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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2626\. Array Reduce Transformation
Easy
Given an integer array `nums`, a reducer function `fn`, and an initial value `init`, return a **reduced** array.
A **reduced** array is created by applying the following operation: `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 final value of `val` is returned.
If the length of the array is 0, it 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]^2 = 101
(101) + nums[1]^2 = 105
(105) + nums[2]^2 = 114
(114) + nums[3]^2 = 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 <= 1000`
* `0 <= nums[i] <= 1000`
* `0 <= init <= 1000`