leetcode(easy)-Array-Reduce-Transformation.js

// 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.




// var reduce = function(nums, fn, init) {
//
// };


fn = function sum(accum, curr) { return accum + curr; }

const reduce = (nums, fn, init) => {
    if (Array.isArray(nums) && nums.length === 0) {
        return init
    }
    let initial = init ?? 0;
    for (let i = 0; i < nums.length; i++) {
        initial = fn(initial, nums[i])
    }
    return initial
}