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How to create a function that returns N number of arrays from a number array with difference between sum of each array is minimal

I am trying to create a function in JS/TS that returns N number of arrays with same length using all the elements in a number array A, with minimal difference in sum of elements in new arrays.
The size of number array A will be always divisible by N. This means we can always create N arrays with equal length.
All I found was this resource which not exactly what I am looking for
Reference: https://www.geeksforgeeks.org/split-array-into-two-subarrays-such-that-difference-of-their-sum-is-minimum/?tab=article but this is just to calculate the difference.
I have tried partion method which gives us N number of arrays with sum of each array is equal but the number of elements in arrays will be different.
function createGroups(baseArray:number[], numberOfGroupsNeeded:number){
// sumOfBaseArray/numberOfGroupsNeeded = average sum of splitted arrays
// return splited arrays with the same length
// and the difference in the sum of each array are minimal
}
Example Input: baseArray:number[] = [5, 4, 3, 2, 2, 1] , numberOfGroupsNeeded = 2
let sumOfBaseArray:number = baseArray.reduce((partialSum, a) => partialSum + a, 0)
let requiredSumForEachNewArrays = sumOfBaseArray/numberOfGroupsNeeded;
Sum of resulted arrys should be closer to requiredSumForEachNewArrays which is 8.5 ≈ 9
Output: [4, 3, 2] & [5, 2, 1]
Can someone help me to solve this problem? thanks

Problem with getting javascript to forget a variables value

I'm trying to solve a coding challenge
it gives me an array and waits for the answer ,then gives me the next array and so on.
https://www.codewars.com/kata/5648b12ce68d9daa6b000099/train/javascript
I am trying to take this input: var busStops = [[10,0],[3,5],[5,8]] and return 5.
the code is supposed to add the first in each pair to a total while subtracting the second in each pair from the total eg: 10 - 0 + 3 - 5 + 5 - 8 = 5
First my code loops through the inner arrays and outer array ,pushing it into myarr as a regular array eg: [10,0,3,5,5,8].
It then adds the value if it is index is 0 or even and subtracts it if the index is odd.
This actually works!
Until it is given a second array eg: [[3,0],[9,1],[4,10],[12,2],[6,1],[7,10]]
It is still calculating the total correctly but is still remembering the total from the first array meaning it is returning 22 instead of 17
Why?
There is a var answer = 0 that is being executed ahead of the second loop
It should forget the value of the previous answer.
Right?
Edit: I figured out my problem. I just needed to empty myarr after the total was calculated!
let myarr = [];
var number = function (busStops) {
for (var i = 0; i < busStops.length; i++) {
for (var j = 0; j < busStops[i].length; j++) {
/*console.log(busStops[i][j]);*/
myarr.push(busStops[i][j]);
}
}
console.log("test");
var answer = 0;
console.log("test again");
for (let t = 0; t < myarr.length; t++) {
if (t == 0 || t % 2 == 0) {
answer = answer + myarr[t];
} else {
answer = answer - myarr[t];
}
}
console.log(answer);
return answer;
};

The task at your hand tries to somehow find a value (an integer) from an array of arrays (multidimensional array). That task seems to be reducing that multidimensional array into a single integer.
Luckily, JavaScript has some powerful array methods and one of them is the reduce method:
The reduce() method executes a user-supplied "reducer" callback function on each element of the array, in order, passing in the return value from the calculation on the preceding element. The final result of running the reducer across all elements of the array is a single value. Source: MDN
The reduce method tries to reduce an array's elements into a single value and in your case we want to reduce that multidimensional array into a single value that is the number persons who are still in the bus.
Before typing some code, let's dig a bit deeper into the reduce method:
It accepts 2 parameters, a callback function that acts as the reducer and the initial value to be used in the first iteration of the reduce method.
The reducer callback function, on its own, accepts 4 parameters that are supplied by the reduce method. You may learn more about those parameters here as am only going to focus on the first 2 parameters the reducer accepts:
previousValue: hold the value from the previous reducer call. On first call, it contains the value you set to the initial value parameter of the reduce method or, if you didn't supply an initial value, previousValue shall hold the value of your array's first element (arr[0]).
currentValue: hold the current reduce iteration's item.
Now, let's get back to the task we have, we need to calculate the number of persons who are still in the bus based on a supplied multidimensional array. Each item in that multidimensional array is an array of two values where the result we need at the end is: the sum of the differences between each array, in the multidimensional array, first and second values (sum = multiDim[i][0] - multiDim[i][1] + multiDim[i + 1][0] + multiDim[i + 1][1] etc...).
To solve this task, we'll reduce that multidimensional array into a single number (let's call it result) by using a simple reducer function that will start by an initial value of 0 (as we're calculating a sum in our case) and will add, to the result, the difference between the first and the second values of the array supplied by the reduce at each iteration.
To illustrate, here's a live demo:
/**
* a function that calculates and returns the number of person who are still in the bus or precisely, the sum of the differences between "c[0]" and "c[1]"
* busArray: the supplied multidimensional array to reduce
* the reducer accepts two parameters:
* r: the result from the last call to the reducer function (initially set to 0, the initial value (see second parameter passed to the "reduce" method))
c: hold the current iteration's array.
*/
const calculateWhoAreInTheBus = busArray => busArray.reduce((r, c) => r += c[0] - c[1], 0);
/** test the created "calculateWhoAreInTheBus" function */
console.log(calculateWhoAreInTheBus([
[10, 0],
[3, 5],
[5, 8]
])); // should print: 5
console.log(calculateWhoAreInTheBus([
[3, 0],
[9, 1],
[4, 10],
[12, 2],
[6, 1],
[7, 10]
])); // should print: 17
console.log(calculateWhoAreInTheBus([
[3, 0],
[9, 1],
[4, 8],
[12, 2],
[6, 1],
[7, 8]
])); // should print: 21
console.log(calculateWhoAreInTheBus([
[0, 0],
[0, 0]
])); // should print: 0

I would advice you to use Array.prototype.reduce instead. For example like this:
const reducer = (previous, current) => previous + current[0] - current[1];
const answer = busStops.reduce(reducer, 0);
It is very brief (although this is not a goal in and of itself) and the reducer function does almost trivial work, so it does not complicate unneccesarily. Best of all it encapsulates the functionality with a minimal need of extra variables.
Othwerwise you could simplify your function a bit but use the let keyword to keep variables locked to scope like:
function number(busStops) {
let answer = 0;
for (let bs of busStops) {
answer += bs[0] - bs[1];
}
return answer;
}

Sum of two multi dimensional array

I have a array in JavaScript like this.
var arr=
[
['A'],[1,2,3,4],
['A'],[4,3,2,1],
['B'],[10,12,3,1],
['B'],[1,2,3,4],
.
.
.
.
['AZ'],[1,2,3,4]
]
and I want the output to summarize the array like -
var output=
[
['A'],[5,5,5,5],
['B'],[11,14,6,5],
['AZ'],[1,2,3,4]
]
Thanks.

Script
You can use the following script to achieve what you want to do.
const arr = [
["A"],
[1, 2, 3, 4],
["A"],
[4, 3, 2, 1],
["B"],
[10, 12, 3, 1],
["B"],
[1, 2, 3, 4],
["AZ"],
[1, 2, 3, 4],
];
/**
* Generator to return key-value pairs with array[i] being the key and array[i+1] being the value
* #param {Array<any>} array
*/
function* keyValue(array) {
// make sure we can build pairs (other ways of handling this are also possible)
if (array.length % 2 !== 0)
throw new RangeError(
"Array length must be dividable by 2 without remainder!"
);
for (let i = 0; i < array.length; i += 2) {
yield [array[i], array[i + 1]];
}
}
// here the created key-value pairs
console.log("Key-value pairs created by keyValue() generator function:");
console.log([...keyValue(arr)]);
// loop over key value pairs and sum up all the individul arrays based on the letter assigned to them
const result = [...keyValue(arr)].reduce((all, [[key], array]) => {
// if we don't have values for this letter, assing copy of the array to that letter
if (!all[key]) all[key] = [...array];
// we have some values for that letter already, sum up each value
else all[key] = all[key].map((prev, idx) => prev + array[idx]);
return all;
}, {});
// this would be a "better" result to my mind as there is no point wrapping single string values in arrays
// When using objects the values can easily be accessed in O(1)
console.log(result);
// now transform JS object to array of arrays
console.log("Result:");
const transformed = Object.entries(result).flatMap(([key, value]) => [[key], value]);
console.log(transformed);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Please note: This implementation assumes that the arrays for a given letter have the same length (as is the case in your example).
Explanation
First of all, I use a generator function keyValue() to always group two consecutive values in the array (a key and a value) together. One could also do this differently but once you understand how generators work that's an easy and elegant approach, I think. For this demo I just throw an error if the array is not dividable by 2 without remainder, but one could also handle this more gracefully.
Then, using reduce(), I iterate over the array created by using keyValue() and for each element in the array I check if I've encountered that value before. If I have not, I create a copy of the array (for immutablility) and assign it to the key i.e. a letter. If I have encountered a certain letter before I add up the values that I have previously saved assigned to that letter with the ones I am currently processing. After iteration all sums are calculated and I have a JavaScript object containing the results.
To my mind, this would be a good output because your output is a bit odd, since there is no point storing single letters in an array or even arrays of arrays. Using a JavaScript object is much more convenient and faster for lookups.
Nevertheless, you can easily deduct your result from the created object using flatMap().

How can division be used to find the product of every value an input array except the value with a corresponding index

In an attempt to improve my general problem solving skills, I recently subscribed to Daily Coding Problem. One of the challenges that came up has the following description:
This problem was asked by Uber.
Given an array of integers, return a new array such that each element
at index i of the new array is the product of all the numbers in the
original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output
would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the
expected output would be [2, 3, 6].
Follow-up: what if you can't use division?
I solved this particular challenge within minutes using the following function:
function solution(_input) {
return _input.map((_number, _index, _list) => {
return _list.reduce((_accumulator, _currentValue, _currentIndex) => {
return _accumulator * ((_index !== _currentIndex) ? _currentValue : 1);
}, 1);
});
}
My function works, matching every expected output perfectly... But this makes me curious about the last line of the challenge.
How could division be used to solve this?

As #Steve alluded in the comments, you would:
first find the product of all the elements in the array:
const product = input.reduce((accumulator, value) => accumulator * value, 1);
then map the array to the product divided by each element.
return input.map(value => product / value);
This reduces operational complexity from O(N2) to O(N) (if I'm not mistaken) because we are removing the nested loop.
const func = input => {
const product = input.reduce((accumulator, value) => accumulator * value, 1);
return input.map(value => product / value);
}
console.log(func([1, 2, 3, 4, 5]));

Javascript:: Find differences between array that contains duplicated values to a unique array of that array

I found many posts on stack overflow about that similar subject but none of them solve this issue here.
<script>
//Array GanginaA contains duplicated values.
//Array GanginaB contains only unique values that have been fetched from GanginaA
GanginaA=[0,1,2,3,4,5,5,6,7,8,9,9];
GanginaB=[0,1,2,3,4,5,6,7,8,9];
var hezi=<!--The Magic Goes Here-->
console.log(hezi);
/*
* Expected Output:
* 5,9
*/
</script>
GanginaA will always be longer or identical to GanginaB so there is no reason to calculate by the value of the longer array length.
GanginaB will always contains unique values that taken from GanginaA so it will always be the shorter array length or identical to GanginaA array.
Now it makes it a lot easier to find doubles.

You can use filter to get the elements like below
GanginaA = [0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 9];
GanginaB = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var hezi = GanginaB.filter(function (item, index) {
return GanginaA.indexOf(item) !== GanginaA.lastIndexOf(item)
});
console.log(hezi.join(" , ")); // 5, 9

the easier I can think of :
var hezi=[];
for (var i=0;i<GanginaA.length;i++){
hezi[GanginaA[i]] = GanginaA[i];
hezi[GanginaB[i]] = GanginaB[i];
}
hezi = hezi.filter (function(el){return el!=undefined;});
does everything in O(n) actions and not O(n^2)

Javascript's objects have hashmap like behaviour, so you can use them kind of like a set. If you iterate over all the values and set them to be keys within an object, you can use the Object.keys method to get an array of unique values out.
function uniqueValues() {
var unique = {};
[].forEach.call(arguments, function(array) {
array.forEach(function(value) {
unique[value] = true;
});
});
return Object.keys(unique);
};
This function will return the unique elements in any number of arrays, passed as arguments.
uniqueValues([1, 2, 3], [ 1, 1, 1], [2, 2, 2], [3, 3, 3]); // [ 1, 2 3 ]
One drawback to this method is that Javascript coerces all keys to strings, you can turn them back into numbers by changing the return statement to:
return Object.keys(unique).map(Number);