LeetCode #2722 — MEDIUM

Join Two Arrays by ID

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

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The Problem

Problem Statement

Given two arrays arr1 and arr2, return a new array joinedArray. All the objects in each of the two inputs arrays will contain an id field that has an integer value. 

joinedArray is an array formed by merging arr1 and arr2 based on their id key. The length of joinedArray should be the length of unique values of id. The returned array should be sorted in ascending order based on the id key.

If a given id exists in one array but not the other, the single object with that id should be included in the result array without modification.

If two objects share an id, their properties should be merged into a single object:

  • If a key only exists in one object, that single key-value pair should be included in the object.
  • If a key is included in both objects, the value in the object from arr2 should override the value from arr1.

Example 1:

Input: 
arr1 = [
    {"id": 1, "x": 1},
    {"id": 2, "x": 9}
], 
arr2 = [
    {"id": 3, "x": 5}
]
Output: 
[
    {"id": 1, "x": 1},
    {"id": 2, "x": 9},
    {"id": 3, "x": 5}
]
Explanation: There are no duplicate ids so arr1 is simply concatenated with arr2.

Example 2:

Input: 
arr1 = [
    {"id": 1, "x": 2, "y": 3},
    {"id": 2, "x": 3, "y": 6}
], 
arr2 = [
    {"id": 2, "x": 10, "y": 20},
    {"id": 3, "x": 0, "y": 0}
]
Output: 
[
    {"id": 1, "x": 2, "y": 3},
    {"id": 2, "x": 10, "y": 20},
    {"id": 3, "x": 0, "y": 0}
]
Explanation: The two objects with id=1 and id=3 are included in the result array without modifiction. The two objects with id=2 are merged together. The keys from arr2 override the values in arr1.

Example 3:

Input: 
arr1 = [
    {"id": 1, "b": {"b": 94},"v": [4, 3], "y": 48}
]
arr2 = [
    {"id": 1, "b": {"c": 84}, "v": [1, 3]}
]
Output: [
    {"id": 1, "b": {"c": 84}, "v": [1, 3], "y": 48}
]
Explanation: The two objects with id=1 are merged together. For the keys "b" and "v" the values from arr2 are used. Since the key "y" only exists in arr1, that value is taken form arr1.

Constraints:

  • arr1 and arr2 are valid JSON arrays
  • Each object in arr1 and arr2 has a unique integer id key
  • 2 <= JSON.stringify(arr1).length <= 106
  • 2 <= JSON.stringify(arr2).length <= 106

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: Given two arrays arr1 and arr2, return a new array joinedArray. All the objects in each of the two inputs arrays will contain an id field that has an integer value. joinedArray is an array formed by merging arr1 and arr2 based on their id key. The length of joinedArray should be the length of unique values of id. The returned array should be sorted in ascending order based on the id key. If a given id exists in one array but not the other, the single object with that id should be included in the result array without modification. If two objects share an id, their properties should be merged into a single object: If a key only exists in one object, that single key-value pair should be included in the object. If a key is included in both objects, the value in the object from arr2 should override the value from arr1.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

[{"id": 1,"x": 1},{"id": 2,"x": 9}]
[{"id": 3,"x": 5}]

Example 2

[{"id": 1,"x": 2,"y": 3},{"id": 2,"x": 3,"y": 6}]
[{"id": 2,"x": 10,"y": 20},{"id": 3,"x": 0,"y": 0}]

Example 3

[{"id":1,"b":{"b": 94},"v":[4,3],"y":48}]
[{"id":1,"b":{"c": 84},"v":[1,3]}]
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 #2722: Join Two Arrays by ID
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2722: Join Two Arrays by ID
// function join(arr1: ArrayType[], arr2: ArrayType[]): ArrayType[] {
//     const r = (acc: Obj, x: ArrayType): Obj => ((acc[x.id] = x), acc);
//     const d = arr1.reduce(r, {});
// 
//     arr2.forEach(x => {
//         if (d[x.id]) {
//             Object.assign(d[x.id], x);
//         } else {
//             d[x.id] = x;
//         }
//     });
//     return Object.values(d);
// }
// 
// type JSONValue = null | boolean | number | string | JSONValue[] | { [key: string]: JSONValue };
// type ArrayType = { id: number } & Record<string, JSONValue>;
// 
// type Obj = Record<number, ArrayType>;
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.