSR scans the stack to see if any pointers on the stack point to this object. If it finds such pointers, it replaces them with a temporary placeholder. During the next garbage collection cycle, these placeholders are updated with the actual promoted object's address. This avoids scanning the entire stack during the promotion phase, significantly improving performance.

Your implementation should take two inputs: a list of objects and a stack (represented as an array). The function should identify pointers on the stack that point to the object being promoted and return a new stack with those pointers replaced by a placeholder value (null).

Key Requirements:

Object Identification: The function must correctly identify pointers on the stack that point to the target object.
Placeholder Replacement: Pointers to the target object must be replaced with null.
Stack Integrity: The rest of the stack should remain unchanged.
No Side Effects: The original stack array should not be modified. A new stack array should be returned.

Expected Behavior:

The function should return a new array representing the modified stack, with pointers to the promoted object replaced by null. If no pointers to the object are found on the stack, the function should return a copy of the original stack.

Edge Cases to Consider:

Empty Stack: The stack array might be empty.
Object Not Found: The target object might not be referenced anywhere on the stack.
Multiple Pointers: There might be multiple pointers to the same object on the stack.
Null Values: The stack might already contain null values.
Object Identity: The comparison should be based on object identity (===), not value equality.

Examples

Example 1:

Input: objects = [{id: 1}, {id: 2}, {id: 3}], stack = [ {id: 1}, {id: 2}, {id: 1}]
Output: [ null, {id: 2}, null ]
Explanation: The object with id 1 is being promoted. Two pointers on the stack point to this object. These pointers are replaced with null.

Example 2:

Input: objects = [{id: 1}, {id: 2}], stack = [ {id: 2}, {id: 3}]
Output: [ {id: 2}, {id: 3} ]
Explanation: The object with id 1 is being promoted, but no pointers to it are found on the stack. The original stack is returned.

Example 3: (Edge Case - Empty Stack)

Input: objects = [{id: 1}], stack = []
Output: []
Explanation: The stack is empty, so no replacements are needed.

Constraints

objects will be an array of objects. Each object will have a unique id property.
stack will be an array of objects or null values.
The length of stack can be up to 1000.
The id property of objects in objects will be a number.
The function should execute in O(n) time, where n is the length of the stack. (This is a soft constraint; prioritize correctness first.)

Notes

You don't need to implement the entire garbage collection process. Focus solely on the OSR logic.
Consider using Array.map() or similar methods for creating a new array without modifying the original.
The id property is used for identifying the object to be promoted. You'll need to iterate through the objects array to find the object with the matching id.
The comparison between objects should be based on strict equality (===) to ensure you're comparing object identities, not just values.
Think about how to efficiently search the objects array to find the target object.