Array Reduce Transformation solution and explanation
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The problem
Given an integer array nums, a reducer function fn, and an initial value init, return the final result obtained by executing the fn function on each element of the array, sequentially, passing in the return value from the calculation on the preceding element.
This result is achieved through the following operations: val = fn(init, nums[0]), val = fn(val, nums[1]), val = fn(val, nums[2]), ... until every element in the array has been processed. The ultimate value of val is then returned.
If the length of the array is 0, the function should return init.
Please solve it without using the built-in Array.reduce method.
Example 1:
Input:
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr; }
init = 0
Output: 10
Explanation:
initially, the value is init=0.
(0) + nums[0] = 1
(1) + nums[1] = 3
(3) + nums[2] = 6
(6) + nums[3] = 10
The final answer is 10.
Example 2:
Input:
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr * curr; }
init = 100
Output: 130
Explanation:
initially, the value is init=100.
(100) + nums[0] * nums[0] = 101
(101) + nums[1] * nums[1] = 105
(105) + nums[2] * nums[2] = 114
(114) + nums[3] * nums[3] = 130
The final answer is 130.
Example 3:
Input:
nums = []
fn = function sum(accum, curr) { return 0; }
init = 25
Output: 25
Explanation: For empty arrays, the answer is always init.
Constraints:
0 <= nums.length <= 10000 <= nums[i] <= 10000 <= init <= 1000
Approach overview
Problem Overview: You receive an array nums, a reducer function fn, and an initial value init. The task is to repeatedly apply the function so that each array element updates an accumulator value, eventually producing a single final result.
This mirrors how the classic reduce operation works in functional programming. Starting with init, the reducer combines the accumulator with each element in order: acc = fn(acc, nums[i]). After processing every element, the final accumulator value is returned.
Approach 1: Iterative Loop Reduction (O(n) time, O(1) space)
The most direct implementation uses a simple loop. Initialize a variable accumulator with init. Then iterate through the array from index 0 to n-1. For every element, call the reducer function and update the accumulator: accumulator = fn(accumulator, nums[i]). After the loop finishes, return the accumulator.
This approach works because reduction is inherently sequential. Each step depends on the previous accumulator value. The algorithm performs exactly one pass over the array, making the time complexity O(n). Only a single variable is maintained regardless of input size, so the space complexity remains O(1). This pattern appears frequently in problems involving arrays and aggregation tasks such as sums, products, or custom accumulations.
Approach 2: Recursive Reduction (O(n) time, O(n) space)
The same reduction logic can be expressed recursively. Define a recursive function that processes the array index by index. At each step, compute a new accumulator using the current element, then call the function for the next index.
The recursion terminates when the index reaches the end of the array. The final accumulator value is returned back through the call stack. While the algorithm still processes each element once for O(n) time complexity, recursion adds stack frames for every call, resulting in O(n) auxiliary space.
This version is conceptually aligned with functional programming patterns and demonstrates how reduction can be defined in terms of smaller subproblems. It’s a good exercise when practicing recursion or understanding functional abstractions used in languages like JavaScript.
Recommended for interviews: The iterative loop reduction is the expected approach. It is simpler, avoids recursion overhead, and clearly demonstrates how reduction accumulates results across an array. The recursive version still shows strong understanding of functional decomposition, but interviewers generally prefer the iterative implementation because it uses constant space and mirrors real-world implementations of reduce.
Approach 1: Iterative Loop Reduction
This approach simply involves directly iterating through the elements of the array using a loop, starting with the initial value. For each element, apply the reducer function and update the accumulated result. This approach mimics the behavior of the JavaScript Array.reduce method.
def reduce(nums, fn, init):
result = init
for num in nums:
result = fn(result, num)
return result
def sum_fn(accum, curr):
return accum + curr
nums = [1, 2, 3, 4]
init = 0
result = reduce(nums, sum_fn, init)
print(result)import java.util.function.IntBinaryOperator;
public class ArrayReducer {
public static int reduce(int[] nums, IntBinaryOperator fn, int init) {
int result = init;
for (int num : nums) {
result = fn.applyAsInt(result, num);
}
return result;
}
public static int sum(int accum, int curr) {
return accum + curr;
}
public static void main(String[] args) {
int[] nums = {1, 2, 3, 4};
int init = 0;
int result = reduce(nums, ArrayReducer::sum, init);
System.out.println(result);
}
}#include <iostream>
#include <vector>
int reduce(std::vector<int>& nums, int (*fn)(int, int), int init) {
int result = init;
for (int num : nums) {
result = fn(result, num);
}
return result;
}
int sum(int accum, int curr) {
return accum + curr;
}
int main() {
std::vector<int> nums = {1, 2, 3, 4};
int init = 0;
int result = reduce(nums, sum, init);
std::cout << result << std::endl;
return 0;
}function reduce(nums, fn, init) {
let result = init;
for (let num of nums) {
result = fn(result, num);
}
return result;
}
function sum(accum, curr) {
return accum + curr;
}
const nums = [1, 2, 3, 4];
const init = 0;
const result = reduce(nums, sum, init);
console.log(result);#include <stdio.h>
int reduce(int* nums, int numsSize, int (*fn)(int, int), int init) {
int result = init;
for (int i = 0; i < numsSize; i++) {
result = fn(result, nums[i]);
}
return result;
}
int sum(int accum, int curr) {
return accum + curr;
}
int main() {
int nums[] = {1, 2, 3, 4};
int init = 0;
int result = reduce(nums, 4, sum, init);
printf("%d\n", result);
return 0;
}using System;
public class ArrayReducer {
public static int Reduce(int[] nums, Func<int, int, int> fn, int init) {
int result = init;
foreach (var num in nums) {
result = fn(result, num);
}
return result;
}
public static int Sum(int accum, int curr) {
return accum + curr;
}
public static void Main() {
int[] nums = {1, 2, 3, 4};
int init = 0;
int result = Reduce(nums, Sum, init);
Console.WriteLine(result);
}
}This Python implementation defines a reduce function that iterates through the list nums, applying the provided function fn to the accumulated result. The sum_fn helper function adds two numbers.
Time Complexity: O(n), as each element of the array is processed once.
Space Complexity: O(1), since space usage is minimal and constant.
Approach 2: Recursive Reduction
This approach employs a recursive strategy to apply the reducer function on each element of the array. By defining a base case for the recursion (i.e., an empty array returns the initial value), the recursion continues until all elements are processed. Care must be taken with recursion due to stack size limitations for large inputs.
def reduce_recursive(nums, fn, init, index=0):
if index == len(nums):
return init
return fn(reduce_recursive(nums, fn, init, index + 1), nums[index])
def sum_fn(accum, curr):
return accum + curr
nums = [1, 2, 3, 4]
init = 0
result = reduce_recursive(nums, sum_fn, init)
print(result)import java.util.function.IntBinaryOperator;
public class ArrayReducerRecursive {
public static int reduceRecursive(int[] nums, int index, IntBinaryOperator fn, int init) {
if (index == nums.length) return init;
return fn.applyAsInt(reduceRecursive(nums, index + 1, fn, init), nums[index]);
}
public static int sum(int accum, int curr) {
return accum + curr;
}
public static void main(String[] args) {
int[] nums = {1, 2, 3, 4};
int init = 0;
int result = reduceRecursive(nums, 0, ArrayReducerRecursive::sum, init);
System.out.println(result);
}
}#include <iostream>
#include <vector>
int reduce_recursive(std::vector<int>::iterator begin, std::vector<int>::iterator end, int (*fn)(int, int), int init) {
if (begin == end) return init;
return fn(reduce_recursive(begin + 1, end, fn, init), *begin);
}
int sum(int accum, int curr) {
return accum + curr;
}
int main() {
std::vector<int> nums = {1, 2, 3, 4};
int init = 0;
int result = reduce_recursive(nums.begin(), nums.end(), sum, init);
std::cout << result << std::endl;
return 0;
}function reduceRecursive(nums, fn, init, index = 0) {
if (index >= nums.length) return init;
return fn(reduceRecursive(nums, fn, init, index + 1), nums[index]);
}
function sum(accum, curr) {
return accum + curr;
}
const nums = [1, 2, 3, 4];
const init = 0;
const result = reduceRecursive(nums, sum, init);
console.log(result);#include <stdio.h>
int reduce_recursive(int* nums, int numsSize, int (*fn)(int, int), int init) {
if (numsSize == 0) return init;
return fn(reduce_recursive(nums + 1, numsSize - 1, fn, init), nums[0]);
}
int sum(int accum, int curr) {
return accum + curr;
}
int main() {
int nums[] = {1, 2, 3, 4};
int init = 0;
int result = reduce_recursive(nums, 4, sum, init);
printf("%d\n", result);
return 0;
}using System;
public class ArrayReducerRecursive {
public static int ReduceRecursive(int[] nums, int index, Func<int, int, int> fn, int init) {
if (index == nums.Length) return init;
return fn(ReduceRecursive(nums, index + 1, fn, init), nums[index]);
}
public static int Sum(int accum, int curr) {
return accum + curr;
}
public static void Main() {
int[] nums = {1, 2, 3, 4};
int init = 0;
int result = ReduceRecursive(nums, 0, Sum, init);
Console.WriteLine(result);
}
}The Python version of recursive reduction calls itself incrementing the index. It terminates once all elements are processed, returning the accumulated value for an empty slice as the base result.
Time Complexity: O(n), for visiting each element.
Space Complexity: O(n), implied by the recursive depth of calls.
Approach 3: Default Approach
type Fn = (accum: number, curr: number) => number;
function reduce(nums: number[], fn: Fn, init: number): number {
let acc: number = init;
for (const x of nums) {
acc = fn(acc, x);
}
return acc;
}Complexity comparison
| Approach | Complexity |
|---|---|
| Iterative Loop Reduction | Time Complexity: O(n), as each element of the array is processed once. |
| Recursive Reduction | Time Complexity: O(n), for visiting each element. |
| Default Approach | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Loop Reduction | O(n) | O(1) | Best general solution. Single pass through the array with constant extra memory. |
| Recursive Reduction | O(n) | O(n) | Useful for learning recursion or functional programming patterns. |
Frequently asked questions
Is Array Reduce Transformation easy or hard?
Array Reduce Transformation is classified as an Easy problem. The main requirement is understanding how accumulation works across an array and implementing the reducer logic with either a loop or recursion.
Array Reduce Transformation Python/Java solution
The same logic works across languages. Use a loop that updates an accumulator variable for each element. Python, Java, C++, C#, and JavaScript implementations all follow the pattern of initializing a value and updating it with a reducer function during iteration.
How to solve Array Reduce Transformation in O(n)?
Initialize an accumulator with the provided initial value. Iterate through the array and repeatedly apply the reducer function using the pattern accumulator = fn(accumulator, nums[i]). After processing all elements, return the accumulator. This single-pass iteration guarantees O(n) time complexity.
What is the best approach for Array Reduce Transformation?
The iterative loop reduction approach is the best solution. It processes the array once, updating an accumulator with a reducer function for each element. This method runs in O(n) time and uses O(1) extra space, which matches how reduce operations are typically implemented in production libraries.
Is Array Reduce Transformation asked at Google/Amazon/Meta?
Problems involving array reduction and functional transformations appear in interviews at companies like Google, Amazon, and Meta, especially for JavaScript roles. Interviewers often test understanding of array operations such as map, filter, and reduce along with iteration fundamentals.
What data structure is used in Array Reduce Transformation?
The problem primarily uses arrays. The algorithm maintains a single accumulator variable that stores intermediate results while iterating through the array elements.
What is the time complexity of Array Reduce Transformation?
The time complexity is O(n) because every element of the array must be processed exactly once to update the accumulator. The iterative approach uses O(1) space, while the recursive implementation requires O(n) stack space due to function calls.