LeetCode 2626 - Array Reduce Transformation

The problem is asking us to simulate a "reduce" operation on an array of integers without using the built-in Array.reduce method. The input consists of three elements: an array nums, a function fn, and an initial value init.

leetcodeeasy

LeetCode Problem 2626

Difficulty: 🟢 Easy
Topics: —

Solution

Problem Understanding

The problem is asking us to simulate a "reduce" operation on an array of integers without using the built-in Array.reduce method. The input consists of three elements: an array nums, a function fn, and an initial value init. The fn function takes two arguments: the accumulator (the result computed so far) and the current element of the array. Starting with init as the initial accumulator, we apply fn sequentially to each element in nums. The final accumulator value after processing all elements is returned.

In simpler terms, this is like starting with a running total (init) and updating it by applying a rule (fn) to each element in order. For empty arrays, no operation is performed and init is returned.

The constraints tell us that nums can have up to 1000 elements, each element can be up to 1000, and init can also be up to 1000. This indicates that a simple iterative solution is sufficient because the input size is small. Important edge cases include an empty array, an array with one element, and functions that might not depend linearly on the elements (e.g., squaring them).

Approaches

The brute-force approach is to manually iterate over each element of the array and apply the function fn in order. We start with init as the accumulator and update it for every element. This approach is straightforward and correct because it directly follows the problem definition. It is also efficient enough given the constraints, as nums has at most 1000 elements.

The key observation is that no fancy data structures or recursion are required. Since each step only depends on the previous accumulator and the current element, a simple loop suffices. There is no need for extra memory beyond storing the accumulator, making this approach optimal.

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(1) Iterate through nums, updating accumulator sequentially
Optimal O(n) O(1) Same as brute force, no extra space needed

Algorithm Walkthrough

  1. Initialize a variable accum with the value of init. This variable will store the running result of applying fn.
  2. Iterate through each element num in the array nums.
  3. For each element, update accum by applying fn(accum, num). This ensures the previous result is combined with the current element according to the reducer function.
  4. After processing all elements, return accum as the final result.

Why it works: The algorithm maintains an invariant that accum always represents the result of applying fn to all processed elements. By sequentially updating accum for each element, the final value correctly represents the reduction over the entire array.

Python Solution

from typing import List, Callable

class Solution:
    def reduce(self, nums: List[int], fn: Callable[[int, int], int], init: int) -> int:
        accum = init
        for num in nums:
            accum = fn(accum, num)
        return accum

In this implementation, we first store init in accum. Then we iterate through the list nums and update accum by calling fn(accum, num) for each element. Finally, we return accum. The Python Callable type ensures that fn is a function taking two integers and returning an integer.

Go Solution

package main

func reduce(nums []int, fn func(int, int) int, init int) int {
    accum := init
    for _, num := range nums {
        accum = fn(accum, num)
    }
    return accum
}

The Go solution mirrors the Python approach. We initialize accum to init and loop through the slice nums, updating accum by calling the provided function fn. Go uses slices instead of arrays for dynamic length, and the range loop ensures we access each element sequentially.

Worked Examples

Example 1:

nums = [1,2,3,4], init = 0, fn = sum
Iteration | accum
0         | 0
1         | 0 + 1 = 1
2         | 1 + 2 = 3
3         | 3 + 3 = 6
4         | 6 + 4 = 10
Result = 10

Example 2:

nums = [1,2,3,4], init = 100, fn = sum of squares
Iteration | accum
0         | 100
1         | 100 + 1*1 = 101
2         | 101 + 2*2 = 105
3         | 105 + 3*3 = 114
4         | 114 + 4*4 = 130
Result = 130

Example 3:

nums = [], init = 25, fn = anything
Since the array is empty, no operation occurs.
Result = 25

Complexity Analysis

Measure Complexity Explanation
Time O(n) We iterate once over the array of size n.
Space O(1) Only a single accumulator variable is used, independent of n.

The complexity is efficient because we perform only a single pass and do not allocate additional data structures.

Test Cases

# provided examples
assert Solution().reduce([1,2,3,4], lambda a,b: a+b, 0) == 10  # simple sum
assert Solution().reduce([1,2,3,4], lambda a,b: a+b*b, 100) == 130  # sum of squares
assert Solution().reduce([], lambda a,b: 0, 25) == 25  # empty array

# additional tests
assert Solution().reduce([5], lambda a,b: a*b, 2) == 10  # single element, multiplication
assert Solution().reduce([1,1,1,1], lambda a,b: a-b, 10) == 6  # subtraction
assert Solution().reduce([0,0,0], lambda a,b: a+b, 0) == 0  # zeros only
assert Solution().reduce([1,2,3], lambda a,b: a*b, 1) == 6  # product
Test Why
[1,2,3,4], sum, 0 validates normal accumulation
[1,2,3,4], sum of squares, 100 validates non-linear reducer and non-zero init
[], any fn, 25 empty array returns init
[5], multiply, 2 single element edge case
[1,1,1,1], subtract, 10 ensures order matters for non-commutative fn
[0,0,0], sum, 0 all zeros handled correctly
[1,2,3], multiply, 1 product accumulation works

Edge Cases

First, an empty array is an important edge case because no iteration occurs. The algorithm correctly returns init directly without attempting to access elements, preventing out-of-bounds errors.

Second, a single-element array tests that the function correctly applies fn once. This ensures that the algorithm does not skip the first element or accidentally iterate multiple times.

Third, functions that are not commutative, such as subtraction or division, are edge cases because the order of application affects the result. By updating the accumulator sequentially as defined, the implementation preserves the correct order.

This solution handles all these edge cases robustly by relying on a simple loop that applies the reducer function in sequence, starting with the initial value.