LeetCode #2623 — MEDIUM

Memoize

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

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.

You can assume there are 3 possible input functions: sum, fib, and factorial.

sum accepts two integers a and b and returns a + b. Assume that if a value has already been cached for the arguments (b, a) where a != b, it cannot be used for the arguments (a, b). For example, if the arguments are (3, 2) and (2, 3), two separate calls should be made.
fib accepts a single integer n and returns 1 if n <= 1 or fib(n - 1) + fib(n - 2) otherwise.
factorial accepts a single integer n and returns 1 if n <= 1 or factorial(n - 1) * n otherwise.

Example 1:

Input:
fnName = "sum"
actions = ["call","call","getCallCount","call","getCallCount"]
values = [[2,2],[2,2],[],[1,2],[]]
Output: [4,4,1,3,2]
Explanation:
const sum = (a, b) => a + b;
const memoizedSum = memoize(sum);
memoizedSum(2, 2); // "call" - returns 4. sum() was called as (2, 2) was not seen before.
memoizedSum(2, 2); // "call" - returns 4. However sum() was not called because the same inputs were seen before.
// "getCallCount" - total call count: 1
memoizedSum(1, 2); // "call" - returns 3. sum() was called as (1, 2) was not seen before.
// "getCallCount" - total call count: 2

Example 2:

Input:
fnName = "factorial"
actions = ["call","call","call","getCallCount","call","getCallCount"]
values = [[2],[3],[2],[],[3],[]]
Output: [2,6,2,2,6,2]
Explanation:
const factorial = (n) => (n <= 1) ? 1 : (n * factorial(n - 1));
const memoFactorial = memoize(factorial);
memoFactorial(2); // "call" - returns 2.
memoFactorial(3); // "call" - returns 6.
memoFactorial(2); // "call" - returns 2. However factorial was not called because 2 was seen before.
// "getCallCount" - total call count: 2
memoFactorial(3); // "call" - returns 6. However factorial was not called because 3 was seen before.
// "getCallCount" - total call count: 2

Example 3:

Input:
fnName = "fib"
actions = ["call","getCallCount"]
values = [[5],[]]
Output: [8,1]
Explanation:
fib(5) = 8 // "call"
// "getCallCount" - total call count: 1

Constraints:

0 <= a, b <= 10⁵
1 <= n <= 10
1 <= actions.length <= 10⁵
actions.length === values.length
actions[i] is one of "call" and "getCallCount"
fnName is one of "sum", "factorial" and "fib"

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. You can assume there are 3 possible input functions: sum, fib, and factorial. sum accepts two integers a and b and returns a + b. Assume that if a value has already been cached for the arguments (b, a) where a != b, it cannot be used for the arguments (a, b). For example, if the arguments are (3, 2) and (2, 3), two separate calls should be made. fib accepts a single integer n and returns 1 if n <= 1 or fib(n - 1) + fib(n - 2) otherwise. factorial accepts a single integer n and returns 1 if n <= 1 or factorial(n - 1) * n otherwise.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"sum"
["call","call","getCallCount","call","getCallCount"]
[[2,2],[2,2],[],[1,2],[]]

Example 2

"factorial"
["call","call","call","getCallCount","call","getCallCount"]
[[2],[3],[2],[],[3],[]]

Example 3

"fib"
["call","getCallCount"]
[[5],[]]

Core Insight

What unlocks the optimal approach

You can create copy of a function by spreading function parameters. function outerFunction(passedFunction) { return newFunction(...params) { return passedFunction(...params); }; }
params is an array. Since you know all values in the array are numbers, you can turn it into a string with JSON.stringify().
In the outerFunction, you can declare a Map or Object. In the inner function you can avoid executing the passed function if the params have already been passed before.

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 #2623: Memoize
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2623: Memoize
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const cache: Record<any, any> = {};
// 
//     return function (...args) {
//         if (args in cache) {
//             return cache[args];
//         }
//         const result = fn(...args);
//         cache[args] = result;
//         return result;
//     };
// }
// 
// /**
//  * 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 #2623: Memoize
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2623: Memoize
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const cache: Record<any, any> = {};
// 
//     return function (...args) {
//         if (args in cache) {
//             return cache[args];
//         }
//         const result = fn(...args);
//         cache[args] = result;
//         return result;
//     };
// }
// 
// /**
//  * 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 #2623: Memoize
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2623: Memoize
# type Fn = (...params: any) => any;
# 
# function memoize(fn: Fn): Fn {
#     const cache: Record<any, any> = {};
# 
#     return function (...args) {
#         if (args in cache) {
#             return cache[args];
#         }
#         const result = fn(...args);
#         cache[args] = result;
#         return result;
#     };
# }
# 
# /**
#  * 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 #2623: Memoize
// 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 #2623: Memoize
// type Fn = (...params: any) => any;
// 
// function memoize(fn: Fn): Fn {
//     const cache: Record<any, any> = {};
// 
//     return function (...args) {
//         if (args in cache) {
//             return cache[args];
//         }
//         const result = fn(...args);
//         cache[args] = result;
//         return result;
//     };
// }
// 
// /**
//  * 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 #2623: Memoize
type Fn = (...params: any) => any;

function memoize(fn: Fn): Fn {
    const cache: Record<any, any> = {};

    return function (...args) {
        if (args in cache) {
            return cache[args];
        }
        const result = fn(...args);
        cache[args] = result;
        return result;
    };
}

/**
 * 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.