LeetCode #2630 — HARD

Memoize II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a function fn, return a memoized version of that function.

A memoized function is a function that will never be called twice with the same inputs. Instead it will return a cached value.

fn can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are === to each other.

Example 1:

Input: 
getInputs = () => [[2,2],[2,2],[1,2]]
fn = function (a, b) { return a + b; }
Output: [{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]
Explanation:
const inputs = getInputs();
const memoized = memoize(fn);
for (const arr of inputs) {
  memoized(...arr);
}

For the inputs of (2, 2): 2 + 2 = 4, and it required a call to fn().
For the inputs of (2, 2): 2 + 2 = 4, but those inputs were seen before so no call to fn() was required.
For the inputs of (1, 2): 1 + 2 = 3, and it required another call to fn() for a total of 2.

Example 2:

Input: 
getInputs = () => [[{},{}],[{},{}],[{},{}]] 
fn = function (a, b) { return ({...a, ...b}); }
Output: [{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]
Explanation:
Merging two empty objects will always result in an empty object. It may seem like there should only be 1 call to fn() because of cache-hits, however none of those objects are === to each other.

Example 3:

Input: 
getInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; }
fn = function (a, b) { return ({...a, ...b}); }
Output: [{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]
Explanation:
Merging two empty objects will always result in an empty object. The 2nd and 3rd third function calls result in a cache-hit. This is because every object passed in is identical.

Constraints:

1 <= inputs.length <= 10⁵
0 <= inputs.flat().length <= 10⁵
inputs[i][j] != NaN

Step 01

Brute Force Baseline

Problem summary: Given a function fn, return a memoized version of that function. A memoized function is a function that will never be called twice with the same inputs. Instead it will return a cached value. fn can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are === to each other.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

() => [[2,2],[2,2],[1,2]]
function (a, b) { return a + b; }

Example 2

() => [[{},{}],[{},{}],[{},{}]]
function (a, b) { return ({...a, ...b}); }

Example 3

() => { const o = {}; return [[o,o],[o,o],[o,o]]; }
function (a, b) { return ({...a, ...b}); }

Core Insight

What unlocks the optimal approach

Just because JSON.stringify(obj1) === JSON.stringify(obj2), doesn't necessarily mean obj1 === obj2.
You could iterate over all previously passed inputs to check if there has been a match. However, that will be very slow.
Javascript Maps are a could way to associate arbitrary data.
Make a tree structure of Maps. The depth of the tree should match the number of input parameters.

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

Largest constraint values

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 #2630: Memoize II
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2630: Memoize II
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const idxMap: Map<string, number> = new Map();
//     const cache: Map<string, any> = new Map();
// 
//     const getIdx = (obj: any): number => {
//         if (!idxMap.has(obj)) {
//             idxMap.set(obj, idxMap.size);
//         }
//         return idxMap.get(obj)!;
//     };
// 
//     return function (...params: any) {
//         const key = params.map(getIdx).join(',');
//         if (!cache.has(key)) {
//             cache.set(key, fn(...params));
//         }
//         return cache.get(key)!;
//     };
// }
// 
// /**
//  * let callCount = 0;
//  * const memoizedFn = memoize(function (a, b) {
//  *	 callCount += 1;
//  *   return a + b;
//  * })
//  * memoizedFn(2, 3) // 5
//  * memoizedFn(2, 3) // 5
//  * console.log(callCount) // 1
//  */

// Accepted solution for LeetCode #2630: Memoize II
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2630: Memoize II
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const idxMap: Map<string, number> = new Map();
//     const cache: Map<string, any> = new Map();
// 
//     const getIdx = (obj: any): number => {
//         if (!idxMap.has(obj)) {
//             idxMap.set(obj, idxMap.size);
//         }
//         return idxMap.get(obj)!;
//     };
// 
//     return function (...params: any) {
//         const key = params.map(getIdx).join(',');
//         if (!cache.has(key)) {
//             cache.set(key, fn(...params));
//         }
//         return cache.get(key)!;
//     };
// }
// 
// /**
//  * let callCount = 0;
//  * const memoizedFn = memoize(function (a, b) {
//  *	 callCount += 1;
//  *   return a + b;
//  * })
//  * memoizedFn(2, 3) // 5
//  * memoizedFn(2, 3) // 5
//  * console.log(callCount) // 1
//  */

# Accepted solution for LeetCode #2630: Memoize II
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2630: Memoize II
# type Fn = (...params: any) => any;
# 
# function memoize(fn: Fn): Fn {
#     const idxMap: Map<string, number> = new Map();
#     const cache: Map<string, any> = new Map();
# 
#     const getIdx = (obj: any): number => {
#         if (!idxMap.has(obj)) {
#             idxMap.set(obj, idxMap.size);
#         }
#         return idxMap.get(obj)!;
#     };
# 
#     return function (...params: any) {
#         const key = params.map(getIdx).join(',');
#         if (!cache.has(key)) {
#             cache.set(key, fn(...params));
#         }
#         return cache.get(key)!;
#     };
# }
# 
# /**
#  * let callCount = 0;
#  * const memoizedFn = memoize(function (a, b) {
#  *	 callCount += 1;
#  *   return a + b;
#  * })
#  * memoizedFn(2, 3) // 5
#  * memoizedFn(2, 3) // 5
#  * console.log(callCount) // 1
#  */

// Accepted solution for LeetCode #2630: Memoize II
// 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 #2630: Memoize II
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const idxMap: Map<string, number> = new Map();
//     const cache: Map<string, any> = new Map();
// 
//     const getIdx = (obj: any): number => {
//         if (!idxMap.has(obj)) {
//             idxMap.set(obj, idxMap.size);
//         }
//         return idxMap.get(obj)!;
//     };
// 
//     return function (...params: any) {
//         const key = params.map(getIdx).join(',');
//         if (!cache.has(key)) {
//             cache.set(key, fn(...params));
//         }
//         return cache.get(key)!;
//     };
// }
// 
// /**
//  * let callCount = 0;
//  * const memoizedFn = memoize(function (a, b) {
//  *	 callCount += 1;
//  *   return a + b;
//  * })
//  * memoizedFn(2, 3) // 5
//  * memoizedFn(2, 3) // 5
//  * console.log(callCount) // 1
//  */

// Accepted solution for LeetCode #2630: Memoize II
type Fn = (...params: any) => any;

function memoize(fn: Fn): Fn {
    const idxMap: Map<string, number> = new Map();
    const cache: Map<string, any> = new Map();

    const getIdx = (obj: any): number => {
        if (!idxMap.has(obj)) {
            idxMap.set(obj, idxMap.size);
        }
        return idxMap.get(obj)!;
    };

    return function (...params: any) {
        const key = params.map(getIdx).join(',');
        if (!cache.has(key)) {
            cache.set(key, fn(...params));
        }
        return cache.get(key)!;
    };
}

/**
 * let callCount = 0;
 * const memoizedFn = memoize(function (a, b) {
 *	 callCount += 1;
 *   return a + b;
 * })
 * memoizedFn(2, 3) // 5
 * memoizedFn(2, 3) // 5
 * console.log(callCount) // 1
 */

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.