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arrayReduceTransformation.js
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arrayReduceTransformation.js
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// LeetCode #2626. Array Reduce Transformation
/* Instructions
Given an integer array nums, a reducer function fn, and an intial 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, arr[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
*/
// Solution
var reduce = function(nums, fn, init) {
let answer = init
for (let i = 0; i < nums.length; i++) {
answer = fn(answer, nums[i])
}
return answer
};
// Explanation
// The function `reduce` is defined and takes three arguments, nums, fn, and init.
// Inside the `reduce` function, a variable `answer` is initialized to the init value.
// Then, a for loop is used to iterate through every element in the `nums` array.
// The reducer function `fn` is called with the current `answer` value and the current element in the `nums` array
// The result of the reducer function is reassigned to the `answer` variable.
// Finally, the `answer` variable is returned as the final reduced value.