Problem Description

You need to create a class called ArrayWrapper that takes an array of integers in its constructor. This class must implement two specific behaviors:

1. Addition behavior: When two instances of ArrayWrapper are added together using the + operator, the result should be the sum of all elements from both arrays. For example, if you have obj1 containing [1,2] and obj2 containing [3,4], then obj1 + obj2 should equal 10 (which is 1+2+3+4).

2. String conversion behavior: When the String() function is called on an instance of ArrayWrapper, it should return a string representation of the array with elements separated by commas and enclosed in square brackets. For example, if an instance contains [1,2,3], calling String() on it should return the string "[1,2,3]".

The solution achieves this by:

• Storing the array in a private property nums
• Pre-calculating and storing the sum of all elements in a private property s during construction
• Implementing the valueOf() method to return the sum when the object is used in numeric operations (like addition)
• Implementing the toString() method to return the formatted string representation of the array

The valueOf() method is automatically called by JavaScript/TypeScript when an object is used in a numeric context (like addition), while toString() is called when the object needs to be converted to a string.

Intuition

The key insight here is understanding how JavaScript/TypeScript handles type coercion and object-to-primitive conversion. When we use the + operator between two objects, JavaScript needs to convert them to primitive values. Similarly, when we call String() on an object, JavaScript needs to convert it to a string representation.

JavaScript provides special methods that get called automatically during these conversions:

• valueOf() is called when an object needs to be converted to a primitive value in a numeric context
• toString() is called when an object needs to be converted to a string

By overriding these two methods, we can control exactly what happens when our ArrayWrapper instances are used with the + operator or converted to strings.

For the addition behavior, we need the sum of all array elements. Instead of calculating this sum every time two objects are added, we can optimize by pre-calculating the sum once during object construction and storing it. This way, when valueOf() is called, we simply return the pre-calculated sum.

For the string representation, we need to format the array with brackets and comma separation. The toString() method can leverage JavaScript's default array-to-string conversion (which already provides comma separation) and simply wrap it with brackets using template literals: [${this.nums}].

This approach elegantly solves both requirements by utilizing JavaScript's built-in type conversion mechanisms rather than trying to override operators directly (which isn't possible in JavaScript/TypeScript).

Solution Approach

The implementation uses a class-based approach with private properties and method overriding to achieve the required functionality.

Class Structure and Constructor:

class ArrayWrapper {
    private nums: number[];
    private s: number;
  
    constructor(nums: number[]) {
        this.nums = nums;
        this.s = nums.reduce((a, b) => a + b, 0);
    }
}

The class maintains two private properties:

• nums: Stores the original array for string representation
• s: Stores the pre-calculated sum of all array elements

During construction, we use the reduce() method to calculate the sum once. The reduce function iterates through the array, accumulating the sum by adding each element to the accumulator, starting from an initial value of 0.

Implementing Addition Behavior:

valueOf() {
    return this.s;
}

The valueOf() method returns the pre-calculated sum. When two ArrayWrapper instances are added using the + operator, JavaScript automatically calls valueOf() on each instance to get their numeric values, then adds those values together. This gives us the desired behavior where obj1 + obj2 returns the sum of all elements from both arrays.

Implementing String Conversion:

toString() {
    return `[${this.nums}]`;
}

The toString() method uses template literals to format the output. When ${this.nums} is evaluated inside the template literal, JavaScript automatically converts the array to a comma-separated string. We then wrap this with square brackets to achieve the required format. When String() is called on an instance, it triggers this toString() method.

Time and Space Complexity:

• Constructor: O(n) time where n is the length of the input array (for calculating the sum)
• valueOf(): O(1) time (returns pre-calculated value)
• toString(): O(n) time (for converting array to string)
• Space: O(n) for storing the array

Example Walkthrough

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

Step 1: Create two ArrayWrapper instances

const obj1 = new ArrayWrapper([1, 2]);
const obj2 = new ArrayWrapper([3, 4, 5]);

When obj1 is constructed:

• nums property stores [1, 2]
• s property calculates and stores 1 + 2 = 3 using reduce

When obj2 is constructed:

• nums property stores [3, 4, 5]
• s property calculates and stores 3 + 4 + 5 = 12 using reduce

Step 2: Addition operation

const result = obj1 + obj2;  // result = 15

Here's what happens behind the scenes:

1. JavaScript sees the + operator with two objects
2. It needs to convert both objects to primitive values
3. JavaScript calls valueOf() on obj1, which returns 3
4. JavaScript calls valueOf() on obj2, which returns 12
5. JavaScript performs the addition: 3 + 12 = 15

Step 3: String conversion

const str1 = String(obj1);  // str1 = "[1,2]"
const str2 = String(obj2);  // str2 = "[3,4,5]"

For String(obj1):

1. JavaScript calls the toString() method on obj1
2. Inside toString(), the template literal [${this.nums}] is evaluated
3. this.nums (which is [1, 2]) gets converted to "1,2"
4. The final result is "[1,2]"

For String(obj2):

1. JavaScript calls the toString() method on obj2
2. Inside toString(), the template literal evaluates [${[3, 4, 5]}]
3. The array converts to "3,4,5"
4. The final result is "[3,4,5]"

Complete code example:

class ArrayWrapper {
    constructor(nums) {
        this.nums = nums;
        this.s = nums.reduce((a, b) => a + b, 0);
    }
  
    valueOf() {
        return this.s;  // Returns pre-calculated sum
    }
  
    toString() {
        return `[${this.nums}]`;  // Returns formatted string
    }
}

// Usage
const obj1 = new ArrayWrapper([1, 2]);
const obj2 = new ArrayWrapper([3, 4, 5]);
console.log(obj1 + obj2);    // Output: 15
console.log(String(obj1));   // Output: "[1,2]"
console.log(String(obj2));   // Output: "[3,4,5]"

The elegance of this solution lies in leveraging JavaScript's automatic type conversion mechanisms rather than trying to manually override operators, making the code both efficient and maintainable.

Solution Implementation

Time and Space Complexity

Time Complexity:

• Constructor: O(n) where n is the length of the input array, due to the reduce operation that iterates through all elements to calculate the sum.
• valueOf(): O(1) as it simply returns the pre-computed sum stored in this.s.
• toString(): O(n) where n is the length of the array, as it needs to convert the entire array to a string representation.

Space Complexity:

• Overall: O(n) where n is the length of the input array.
  • The nums array stores a reference to the input array: O(n)
  • The sum s is a single number: O(1)
• valueOf(): O(1) additional space as it returns a primitive value.
• toString(): O(n) additional space for creating the string representation of the array.

Common Pitfalls

1. Mutability of the Stored Array

The most significant pitfall in this implementation is that the stored array reference can be modified externally, leading to inconsistency between the stored sum and the actual array contents.

Problem Example:

original_array = [1, 2, 3]
initializeArrayWrapper(original_array)
print(valueOf())  # 6
original_array.append(4)  # Modifying the original array
print(valueOf())  # Still 6, but array is now [1,2,3,4]
print(toString())  # "[1,2,3,4]" - inconsistent with sum!

Solution: Create a defensive copy of the input array:

def initializeArrayWrapper(inputNums):
    global nums, sum_value
    nums = inputNums.copy()  # Create a copy instead of storing reference
    sum_value = sum(nums)

2. Global Variable State Management

Using global variables makes it impossible to maintain multiple independent ArrayWrapper instances simultaneously.

Problem Example:

initializeArrayWrapper([1, 2])
wrapper1_value = valueOf()  # 3
initializeArrayWrapper([3, 4])  # This overwrites the first wrapper!
wrapper2_value = valueOf()  # 7
# Can't access the first wrapper anymore

Solution: Use a class-based approach to encapsulate state:

class ArrayWrapper:
    def __init__(self, nums):
        self.nums = nums.copy()
        self.sum_value = sum(self.nums)
  
    def valueOf(self):
        return self.sum_value
  
    def toString(self):
        return f"[{','.join(map(str, self.nums))}]"

3. Empty Array Handling

The current implementation doesn't explicitly handle edge cases like empty arrays, which could lead to unexpected behavior.

Problem Example:

initializeArrayWrapper([])
print(valueOf())  # 0 (works, but might not be obvious)
print(toString())  # "[]" (works, but join on empty list needs verification)

Solution: Add explicit validation and documentation:

def initializeArrayWrapper(inputNums):
    global nums, sum_value
    if inputNums is None:
        nums = []
        sum_value = 0
    else:
        nums = inputNums.copy()
        sum_value = sum(nums) if nums else 0

4. Type Safety Issues

The function doesn't validate that all elements in the input are actually numbers, which could cause runtime errors.

Problem Example:

initializeArrayWrapper([1, "2", 3])  # Mixed types
# sum() will fail with TypeError

Solution: Add type validation:

def initializeArrayWrapper(inputNums):
    global nums, sum_value
    # Validate all elements are numbers
    if not all(isinstance(x, (int, float)) for x in inputNums):
        raise ValueError("All elements must be numbers")
    nums = inputNums.copy()
    sum_value = sum(nums)