2626. Array Reduce Transformation
Problem Description
In this problem, we are provided with an array of integers called nums
, a reducer function fn
, and an initial value init
. The task is to produce a single value using a reduction process that applies fn
to each successive element of nums
, starting with init
as the first argument and the first array element as the second argument, and then using the result as the new first argument in the next application with the subsequent array element. This process continues until every element in the array has been processed. The output of the final call to fn
is what we will return as the result. In case nums
is an empty array, we should simply return init
. We are instructed to solve the problem without resorting to the built-in Array.reduce
method available in many programming languages.
Intuition
To approach this solution, we naturally think of iterating through the array, element by element, and at each step applying the reducer function fn
to the current result acc
and the current element x
. Starting with init
as our accumulator acc
, we update acc
with the result of fn(acc, x)
after each array element is processed. The use of a loop for iteration is a fundamental tool in programming, allowing us to apply the same operation for each element in a sequence - this fits perfectly with the task of reducing an array. The strength of the reducer pattern is that it's a general concept; the specific behavior is defined by the provided function fn
, which makes this approach both flexible and powerful.
The given TypeScript solution follows this intuition directly:
- Initialize the accumulator
acc
with the provided initial valueinit
. - Iterate through each element
x
in thenums
array, one by one. - For each element
x
, apply the reducer functionfn
to the accumulatoracc
andx
and updateacc
with the result. - After processing all elements, return
acc
as it now represents the accumulation of the sequential applications offn
.
By following these steps, we adhere to the instructions and compute the reduced value expected by the problem description.
Solution Approach
The implementation follows a straightforward pattern that's commonly used in functional programming, known as reduction or folding. The idea is to maintain a running total or combined result as we process a sequence of values.
Here's a step-by-step explanation of the solution's implementation:
-
Initialization: We start by initializing an accumulator variable
acc
with the value ofinit
, which is the initial value provided to us. In functional reduction, the accumulator is the placeholder for the ongoing result. -
Looping Through Elements: We then enter a loop that will iterate over each element in the
nums
array. This is done using afor...of
statement in TypeScript, a language construct particularly well-suited for iterating over array elements. -
Applying the Reducer Function: Inside the loop, for each element
x
, we call the reducer functionfn
withacc
andx
as arguments. The return value offn(acc, x)
is then assigned back toacc
. In mathematical terms, iffn
is denoted asf
, this step could be written asacc = f(acc, x)
. This updates the accumulator with the "reduced" value that incorporates the contribution of the current array elementx
. -
Returning the Result: After the loop has processed all of the elements, the final value of
acc
holds the reduced result of the entire array. We then exit the loop, and the function returns this final value. -
Handling Edge Cases: As per the problem description, if the array is empty (
nums.length === 0
), we simply return the initial valueinit
. In the provided implementation, this is implicitly handled, as the loop will not iterate at all for an empty array, andacc
, which still equalsinit
, will be returned.
The algorithm runs in O(n) time complexity, where n is the number of elements in the array since it processes each element exactly once. The space complexity is O(1), as we only use a fixed amount of extra space for the accumulator.
The TypeScript implementation provided uses no complex data structures or patterns beyond a for loop, making it very accessible and easily understandable.
The implementation provided is algorithmically efficient and straightforward, which stands as a testament to the elegance of the reduce pattern.
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Start EvaluatorExample Walkthrough
Let's consider a small example to illustrate the solution approach. Suppose we are given the following:
- An array
nums
of integers:[3, 5, 2, 4]
- A reducer function
fn
, which takes two integers and returns their sum. - An initial value
init
of10
.
Our goal is to reduce the array using the function fn
and init
as the starting point. Now, let's walk through the solution step by step:
-
Initialization: We initialize an accumulator
acc
with the value ofinit
which is10
. -
Looping Through Elements: We start an iteration over each element in the
nums
array:- First iteration: Take the first element
3
, apply the reducer functionfn(acc, x) => fn(10, 3)
which returns13
. Updateacc
to13
. - Second iteration: Take the next element
5
, apply the reducer functionfn(acc, x) => fn(13, 5)
which returns18
. Updateacc
to18
. - Third iteration: Take the next element
2
, apply the reducer functionfn(acc, x) => fn(18, 2)
which returns20
. Updateacc
to20
. - Fourth iteration: Take the last element
4
, apply the reducer functionfn(acc, x) => fn(20, 4)
which returns24
. Updateacc
to24
.
- First iteration: Take the first element
-
Returning the Result: Having processed all elements in the array,
acc
now holds the value24
, which is the final reduced result. -
Handling Edge Cases: If
nums
were an empty array, the for loop would not run and we would simply returninit
, which is10
in this case.
In this example, the final output, after reducing the array starting with an initial value of 10
, is 24
. The step-by-step reduction process consolidates all the values into one, applying the reducer function sequentially to each array element and the running accumulation.
Solution Implementation
1from typing import List, Callable
2
3# Type alias for the reducer function which is a callback that will be passed
4# to the reduce function. It takes two numbers, `accumulator` and `current_value`,
5# and returns a number.
6ReducerFunction = Callable[[int, int], int]
7
8def reduce(numbers: List[int], reducer: ReducerFunction, initial_value: int) -> int:
9 """
10 Applies a reducer function on each element of the `numbers` array, resulting in a single output value.
11
12 :param numbers: The list of numbers to be reduced.
13 :param reducer: The function to execute on each element in the list.
14 :param initial_value: The initial value to start the accumulation from.
15 :return: The reduced value after all elements have been processed.
16 """
17 # The variable that will accumulate the result of calling the reducer
18 # function repeatedly.
19 accumulator = initial_value
20
21 # Iterate through each number in the list
22 for current_value in numbers:
23 # Apply the reducer function to the current accumulator and the
24 # current value, then assign the result back to the accumulator.
25 accumulator = reducer(accumulator, current_value)
26
27 # Return the final accumulated value after processing all elements.
28 return accumulator
29
1import java.util.function.BiFunction;
2
3/**
4 * Applies a reducer function on each element of the `numbers` array, resulting in a single output value.
5 *
6 * @param numbers The array of numbers to be reduced.
7 * @param reducer The function to execute on each element in the array.
8 * @param initialValue The initial value to start the accumulation from.
9 * @return The reduced value after all elements have been processed.
10 */
11public static int reduce(int[] numbers, BiFunction<Integer, Integer, Integer> reducer, int initialValue) {
12 // The variable that will accumulate the result of calling the reducer function repeatedly.
13 int accumulator = initialValue;
14
15 // Iterate through each number in the array
16 for (int currentValue : numbers) {
17 // Apply the reducer function to the current accumulator and the current value,
18 // then assign the result back to the accumulator.
19 accumulator = reducer.apply(accumulator, currentValue);
20 }
21
22 // Return the final accumulated value after processing all elements.
23 return accumulator;
24}
25
1#include <vector>
2#include <functional>
3
4// Type alias for the reducer function which is a callback that will be passed to the reduce function.
5// It takes two integers, `accumulator` and `currentValue` and returns an integer.
6using ReducerFunction = std::function<int(int, int)>;
7
8/**
9 * Applies a reducer function on each element of the `numbers` vector, resulting in a single output value.
10 *
11 * @param numbers The vector of integers to be reduced.
12 * @param reducer The function to execute on each element in the vector.
13 * @param initialValue The initial value to start the accumulation from.
14 * @returns The reduced value after all elements have been processed.
15 */
16int reduce(const std::vector<int>& numbers, const ReducerFunction& reducer, int initialValue) {
17 // The variable that will accumulate the result of calling the reducer function repeatedly.
18 int accumulator = initialValue;
19
20 // Iterate through each number in the vector
21 for (int currentValue : numbers) {
22 // Apply the reducer function to the current accumulator and the current value,
23 // then assign the result back to the accumulator.
24 accumulator = reducer(accumulator, currentValue);
25 }
26
27 // Return the final accumulated value after processing all elements.
28 return accumulator;
29}
30
1// Type declaration for the reducer function which is a callback that will be passed to the reduce function.
2// It takes two numbers, `accumulator` and `currentValue` and returns a number.
3type ReducerFunction = (accumulator: number, currentValue: number) => number;
4
5/**
6 * Applies a reducer function on each element of the `numbers` array, resulting in a single output value.
7 *
8 * @param numbers - The array of numbers to be reduced.
9 * @param reducer - The function to execute on each element in the array.
10 * @param initialValue - The initial value to start the accumulation from.
11 * @returns The reduced value after all elements have been processed.
12 */
13function reduce(numbers: number[], reducer: ReducerFunction, initialValue: number): number {
14 // The variable that will accumulate the result of calling the reducer function repeatedly.
15 let accumulator: number = initialValue;
16
17 // Iterate through each number in the array
18 for (const currentValue of numbers) {
19 // Apply the reducer function to the current accumulator and the current value,
20 // then assign the result back to the accumulator.
21 accumulator = reducer(accumulator, currentValue);
22 }
23
24 // Return the final accumulated value after processing all elements.
25 return accumulator;
26}
27
Time and Space Complexity
Time Complexity
The time complexity of the reduce
function is O(n)
, where n
is the number of elements in the nums
array. This is because the function iterates through each element in the array exactly once.
Space Complexity
The space complexity of the reduce
function is O(1)
as it uses a fixed amount of space. The acc
variable is updated during each iteration, but no additional space that grows with the input size is utilized.
Which of the following array represent a max heap?
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