DailyCodingProblem, find pair in array that matches a given value - javascript

I'm fairly new to coding and enlisted for the daily coding problem mailing list and got this question:
Given a list of numbers and a number k, return whether any two numbers
from the list add up to k.
My solution (after some stackoverflow digging) looks like this;
function problemOne_Solve()
{
const k = 17;
const values = [11, 15, 3, 8, 2];
for (i=0; i < values.length; i++) {
if ( values.find( (sum) => { return k-values[i] === sum} ) ) return true;
}
return false;
}
I'm wondering why it works. To me it looks like the part with the fat-arrow function closes the brackets inside the if statements conditional logic. And there is no such brackets after the if statement, which I thought was required.
I was also wondering how i would go about outputting the pair or pairs that sums up to "k," to build further on the solution. I would like to be able to display the pairs on the page for example.


.find takes a callback, which is invoked for every item in the array (or, up until a match is found). The first argument to the callback is the item being iterated over. If a match is found (if the return value from the callback was truthy for any element), the .find returns the item that resulted in a truthy return value.
So, on the first i = 0 iteration, and values[i] is 11, (sum) => { return k-values[i] === sum} will first check whether 17 - 11 === 11, and then whether 17 - 11 === 15, and then whether 17 - 11 = 3, etc.
This condition will generally be fulfilled if two numbers in the array add up to the k, but the algorithm is buggy. For example, an array composed of [1] will check the 1 against itself on the first iteration, adding up to 2:
function problemOne_Solve() {
const k = 2;
const values = [1];
for (i=0; i < values.length; i++) {
if ( values.find( (sum) => { return k-values[i] === sum} ) ) return true;
}
return false;
}
console.log(problemOne_Solve());
That is wrong. Another problem is that .find returns the found value. But, if the array is an array of numbers, the found value may be 0, and 0 is falsey. So the below example should return true because two elements sum up to 0 (0 and 0), but it returns false:
function problemOne_Solve() {
const k = 0;
const values = [0, 0];
for (i=0; i < values.length; i++) {
if ( values.find( (sum) => { return k-values[i] === sum} ) ) return true;
}
return false;
}
console.log(problemOne_Solve());
To get it right and decrease the computational complexity from O(n ^ 2) to O(n), iterate over the array once. Create an object whose keys are the numbers being iterated over, and on each iteration, check to see if a key of target - currNum exists on the object (where target is the target sum, and currNum is the current number from the array):
function problemOne_Solve() {
const target = 17;
const values = [11, 15, 3, 8, 2];
const obj = {};
for (const currNum of values) {
if (obj.hasOwnProperty(target - currNum)) {
return true;
}
obj[currNum] = true;
}
return false;
}
console.log(problemOne_Solve());
I was also wondering how i would go about outputting the pair or pairs that sums up to "k," to build further on the solution. I would like to be able to display the pairs on the page for example.
Instead of returning immediately when a match is found, push to an array and then return that array at the end of the function. Also, instead of setting the object values to true (or false), set them to the number of occurrences the number has been found so far (and decrement the matching number when a match is found):
function problemOne_Solve() {
const target = 17;
const values = [11, 15, 3, 8, 2, 17, 0, 0, 17];
const obj = {};
const matches = [];
for (const currNum of values) {
const otherNum = target - currNum;
if (obj[otherNum]) {
obj[otherNum]--;
matches.push([currNum, otherNum]);
}
obj[currNum] = (obj[currNum] || 0) + 1;
}
return matches;
}
console.log(problemOne_Solve());
And there is no such brackets after the if statement, which I thought was required.
Brackets are not required when there's a single statement after an if (or else if or else), eg:
if (true) console.log('true');
else console.log('this will not log');


And there is no such brackets after the if statement, which I thought was required.
If there is only one statement after if else the brackets becomes optional. Ideally you shouldn't write the if block is one line to make your code clean
As you are a beginner I would recommend you to use simple for loops instead of these fancy methods like find.
You can do that in following steps:
Its clear that you need sum of each element with every other element of array. So you will need a nested loop structure
The outer loop or main loop starts from 0 and loop till end of array.
You need to create a inner loop or nested loop which starts from index after the current index.
In the each iteration of nested loop you need to check if the sum of two elements is equal to requiredSum or not.
function pairWithSum(givenArray, requiredSum){
for(let i = 0; i < givenArray.length; i++){
for(let j = i + 1; j < givenArray.length; j++){
let sum = givenArray[i] + givenArray[j];
if(sum === requiredSum){
return [givenArray[i], givenArray[j]];
}
}
}
return false
}
console.log(pairWithSum([1, 4, 5, 8], 12));
console.log(pairWithSum([1, 4, 5, 8], 15));


I'm wondering why it works
That is because, if expects an expression/ statement to validate.
values.find( (sum) => { return k-values[i] === sum} )
This is a statement and it will be evaluated before and its output will be passed to if for condition.
Now Array.find has a return type: T|<null> where T is any value array is made of. So in second iteration, when values[i] refers to 15, it will return 2.
Now in JS, 2 is a truthy value and hence it goes inside if block. Foe more reference, check All falsey values in JavaScript. Any value that is not in this list will be considered as true.


My Javascript solution using JS object. This solution memories the elements when we go through the array which can be memory expensive. But the complexity will stay as O(n).
const checkTwoSum = (arr, sum) => {
const obj = {};
const found = arr?.find(item => {
const target = sum - item;
if (obj[target]) return true;
else {
obj[item] = 1;
}
});
return !!(found || found === 0);
}

Related

Write a function getDuplicates

Write a function getDuplicates that returns an array of all the elements that appear more than once in the initial items array (keeping the order). If an element appears many times, it should still be added to the result once.
This is my code
function getDuplicates(items) {
let result = [];
if (items === [0,0,0,0]) {return [0]}
for (let i = 0; i < items.length; i++) {
for (let j = i + 1; j < items.length; j++) {
if (items[i] === items[j]) {
result.push(items[i])
}
}
}
return result
}
I get an error:
input: [0, 0, 0, 0]
Hide details
Expected:
[0]
Received:
[0,0,0,0,0,0]

In JavaScript, arrays are objects, so when you use the === operator to compare two arrays, it will only return true if they are the exact same object in memory.
Use a Set to track duplicates: Instead of using an array to store the duplicate elements, we can use a Set to make sure we don't add duplicates to the result array. A Set is an efficient data structure for checking if an element exists or not, and it also automatically removes duplicates.
Use a single loop: Instead of using two nested loops to compare every element with every other element, we can use a single loop to keep track of the elements we've seen so far, and add them to the result if we see them again.
function getDuplicates(items) {
const result = [];
const seen = new Set();
for (const item of items) {
if (seen.has(item) && !result.includes(item)) {
result.push(item);
} else {
seen.add(item);
}
}
return result;
}
console.log(getDuplicates([0, 1, 0, 1, 2]))

a modified version of yours
function getDuplicates(items) {
let result = [];
let added = {};
for (let i = 0; i < items.length; i++) {
if (!added[items[i]] && items.indexOf(items[i], i + 1) !== -1) {
result.push(items[i]);
added[items[i]] = true;
}
}
return result;
}
console.log(getDuplicates([0, 1, 0, 1, 2]))
or in short doing the same
const getDuplicates = items => items.filter((item, index) => items.indexOf(item) !== index && items.lastIndexOf(item) === index);
console.log(getDuplicates([0, 1, 0, 1, 2]))

The best way to filter out the unique elements in an array is JavaScript Set
You cannot compare two arrays just like array1 === array2 because, Arrays have the type Object and you cannot compare two object just with equal to operator. Objects are not compared based on their values but based on the references of the variables. So when you compare two arrays which have same values using array1 === array2, it will compare its memory location only, not its values. So it will be only false.
The best way to achieve your result is to create an Array by checking the number of occurrences of nodes in the parent array, having occurrences count more than one and use a Set to remove the repetitions
function getDuplicates(items) {
return Array.from(new Set(items.filter(node => items.filter(x => node === x).length > 1)))
}
console.log(getDuplicates([0, 1, 0, 1, 2]))

You can try it:
Check if the current number is duplicated by using filter to check the length of an array.
Check if the result array contains duplicates.
function getDuplicates(items) {
let result = [];
for (let i = 0; i < items.length; i++) {
if ((items.filter(item => item == items[i])).length > 1 && !result.includes(items[i])) {
result.push(items[i]);
}
}
return result;
}
console.log(getDuplicates([0, 0, 0, 0]));

So. first of all - comparing 2 array will not work, (Somebody already explained why above).
Your code doesn't work because of if statement. You're checking if an array doesn't have any value except 0.
Try summing all numbers in the array and check if it's 0.
if(arr.reduce((accum, curr) => return accum += curr) == 0) {
return [0];
}

Your code is close, but there are a few issues that need to be addressed. First, you should not use the strict equality operator === to compare arrays, because it checks whether the two arrays have the same reference, not the same elements. Instead, you can use the JSON.stringify() method to compare the string representations of the arrays.
Second, your code only returns [0] if the input array is [0,0,0,0], which is not a general solution for finding duplicates. You can use an object to keep track of the count of each element in the array, and then add the elements that have a count greater than 1 to the result array.
Here's the corrected code:
function getDuplicates(items) {
let result = [];
let count = {};
for (let i = 0; i < items.length; i++) {
if (count[items[i]] === undefined) {
count[items[i]] = 1;
} else {
count[items[i]]++;
}
}
for (let i = 0; i < items.length; i++) {
if (count[items[i]] > 1 && result.indexOf(items[i]) === -1) {
result.push(items[i]);
}
}
return result;
}
This code keeps track of the count of each element in the count object, and then adds the elements that have a count greater than 1 to the result array, while making sure not to add duplicates to the result.

Javascript Detect All Values of Array in another Array

Let's say I have an array const series = [0,4,8] and another array const positions = [0,3,4,7,8]
How would I detect if all of the values of series are included in positions?
My initial reaction was to make a for loop that would cycle through all the values in positions and check to see if they equal the values of series but that seems highly inefficient.

Javascript Set object lacks set-theoretical methods, on the bright side, they are easy to implement:
class RealSet extends Set {
isSuperSet(iterable) {
for (let x of iterable) {
if (!this.has(x))
return false;
}
return true;
}
// #TODO: union, intersect, isSubSet, etc.
}
series = [0,4,8]
positions = [0,3,4,7,8]
console.log(new RealSet(positions).isSuperSet(series))
See also: TC39 proposal for set methods.

worst case complexity will be n * log(n). n for looping all elements in series array and log(n) for binary search.
const positions = [0, 3, 4, 7, 8];
const series = [0, 4, 8];
function bsearch(Arr, value) {
var low = 0,
high = Arr.length - 1,
mid;
while (low <= high) {
mid = Math.floor((low + high) / 2);
if (Arr[mid] == value) return mid;
else if (Arr[mid] < value) low = mid + 1;
else high = mid - 1;
}
return -1;
}
let i;
for (i = 0; i < series.length; ++i) {
const pos = bsearch(positions, series[i]);
if (pos === -1) break;
}
if (i === series.length) {
console.log("All element present");
} else {
console.log("All elmenet not present");
}

My answer to this would be to use Javascript builtins Array.prototype.every and Array.prototype.indexOf.
It would go something like this,
// Returns true if all elements in firstArray are also in secondArray
firstArray.every(element => secondArray.indexOf(element) !== -1)
The every builtin loops through the array, running your callback on each element and returns true if your callback returned true (or truthy values) for all elements, otherwise returns false.
The indexOf builtin searches for the argument you give it in the array with reference equality, returns the index of the argument if it is in the array, or returns -1 if it isn't.
So for every element in firstArray we check if it is in secondArray and then return true if it is by checking if indexOf returns -1 or not.
Combining these, we can solve this problem in a one-liner. For unsorted arrays, this is the best you can get. However, if you assume the array is sorted, you can get much faster using binary search. As the answer before me uses.

If your arrays (both of them) are sorted and they don't contain repeats, you can walk through them this way:
//const series = [0,4,8];
const positions = [0,3,4,7,8];
function check(series,positions){
let pos=0;
for(let item of series){
while(positions[pos]<item && pos<positions.length)
pos++;
if(pos==positions.length || positions[pos]!==item)
return false;
}
return true;
}
console.log("true?",check([0,4,8],positions));
console.log("true?",check([0],positions));
console.log("false?",check([5],positions));
console.log("true?",check([8],positions));
console.log("false?",check([0,4,5],positions));

Create a function to slice without using slice()

I am trying to write a function called slice that accepts an array and two numbers.
The function should return a new array with the elements starting at the index of the first number and going until the index of the second number.
If a third parameter is not passed to the function, it should slice until the end of the array by default.
If the third parameter is greater than the length of the array, it should slice until the end of the array.
function slice(s, n, m) {
let a = [];
a = s.splice(n, m);
if(m === undefined || m > s.length) {
a = s.splice(n, s.length);
}
return a;
}
let s = [1, 2, 3, 4, 5];
slice(s, 1, 7);
output []
UPDATE:
Thanks for everyone's help; I GOT IT!!! happy dance
function slice(arr, start, end) {
let result = [];
from = Math.max(start, 0);
to = Math.min(end);
if((!end) || (end > arr.length)) {
for(let i = from; i<arr.length; i++) {
result.push(arr[i]);}
} else {
for(let i = from; i<to; i++) {
result.push(arr[i]);
}
}
return result;
}
slice([1, 2, 3, 4, 5], 2, 4);

The main problem is that .splice mutates the array, so on your first call to it you remove a part of the array (or in your case the whole array). Therefore if the code enters the if, and you call .splice a second time, the array is already empty. An if / else would work so that .splice gets only called once.
But that would still then not replicate the behaviour of .slice, as .slice does not mutate the original array. Therefore you rather need a loop, that copies one element after the other:
// if "do" doesn't get passed, initialize with array.length (default parameter)
function slice(array, from, to = array.length) {
// make sure the bounds are within the range
from = Math.max(from, 0);
to = Math.min(to, array.length);
// initialize an array we can copy values into
const result = [];
for(let index = from; index < to; index++) {
// left as an exercise :)
}
return result;
}

Answering this since the OP said the time for the homework has passed.
One way to solve this problem is to have two pointers to second and third arguments. If you have all the arguments given, this is how you should start
start = n
end = m
// if m is greater than length of s or if m is not given
if(m == undefined || m > s.length()){
end = s.length() - 1;
}
then it is a simple for loop from start to end, both inclusive.
int[] result = new int[end-start+1];
for(int i = start; i <= end; i++){
result[j] = s[i];
}
Code may not be syntactically correct but you can fix that.

Checking whether the number of unique numbers within array exceeds n

Just as title reads, I need to check whether the number of unique entries within array exceeds n.
Array.prototype.some() seems to fit perfectly here, as it will stop cycling through the array right at the moment, positive answer is found, so, please, do not suggest the methods that filter out non-unique records and measure the length of resulting dataset as performance matters here.
So far, I use the following code, to check if there's more than n=2 unique numbers:
const res = [1,1,2,1,1,3,1,1,4,1].some((e,_,s,n=2) => s.indexOf(e) != s.lastIndexOf(e) ? false : n-- ? false : true);
console.log(res);
.as-console-wrapper { min-height: 100%}
And it returns false while there's, obviously 3 unique numbers (2,3,4).
Your help to figure out what's my (stupid) mistake here is much appreciated.
p.s. I'm looking for a pure JS solution

You can use a Map() with array values as map keys and count as values. Then iterate over map values to find the count of unique numbers. If count exceeds the limit return true, if not return false.
Time complexity is O(n). It can't get better than O(n) because every number in the array must be visited to find the count of unique numbers.
var data = [1, 1, 2, 1, 1, 3, 1, 1, 4, 1];
function exceedsUniqueLimit(limit) {
var map = new Map();
for (let value of data) {
const count = map.get(value);
if (count) {
map.set(value, count + 1);
} else {
map.set(value, 1);
}
}
var uniqueNumbers = 0;
for (let count of map.values()) {
if (count === 1) {
uniqueNumbers++;
}
if (uniqueNumbers > limit) {
return true;
}
}
return false;
}
console.log(exceedsUniqueLimit(2));

To know if a value is unique or duplicate, the whole array needs to be scanned at least once (Well, on a very large array there could be a test to see how many elements there is left to scan, but the overhead for this kind of test will make it slower)
This version uses two Set
function uniqueLimit(data,limit) {
let
dup = new Set(),
unique = new Set(),
value = null;
for (let i = 0, len = data.length; i < len; ++i) {
value = data[i];
if ( dup.has(value) ) continue;
if ( unique.has(value) ) {
dup.add(value);
unique.delete(value);
continue;
}
unique.add(value);
}
return unique.size > limit;
}
I also tried this version, using arrays:
function uniqueLimit(data, limit) {
let unique=[], dup = [];
for (let idx = 0, len = data.length; idx < len; ++idx) {
const value = data[idx];
if ( dup.indexOf(value) >= 0 ) continue;
const pos = unique.indexOf(value); // get position of value
if ( pos >= 0 ) {
unique.splice(pos,1); // remove value
dup.push(value);
continue;
}
unique.push(value);
}
return unique.length > limit;
};
I tested several of the solutions in this thread, and you can find the result here. If there are only a few unique values, the method by using arrays is the fastest, but if there are many unique values it quickly becomes the slowest, and on large arrays slowest by several magnitudes.
More profiling
I did some more tests with node v12.10.0. The results are normalized after the fastest method for each test.
Worst case scenario: 1000000 entries, all unique:
Set 1.00 // See this answer
Map 1.26 // See answer by Nikhil
Reduce 1.44 // See answer by Bali Balo
Array Infinity // See this answer
Best case scenario: 1000000 entries, all the same:
Array 1.00
Set 1.16
Map 2.60
Reduce 3.43
Question test case: [1, 1, 2, 1, 1, 3, 1, 1, 4, 1]
Array 1.00
Map 1.29
Set 1.47
Reduce 4.25
Another test case: [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,
1,1,1,1,1,1,1,3,4,1,1,1,1,1,1,1,2,1,1,1,
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,5 ]
Array 1.00
Set 1.13
Map 2.24
Reduce 2.39
Conclusion
The method that uses Set works for both small and large arrays, and performs well regardless of if there are many unique values or not. The version that are using arrays can be faster if there are few unique values, but quickly becomes very slow if there are many unique values.

Using sets, We count hypothetical unique set size and duplicateSet size and delete unique set element for each duplicate found. If unique set size goes below n, we stop iterating.
function uniqueGtN(res, n) {
let uniqSet = new Set(res);
let max = uniqSet.size;
if (max <= n) return false;
let dupSet = new Set();
return !res.some(e => {
if (dupSet.has(e)) {
if (uniqSet.has(e)) {
uniqSet.delete(e);
console.log(...uniqSet);
return (--max <= n);
}
} else {
dupSet.add(e);
}
});
}
console.log(uniqueGtN([1, 1, 2, 1, 1, 3, 3, 1], 2));

From your original solution, I have changed few things, it seems to be working fine:
(function() {
const array = [1,1,2,1,1,3,1,1,4,1];
function hasExceedingUniqueNumber(array, number) {
return array.some((e,_,s,n=number) => {
let firstIndex = s.indexOf(e);
let lastIndex = s.lastIndexOf(e);
// NOT unique
if (firstIndex != lastIndex) {
return false;
}
// unique
return e > n;
});
}
console.log('1', hasExceedingUniqueNumber(array, 1));
console.log('2', hasExceedingUniqueNumber(array, 2));
console.log('3', hasExceedingUniqueNumber(array, 3));
console.log('4', hasExceedingUniqueNumber(array, 4));
})();
So the shorter version looks like this:
(function() {
const array = [1,1,2,1,1,3,1,1,4,1];
function hasExceedingUniqueNumber(array, number) {
return array.some((e,_,s,n=number) => s.indexOf(e) != s.lastIndexOf(e) ? false : e > n);
}
console.log('1', hasExceedingUniqueNumber(array, 1));
console.log('2', hasExceedingUniqueNumber(array, 2));
console.log('3', hasExceedingUniqueNumber(array, 3));
console.log('4', hasExceedingUniqueNumber(array, 4));
})();

The code listed in your question does not work because m is not shared across the calls to the some callback function. It is a parameter, and its value is 2 at each iteration.
To fix this, either put m outside, or use the thisArg of the some function (but that means you can't use an arrow function)
let m = 2;
const res = [1,1,1,2,1,1,3,1,1,1,4,1,1]
.sort((a,b) => a-b)
.some((n,i,s) => i > 0 && n == s[i-1] ? !(m--) : false);
// ----- or -----
const res = [1,1,1,2,1,1,3,1,1,1,4,1,1]
.sort((a,b) => a-b)
.some(function(n,i,s) { return i > 0 && n == s[i-1] ? !(this.m--) : false; }, { m: 2 });
Note: this code seems to count if the number of duplicates exceeds a certain value, not the number of unique values.
As another side note, I know you mentioned you did not want to use a duplicate removal algorithm, but performant ones (for example hash-based) would result in something close to O(n).
Here is a solution to count all the values appearing exactly once in the initial array. It is a bit obfuscated and hard to read, but you seem to be wanting something concise. It is the most performant I can think of, using 2 objects to store values seen at least once and the ones seen multiple times:
let res = [1,1,2,3,4].reduce((l, e) => (l[+!l[1][e]][e] = true, l), [{},{}]).map(o => Object.keys(o).length).reduce((more,once) => once-more) > 2;
Here is the less minified version for people who don't like the short version:
let array = [1,1,2,3,4];
let counts = array.reduce((counts, element) => {
if (!counts.atLeastOne[element]) {
counts.atLeastOne[element] = true;
} else {
counts.moreThanOne[element] = true;
}
return counts;
}, { atLeastOne: {}, moreThanOne: {} });
let exactlyOnceCount = Object.keys(counts.atLeastOne).length - Object.keys(counts.moreThanOne).length;
let isOverLimit = exactlyOnceCount > 2;

Whenever I have a type of problem like this, I always like to peek at how the underscore JS folks have done it.
[Ed again: removed _.countBy as it isn't relevant to the answer]
Use the _.uniq function to return a list of unique values in the array:
var u = _.uniq([1,1,2,2,2,3,4,5,5]); // [1,2,3,4,5]
if (u.length > n) { ...};
[ed:] Here's how we might use that implementation to write our own, opposite function that returns only non-unique collection items
function nonUnique(array) {
var result = [];
var seen = [];
for (var i = 0, length = array.length; i < length; i++) {
var value = array[i];
if (seen.indexOf(value) === -1) { // warning! naive assumption
seen.push(value);
} else {
result.push(value);
}
}
console.log("non-unique result", result);
return result;
};
function hasMoreThanNUnique(array, threshold) {
var uArr = nonUnique(array);
var accum = 0;
for (var i = 0; i < array.length; i++) {
var val = array[i];
if (uArr.indexOf(val) === -1) {
accum++;
}
if (accum > threshold) return true;
}
return false;
}
var testArrA = [1, 1, 2, 2, 2, 3, 4, 5]; // unique values: [3, 4, 5]
var testArrB = [1, 1, 1, 1, 4]; // [4]
var testResultsA = hasMoreThanNUnique(testArrA, 3)
console.log("testArrA and results", testResultsA);
var testResultsB = hasMoreThanNUnique(testArrB, 3);
console.log("testArrB and results", testResultsB);

So far, I came up with the following:
const countNum = [1,1,1,2,1,1,3,1,1,1,4,1,1].reduce((r,n) => (r[n]=(r[n]||0)+1, r), {});
const res = Object.entries(countNum).some(([n,q]) => q == 1 ? !(m--) : false, m=2);
console.log(res);
.as-console-wrapper{min-height:100%}
But I don't really like array->object->array conversion about that. Is there a faster and (at the same time compact) solution?

How to adjust return values of map() function?

I have been trying to make a excercise in the course I am taking. At the end, I did what was asked, but I personally think I overdid too much and the output is not convenient -- it's a nested array with some blank arrays inside...
I tried to play with return, but then figured out the problem was in the function I used: map always returns an array. But all other functions, which are acceptable for arrays (in paticular forEach and I even tried filter) are not giving the output at all, only undefined. So, in the end, I have to ask you how to make code more clean with normal output like array with just 2 needed numbers in it (I can only think of complex way to fix this and it'll add unneeded junk to the code).
Information
Task:
Write a javascript function that takes an array of numbers and a target number. The function should find two different numbers in the array that, when added together, give the target number. For example: answer([1,2,3], 4) should return [1,3]
Code
const array1 = [1, 2, 3];
const easierArray = [1, 3, 5] //Let's assume number we search what is the sum of 8
const findTwoPartsOfTheNumber = ((arr, targetNum) => {
const correctNumbers = arr.map((num, index) => {
let firstNumber = num;
// console.log('num',num,'index',index);
const arrayWeNeed = arr.filter((sub_num, sub_index) => {
// console.log('sub_num',sub_num,'sub_index',sub_index);
if (index != sub_index && (firstNumber + sub_num) === targetNum) {
const passableArray = [firstNumber, sub_num] //aka first and second numbers that give the targetNum
return sub_num; //passableArray gives the same output for some reason,it doesn't really matter.
}
})
return arrayWeNeed
})
return correctNumbers;
// return `there is no such numbers,that give ${targetNum}`;
})
console.log(findTwoPartsOfTheNumber(easierArray, 8));
console.log(findTwoPartsOfTheNumber(array1, 4));
Output
[[],[5],[3]]
for the first one

You can clean up the outpu by flatting the returned arrays :
return arrayWeNeed.flat();
and
return correctNumbers.flat();
const array1 = [1, 2, 3];
const easierArray = [1, 3, 5] //Let's assume number we search what is the sum of 8
const findTwoPartsOfTheNumber = ((arr, targetNum) => {
const correctNumbers = arr.map((num, index) => {
let firstNumber = num;
// console.log('num',num,'index',index);
const arrayWeNeed = arr.filter((sub_num, sub_index) => {
// console.log('sub_num',sub_num,'sub_index',sub_index);
if (index != sub_index && (firstNumber + sub_num) === targetNum) {
const passableArray = [firstNumber, sub_num] //aka first and second numbers that give the targetNum
return sub_num; //passableArray gives the same output for some reason,it doesn't really matter.
}
})
return arrayWeNeed.flat();
})
return correctNumbers.flat();
// return `there is no such numbers,that give ${targetNum}`;
})
console.log(findTwoPartsOfTheNumber(easierArray, 8));
console.log(findTwoPartsOfTheNumber(array1, 4));
However, using a recursive function could be simpler :
const answer = (arr, num) => {
if (arr.length < 1) return;
const [first, ...rest] = arr.sort();
for (let i = 0; i < rest.length; i++) {
if (first + rest[i] === num) return [first, rest[i]];
}
return answer(rest, num);
};
console.log(answer([1, 2, 3], 4));
console.log(answer([1, 3, 5], 8));

It looks like you are trying to leave .map() and .filter() beforehand, which you can't (without throwing an error). So I suggest a normal for approach for this kind of implementation:
const array1 = [1,2,3];
const easierArray = [1,3,5] //Let's assume number we search what is the sum of 8
const findTwoPartsOfTheNumber = (arr,targetNum) =>{
for(let index = 0; index < arr.length; index++) {
let firstNumber = arr[index];
// console.log('num',num,'index',index);
for(let sub_index = 0; sub_index < arr.length; sub_index++){
const sub_num = arr[sub_index];
// console.log('sub_num',sub_num,'sub_index',sub_index);
if (index != sub_index && (firstNumber + sub_num) === targetNum){
const passableArray = [firstNumber,sub_num]//aka first and second numbers that give the targetNum
return passableArray; //passableArray gives the same output for some reason,it doesn't really matter.
}
}
}
return `there is no such numbers,that give ${targetNum}`;
}
console.log(findTwoPartsOfTheNumber(easierArray,8));
console.log(findTwoPartsOfTheNumber(array1,4));
console.log(findTwoPartsOfTheNumber(array1,10));
I've just grab your code and changed map and filter to for implementation.

There doesn't appear to be any requirement for using specific array functions (map, forEach, filter, etc) in the problem statement you listed, so the code can be greatly simplified by using a while loop and the fact that you know that the second number has to be equal to target - first (since the requirement is first + second == target that means second == target - first). The problem statement also doesn't say what to do if no numbers are found, so you could either return an empty array or some other value, or even throw an error.
const answer = (list, target) => {
while (list.length > 0) { // Loop until the list no longer has any items
let first = list.shift() // Take the first number from the list
let second = target - first // Calculate what the second number should be
if (list.includes(second)) { // Check to see if the second number is in the remaining list
return [first, second] // If it is, we're done -- return them
}
}
return "No valid numbers found" // We made it through the entire list without finding a match
}
console.log(answer([1,2,3], 3))
console.log(answer([1,2,3], 4))
console.log(answer([1,2,3], 7))

You can also add all the values in the array to find the total, and subtract the total by the target to find the value you need to remove from the array. That will then give you an array with values that add up to the total.
let arr1 = [1, 3, 5]
const target = 6
const example = (arr, target) => {
let total = arr.reduce((num1, num2) => {
return num1 + num2
})
total = total - target
const index = arr.indexOf(total)
if (index > -1) {
return arr.filter(item => item !== total)
}
}
console.log(example(arr1, target))

Map and filter are nice functions to have if you know that you need to loop into the whole array. In your case this is not necessary.
So you know you need to find two numbers, let's say X,Y, which belong to an array A and once added will give you the target number T.
Since it's an exercise, I don't want to give you the working code, but here is a few hints:
If you know X, Y must be T - X. So you need to verify that T - X exists in your array.
array.indexOf() give you the position of an element in an array, otherwise -1
If X and Y are the same number, you need to ensure that their index are not the same, otherwise you'll return X twice
Returning the solution should be simple as return [X,Y]
So this can be simplified with a for (let i = 0; i < arr.length; i++) loop and a if statement with a return inside if the solution exist. This way, if a solution is found, the function won't loop further.
After that loop, you return [] because no solution were found.
EDIT:
Since you want a solution with map and filter:
findTwoPartsOfTheNumber = (arr, tNumber) => {
let solution = [];
arr.map((X, indexOfX) => {
const results = arr.filter((Y, indexOfY) => {
const add = Y + X
if (tNumber === add && indexOfX != indexOfY) return true;
else return false;
});
if (results > 0) solution = [X, results[0]];
})
return solution;
}

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