Solution Approach

The implementation uses a recursive function that handles different data types systematically. Let's walk through the solution step by step:

1. Handle Primitive Values and Null

if (o1 === null || typeof o1 !== 'object') {  return o1 === o2; }

First, we check if o1 is a primitive value or null. In JavaScript, typeof null returns 'object', so we explicitly check for null. If o1 is primitive or null, we simply return o1 === o2 to check strict equality.

2. Type Mismatch Check

if (typeof o1 !== typeof o2) {  return false; }

If the types of o1 and o2 don't match, they can't be deeply equal, so we return false immediately.

3. Array vs Object Distinction

if (Array.isArray(o1) !== Array.isArray(o2)) {  return false; }

Since arrays are objects in JavaScript, we need to explicitly check if one value is an array and the other is a plain object. If they differ, return false.

4. Array Comparison

if (Array.isArray(o1)) {  if (o1.length !== o2.length) {  return false;  }  for (let i = 0; i < o1.length; i++) {  if (!areDeeplyEqual(o1[i], o2[i])) {  return false;  }  }  return true; }

For arrays:

• First check if lengths are equal - different lengths mean not deeply equal
• Iterate through each index and recursively call areDeeplyEqual on corresponding elements
• If any pair of elements isn't deeply equal, return false
• If all elements match, return true

5. Object Comparison

const keys1 = Object.keys(o1); const keys2 = Object.keys(o2); if (keys1.length !== keys2.length) {  return false; } for (const key of keys1) {  if (!areDeeplyEqual(o1[key], o2[key])) {  return false;  } } return true;

For objects:

• Extract keys from both objects using Object.keys()
• Compare the number of keys - different counts mean not deeply equal
• Iterate through keys of the first object
• For each key, recursively compare the values in both objects
• If any value doesn't match or if o2 doesn't have a key that o1 has (which would make o2[key] undefined), the recursive call will return false
• If all key-value pairs match, return true

Time Complexity: O(n) where n is the total number of primitive values in the nested structure, as we visit each value once.

Space Complexity: O(d) where d is the maximum depth of nesting, due to the recursive call stack.

Example Walkthrough

Let's walk through a concrete example to see how the solution works:

Example: Check if o1 = {a: [1, 2], b: {c: 3}} and o2 = {a: [1, 2], b: {c: 3}} are deeply equal.

Step-by-step execution:

1. Initial call: areDeeplyEqual(o1, o2)

   • Both are objects (not null, not primitive)
   • Both have same type (object)
   • Neither is an array
   • Extract keys: keys1 = ['a', 'b'], keys2 = ['a', 'b']
   • Same length (2 keys each) ✓
   • Now check each key...

2. Check key 'a': areDeeplyEqual([1, 2], [1, 2])

   • Both are arrays ✓
   • Same length (2) ✓
   • Compare index 0: areDeeplyEqual(1, 1)
     • Both are primitives
     • Return 1 === 1 = true ✓
   • Compare index 1: areDeeplyEqual(2, 2)
     • Both are primitives
     • Return 2 === 2 = true ✓
   • All array elements match, return true ✓

3. Check key 'b': areDeeplyEqual({c: 3}, {c: 3})

   • Both are objects (not arrays)
   • Extract keys: keys1 = ['c'], keys2 = ['c']
   • Same length (1 key each) ✓
   • Check key 'c': areDeeplyEqual(3, 3)
     • Both are primitives
     • Return 3 === 3 = true ✓
   • All object properties match, return true ✓

4. Final result: Since both keys 'a' and 'b' have deeply equal values, the function returns true.

Counter-example: If we had o2 = {a: [1, 2], b: {c: 4}} instead, the execution would differ at step 3:

• When checking key 'c': areDeeplyEqual(3, 4) would return 3 === 4 = false
• This would cause the check for key 'b' to return false
• The overall function would return false

Time and Space Complexity

Time Complexity: O(n) where n is the total number of primitive values (leaves) in both objects combined.

The algorithm performs a depth-first traversal through both objects simultaneously. Each primitive value is visited exactly once during the comparison. For nested objects and arrays, the algorithm recursively explores all properties/elements, but each property or element is only checked once. The operations at each node (type checking, array checking, key comparisons) are all O(1) operations. Therefore, the overall time complexity is linear with respect to the total number of primitive values that need to be compared.

Space Complexity: O(d) where d is the maximum depth of nesting in the objects.

The space complexity is determined by the recursive call stack. In the worst case, when dealing with deeply nested structures, the recursion depth equals the maximum nesting level of the objects. Each recursive call adds a new frame to the call stack with constant space for local variables (keys1, keys2, loop variables). The algorithm doesn't create copies of the objects being compared, only storing references and keys. For a completely flat object with no nesting, the space complexity would be O(1) (excluding the space for storing the keys array which is O(k) where k is the number of keys at that level).

Common Pitfalls

1. Not Checking Key Existence in Second Object

The Pitfall: In the Python implementation's dictionary comparison section, we iterate through keys1 and compare values, but we don't explicitly verify that all keys from o2 exist in o1. While we check the length of keys, this alone doesn't guarantee both objects have the exact same set of keys.

Consider this edge case:

o1 = {"a": 1, "b": 2} o2 = {"a": 1, "c": 2}

Both have 2 keys, but the keys are different. The current implementation would check o1["c"] which would raise a KeyError or return an incorrect result if we're not careful.

The Solution: Add an explicit check to ensure the second object doesn't have extra keys:

def areDeeplyEqual(o1, o2):  # ... previous code ...   if isinstance(o1, list):  # ... list handling ...  else:  # Handle dictionary comparison  keys1 = set(o1.keys())  keys2 = set(o2.keys())   # Check if the key sets are identical  if keys1 != keys2:  return False   # Compare each property recursively  for key in keys1:  if not areDeeplyEqual(o1[key], o2[key]):  return False   return True

2. Type Checking Inconsistency Between Python and JavaScript

The Pitfall: The Python code uses isinstance(o1, (dict, list)) to check for non-primitive types, but this doesn't perfectly mirror JavaScript's behavior. In Python, other iterable types like tuples or sets might pass through incorrectly, and the type checking logic could fail for edge cases.

The Solution: Be more explicit about the types we're handling:

def areDeeplyEqual(o1, o2):  # Define what we consider as primitive types  primitive_types = (int, float, str, bool, type(None))   # Handle primitives explicitly  if isinstance(o1, primitive_types):  return o1 == o2   # Ensure both are either list or dict (the only non-primitive JSON types)  if not isinstance(o1, (list, dict)) or not isinstance(o2, (list, dict)):  return False   # Continue with type-specific comparison...

3. Float Comparison Precision Issues

The Pitfall: When comparing floating-point numbers, precision issues can cause unexpected results:

o1 = {"value": 0.1 + 0.2} o2 = {"value": 0.3} # These might not be considered equal due to floating-point arithmetic

The Solution: Since the problem states inputs are valid JSON output from JSON.parse, and JSON numbers are handled consistently, this is less of an issue. However, if you need to handle floating-point comparison more robustly:

def areDeeplyEqual(o1, o2, epsilon=1e-9):  # For float comparison  if isinstance(o1, float) and isinstance(o2, float):  return abs(o1 - o2) < epsilon   # ... rest of the implementation

Note: This modification should only be used if the problem specifically requires handling floating-point precision issues, as it changes the strict equality semantics.