LeetCode #2695 — EASY

Array Wrapper

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

Solve on LeetCode
The Problem

Problem Statement

Create a class ArrayWrapper that accepts an array of integers in its constructor. This class should have two features:

  • When two instances of this class are added together with the + operator, the resulting value is the sum of all the elements in both arrays.
  • When the String() function is called on the instance, it will return a comma separated string surrounded by brackets. For example, [1,2,3].

Example 1:

Input: nums = [[1,2],[3,4]], operation = "Add"
Output: 10
Explanation:
const obj1 = new ArrayWrapper([1,2]);
const obj2 = new ArrayWrapper([3,4]);
obj1 + obj2; // 10

Example 2:

Input: nums = [[23,98,42,70]], operation = "String"
Output: "[23,98,42,70]"
Explanation:
const obj = new ArrayWrapper([23,98,42,70]);
String(obj); // "[23,98,42,70]"

Example 3:

Input: nums = [[],[]], operation = "Add"
Output: 0
Explanation:
const obj1 = new ArrayWrapper([]);
const obj2 = new ArrayWrapper([]);
obj1 + obj2; // 0

Constraints:

  • 0 <= nums.length <= 1000
  • 0 <= nums[i] <= 1000
  • Note: nums is the array passed to the constructor

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Create a class ArrayWrapper that accepts an array of integers in its constructor. This class should have two features: When two instances of this class are added together with the + operator, the resulting value is the sum of all the elements in both arrays. When the String() function is called on the instance, it will return a comma separated string surrounded by brackets. For example, [1,2,3].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

[[1,2],[3,4]]
"Add"

Example 2

[[23,98,42,70]]
"String"

Example 3

[[],[]]
"Add"
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2695: Array Wrapper
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2695: Array Wrapper
// class ArrayWrapper {
//     private nums: number[];
//     private s: number;
// 
//     constructor(nums: number[]) {
//         this.nums = nums;
//         this.s = nums.reduce((a, b) => a + b, 0);
//     }
// 
//     valueOf() {
//         return this.s;
//     }
// 
//     toString() {
//         return `[${this.nums}]`;
//     }
// }
// 
// /**
//  * const obj1 = new ArrayWrapper([1,2]);
//  * const obj2 = new ArrayWrapper([3,4]);
//  * obj1 + obj2; // 10
//  * String(obj1); // "[1,2]"
//  * String(obj2); // "[3,4]"
//  */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.