LeetCode #2631 — MEDIUM

Group By

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

The Problem

Problem Statement

Write code that enhances all arrays such that you can call the array.groupBy(fn) method on any array and it will return a grouped version of the array.

A grouped array is an object where each key is the output of fn(arr[i]) and each value is an array containing all items in the original array which generate that key.

The provided callback fn will accept an item in the array and return a string key.

The order of each value list should be the order the items appear in the array. Any order of keys is acceptable.

Please solve it without lodash's _.groupBy function.

Example 1:

Input: 
array = [
  {"id":"1"},
  {"id":"1"},
  {"id":"2"}
], 
fn = function (item) { 
  return item.id; 
}
Output: 
{ 
  "1": [{"id": "1"}, {"id": "1"}],   
  "2": [{"id": "2"}] 
}
Explanation:
Output is from array.groupBy(fn).
The selector function gets the "id" out of each item in the array.
There are two objects with an "id" of 1. Both of those objects are put in the first array.
There is one object with an "id" of 2. That object is put in the second array.

Example 2:

Input: 
array = [
  [1, 2, 3],
  [1, 3, 5],
  [1, 5, 9]
]
fn = function (list) { 
  return String(list[0]); 
}
Output: 
{ 
  "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]] 
}
Explanation:
The array can be of any type. In this case, the selector function defines the key as being the first element in the array. 
All the arrays have 1 as their first element so they are grouped together.
{
  "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
}

Example 3:

Input: 
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
fn = function (n) { 
  return String(n > 5);
}
Output:
{
  "true": [6, 7, 8, 9, 10],
  "false": [1, 2, 3, 4, 5]
}
Explanation:
The selector function splits the array by whether each number is greater than 5.

Constraints:

0 <= array.length <= 10⁵
fn returns a string

Step 01

Brute Force Baseline

Problem summary: Write code that enhances all arrays such that you can call the array.groupBy(fn) method on any array and it will return a grouped version of the array. A grouped array is an object where each key is the output of fn(arr[i]) and each value is an array containing all items in the original array which generate that key. The provided callback fn will accept an item in the array and return a string key. The order of each value list should be the order the items appear in the array. Any order of keys is acceptable. Please solve it without lodash's _.groupBy function.

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"},{"id":"1"},{"id":"2"}]
function (item) { return item.id; }

Example 2

[[1,2,3],[1,3,5],[1,5,9]]
function (list) { return String(list[0]); }

Example 3

[1,2,3,4,5,6,7,8,9,10]
function (n) { return String(n > 5); }

Core Insight

What unlocks the optimal approach

First declare an object that will eventually be returned.
Iterate of each element in the array. You can access the array with the "this" keyword.
The key is fn(arr[i]). If the key already exists on the object, set the value to be an empty array. Then push the value onto the array at the key.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #2631: Group By
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2631: Group By
// declare global {
//     interface Array<T> {
//         groupBy(fn: (item: T) => string): Record<string, T[]>;
//     }
// }
// 
// Array.prototype.groupBy = function (fn) {
//     return this.reduce((acc, item) => {
//         const key = fn(item);
//         if (acc[key]) {
//             acc[key].push(item);
//         } else {
//             acc[key] = [item];
//         }
//         return acc;
//     }, {});
// };
// 
// /**
//  * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
//  */

// Accepted solution for LeetCode #2631: Group By
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2631: Group By
// declare global {
//     interface Array<T> {
//         groupBy(fn: (item: T) => string): Record<string, T[]>;
//     }
// }
// 
// Array.prototype.groupBy = function (fn) {
//     return this.reduce((acc, item) => {
//         const key = fn(item);
//         if (acc[key]) {
//             acc[key].push(item);
//         } else {
//             acc[key] = [item];
//         }
//         return acc;
//     }, {});
// };
// 
// /**
//  * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
//  */

# Accepted solution for LeetCode #2631: Group By
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2631: Group By
# declare global {
#     interface Array<T> {
#         groupBy(fn: (item: T) => string): Record<string, T[]>;
#     }
# }
# 
# Array.prototype.groupBy = function (fn) {
#     return this.reduce((acc, item) => {
#         const key = fn(item);
#         if (acc[key]) {
#             acc[key].push(item);
#         } else {
#             acc[key] = [item];
#         }
#         return acc;
#     }, {});
# };
# 
# /**
#  * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
#  */

// Accepted solution for LeetCode #2631: Group By
// Rust example auto-generated from ts reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (ts):
// // Accepted solution for LeetCode #2631: Group By
// declare global {
//     interface Array<T> {
//         groupBy(fn: (item: T) => string): Record<string, T[]>;
//     }
// }
// 
// Array.prototype.groupBy = function (fn) {
//     return this.reduce((acc, item) => {
//         const key = fn(item);
//         if (acc[key]) {
//             acc[key].push(item);
//         } else {
//             acc[key] = [item];
//         }
//         return acc;
//     }, {});
// };
// 
// /**
//  * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
//  */

// Accepted solution for LeetCode #2631: Group By
declare global {
    interface Array<T> {
        groupBy(fn: (item: T) => string): Record<string, T[]>;
    }
}

Array.prototype.groupBy = function (fn) {
    return this.reduce((acc, item) => {
        const key = fn(item);
        if (acc[key]) {
            acc[key].push(item);
        } else {
            acc[key] = [item];
        }
        return acc;
    }, {});
};

/**
 * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
 */

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.