Join Two Arrays by ID - Practice Coding Problems

Join Two Arrays by ID - Problem

JavaScript Medium

Imagine you're working as a data analyst for a social media company. You have two datasets containing user information, and you need to combine them into a single, comprehensive view of your users.

Given two arrays arr1 and arr2, where each object contains an id field with an integer value, your task is to join these arrays based on their IDs, similar to how databases perform JOIN operations.

Key Requirements:

🔗 Merge by ID: Objects with the same id should be combined into one
📊 Property Override: When properties conflict, values from arr2 take precedence
🎯 Preserve Unique: Objects that exist in only one array remain unchanged
📈 Sort Result: Final array must be sorted by id in ascending order

This problem tests your understanding of hash tables, object merging, and data processing - essential skills for backend development and data engineering roles!

Input & Output

example_1.py — Basic merge with overlapping IDs

$ 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}]

💡 Note: No overlapping IDs, so simply combine all objects and sort by ID (1, 2, 3).

example_2.py — Property override from arr2

$ Input: arr1 = [{"id": 1, "name": "Alice", "age": 25}] arr2 = [{"id": 1, "age": 26, "city": "NYC"}]

› Output: [{"id": 1, "name": "Alice", "age": 26, "city": "NYC"}]

💡 Note: Same ID (1) exists in both arrays. Properties are merged: 'name' from arr1, 'age' overridden by arr2 value (26), and 'city' added from arr2.

example_3.py — Complex merge scenario

$ Input: arr1 = [{"id": 2, "a": 1, "b": 2}, {"id": 3, "a": 5, "c": 6}] arr2 = [{"id": 1, "b": 3, "c": 4}, {"id": 2, "a": 2}]

› Output: [{"id": 1, "b": 3, "c": 4}, {"id": 2, "a": 2, "b": 2}, {"id": 3, "a": 5, "c": 6}]

💡 Note: ID 1: only in arr2, kept as-is. ID 2: exists in both, 'a' overridden (1→2), 'b' kept from arr1. ID 3: only in arr1, kept as-is. Final result sorted by ID.

Constraints

1 ≤ arr1.length, arr2.length ≤ 1000
Each object guaranteed to have an integer id field
0 ≤ id ≤ 10⁵
All IDs within each array are unique
Objects may contain various property types (string, number, boolean, null)

Visualization

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Understanding the Visualization

Create Index

Build a hash map (like a filing system) with employee IDs as keys

Add Base Records

Insert all records from the first table into your index

Merge Updates

For each record in the second table, either merge with existing or add new

Generate Report

Extract all records from index and sort by employee ID

Key Takeaway

🎯 Key Insight: Hash maps provide O(1) lookups, making the join operation efficient. Like having a perfectly organized filing system where you can instantly find and update any employee record by their ID!

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28

The optimal solution uses a hash map to achieve O(n + m) time complexity. First, add all objects from arr1 to the hash map using ID as key. Then iterate through arr2: if an ID exists, merge properties (arr2 overrides arr1); otherwise add new object. Finally, extract values from hash map, sort by ID, and return the result. This approach efficiently handles both merging and uniqueness in a single pass through each array.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(n)	For each object in arr1, scan through all of arr2 to find matches
One-Pass Hash Table (Optimal)	O(n + m + k log k)	O(k)	Use hash map to merge arrays efficiently in a single pass
Two Pointers (Sorted Arrays)	O(n log n + m log m)	O(n + m)	Sort both arrays first, then merge using two pointers technique

Brute Force (Nested Loops) — Algorithm Steps

Initialize result array with all objects from arr1
For each object in arr2, scan through result array to find matching ID
If match found, merge properties (arr2 values override arr1)
If no match found, add arr2 object to result
Sort final result by ID

Visualization

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Step-by-Step Walkthrough

Initialize

Copy all objects from arr1 to result array

Scan for Match

For each object in arr2, scan through entire result array

Merge or Add

If match found, merge properties; otherwise add new object

Sort

Finally sort the result array by ID

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// Simplified C implementation
// In practice, would need JSON parsing library

typedef struct {
    int id;
    char data[1000]; // JSON string
} JsonObject;

int compare(const void *a, const void *b) {
    JsonObject *objA = (JsonObject *)a;
    JsonObject *objB = (JsonObject *)b;
    return objA->id - objB->id;
}

int joinArrays(JsonObject arr1[], int len1, JsonObject arr2[], int len2, JsonObject result[]) {
    int resultLen = 0;
    
    // Copy arr1 to result
    for (int i = 0; i < len1; i++) {
        result[resultLen++] = arr1[i];
    }
    
    // Process arr2
    for (int i = 0; i < len2; i++) {
        int found = 0;
        for (int j = 0; j < resultLen; j++) {
            if (result[j].id == arr2[i].id) {
                // Merge (simplified - just replace)
                strcpy(result[j].data, arr2[i].data);
                found = 1;
                break;
            }
        }
        if (!found) {
            result[resultLen++] = arr2[i];
        }
    }
    
    // Sort by ID
    qsort(result, resultLen, sizeof(JsonObject), compare);
    return resultLen;
}

int main() {
    printf("See JavaScript/Python for complete implementation\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n objects in arr2, we potentially scan through n objects in the result array

⚠ Quadratic Growth

Space Complexity

O(n)

Space needed for the result array containing merged objects

⚡ Linearithmic Space

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High Frequency

~18 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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