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Find max In array where items may be numbers or arrays

I want to find the maximum value in an array where each value may be either a positive number or another array of positive numbers. I have this, but I'm not sure how to deal with the nested arrays.
var array = [5, 8, 15, [10, 2, [6, 20, 1], 4, 13, [2, 11]]];
function findMax(arr) {
var max = 0;
for (var i = 0; i < arr.length; i++) {
// what if it's an array?
if (arr[i] > max) {
max= arr[i];
}
}
return max;
}
console.log(findLargest(array));
// answer should be 20

function findMax(arr) {
return arr.reduce((prev, curr) => Math.max(prev, Array.isArray(curr) ? findMax(curr) : curr), 0);
}
console.log(findMax(array));

Just flatten it first?
var array = [5, 8, 15, [10, 2, [6, 20, 1], 4, 13, [2, 11]]];
const findMax = arr => Math.max(...array.flat(Infinity));
console.log(findMax(array));

removing duplicate ids from array of arrays from a specific index

I want to remove duplicate ids from pureIDs[2]. so the index here is 2. the only condition is the arrays I want to search for duplicates are the ones with lower indices so I want to search 0 and 1.
If we want to pureIDs[3] then we should search 0 , 1 , 2 (the lower indices)
My attempt is a total failure Here:
The desired result is : [22, 9] here.
const pureIDs = [
[0, 1, 23, 5, 11],
[2, 15, 23, 25, 10],
[2, 10, 22, 9, 11], // Modify this index (2)
[20, 24],
];
// I give the index here
const ids = getModifiedIDs(pureIDs[2], 2);
console.log(ids); // This should result in : [22, 9]
function getModifiedIDs(ids, from) {
let sets = [];
for(let b = from - 1; b > -1; b--){
console.log(b)
set = ids.filter((el) => {
return !pureIDs[b].includes(el);
});
sets.push(set)
}
return sets;
}

Just a slight modification gives you the result needed.
Change your function to:
function getModifiedIDs(ids, from) {
const set = ids.filter((el) => {
for(let b = from - 1; b > -1; b--){
if (pureIDs[b].includes(el)) {
return false;
}
}
return true;
});
return set;
}

Build a set from the arrays with deletable indexes. Than you get with Array#filter and Set#has your result.
function getModifiedIDs(ids, from) {
let dels = [];
for (i=0; i<from-1; i++)
dels = dels.concat(pureIDs[i]);
let delSet = new Set(dels);
return ids.filter(id => !delSet.has(id) );
}
const pureIDs = [
[0, 1, 23, 5, 11],
[2, 15, 23, 25, 10],
[2, 10, 22, 9, 11], // Modify this index (2)
[20, 24],
];
console.log(getModifiedIDs(pureIDs[2], 3));

Delete numbers below a certain number in an array

Today i'm facing a really weird problem.
I'm actually trying to delete numbers below a certain number in an array.
I have this array [1, 7, 2, 3, 90, 4, 70, 20] and I only want the numbers greater than 20 !
So I should have that in output : [90, 70]
But Instead I have this array : [7, 3, 90, 70, 20] ???
Here is my code :
function superior() {
var arr = [1, 7, 2, 3, 90, 4, 70, 20]
for (var i = 0; i < arr.length; i++) {
if (arr[i] < 20) {
arr.splice(arr.indexOf(arr[i]), 1);
} else {
break;
}
}
return arr;
}
console.log(superior());

Mutating an array while you're looping through it is always going to be tricky. If you shrink the size of the array, but don't change i, then you're can end up interacting with the wrong element of the array the next time through the loop. Also, using break will stop the loop from running entirely, and so skip the rest of the array. You may have meant to use continue, but since you're on the last line of the loop anyway, continue isn't needed.
Rather than trying to change the array and loop through it at the same time, i'd recommend creating a new array. Then you can loop through the old one unhindered:
const arr = [1, 7, 2, 3, 90, 4, 70, 20]
const newArr = []
for (const i = 0; i < arr.length; i++) {
if (arr[i] >= 20) {
newArr.push(arr[i]);
}
}
console.log(newArr)
Filtering an array like this is a very common thing to do, so there are built in tools to do it for you. Every array has a .filter method. You pass into it a function describing what elements you want to keep, and it will produce the new array for you:
const arr = [1, 7, 2, 3, 90, 4, 70, 20]
const newArr = arr.filter(element => element >= 20);
console.log(newArr)

You can filter them out according to a condition.
And replace your existing array with the filtered one. Or if you don't want to replace it, use another variable and assign the filtered value to that variable.
var newArray = array.filter(item => item > 20)
Check .filter()
var array = [1, 7, 2, 3, 90, 4, 70, 20];
array = array.filter(item => item > 20)
console.log(array)

You can use Array.filter()
var arr = [1, 7, 2, 3, 90, 4, 70, 20]
var filteredArr = arr.filter(item => item > 20);
console.log(filteredArr);

You could iterate from the end and omit indexOf, because you have already the index i.
This appoach loops from the end and after splicing, the remaining lower indices remains.
function superior() {
var array = [1, 7, 2, 3, 90, 4, 70, 20],
i = array.length;
while (i--) if (array[i] < 20) array.splice(i, 1);
return array;
}
console.log(superior());

Use temporary array to push new values.
function superior() {
var arr = [1, 7, 2, 3, 90, 4, 70, 20];
temp_arr = [];
for (var i = 0; i < arr.length; i++) {
if (arr[i] > 20) {
temp_arr.push(arr[i]);
}
}
return temp_arr;
}
console.log(superior());

Javascript Nested Array Return Even Numbers

I am trying to write a function that prints only the even numbers from a nested array. Here is my attempt and it keeps returning "undefined".
function printEvents(){
var nestedArr = [[1,2,3],[4,5,6],[7,8],[9,10,11,12]];
for (var i = 0; i<nestedArr.length; i++) {
for (var j = 0; j<nestedArr[i]; j++) {
var evenNumbers = nestedArr[i][j]
}
}
if (evenNumbers % 2 == 0) {
console.log(evenNumbers)
}
}
printEvents();

You could use a recursive approach, if the item is an array. You need to move the test for evenness inside of the for loop.
function printEvents(array) {
var i;
for (i = 0; i < array.length; i++) {
if (Array.isArray(array[i])) {
printEvents(array[i]);
continue;
}
if (array[i] % 2 == 0) {
console.log(array[i]);
}
}
}
printEvents([[1, 2, 3], [4, 5, 6], [7, 8], [9, 10, 11, 12], [[[13, [14]]]]]);
Solution with a callback
function getEven(a) {
if (Array.isArray(a)) {
a.forEach(getEven);
return;
}
if (a % 2 == 0) {
console.log(a);
}
}
getEven([[1, 2, 3], [4, 5, 6], [7, 8], [9, 10, 11, 12], [[[13, [14]]]]]);

You just have a couple issues. You need to check the length of your nested arrays and you need to move your code that checks whether a number is even or not inside the array.
var nestedArr = [[1,2,3],[4,5,6],[7,8],[9,10,11,12]];
for (var i = 0; i<nestedArr.length; i++) {
for (var j = 0; j<nestedArr[i].length; j++) {
var evenNumbers = nestedArr[i][j]
if (evenNumbers % 2 == 0) {
console.log(evenNumbers)
}
}
}

You can do this more easy using filter method which accepts a callback method.
var nestedArr = [[1,2,3],[4,5,6],[7,8],[9,10,11,12]];
var mergedArray=[].concat.apply([], nestedArr);
console.log(mergedArray.filter(a => a%2 == 0));

You can use reduce with every for something quick.
var nestedArr = [ [1, 2, 3],[4, 5, 6],[7, 8],[9, 10, 11, 12]];
var sift = nestedArr.reduce(function(r,o) {
o.every(i => i % 2 === 0 ? r.push(i) : true)
return r;
}, []);
console.log(sift);
If you want a one-liner you can use ReduceRight with Filter
var nestedArr = [[1, 2, 3],[4, 5, 6],[7, 8],[9, 10, 11, 12]];
var sift = nestedArr.reduceRight((p, b) => p.concat(b).filter(x => x % 2 === 0), []);
console.log(sift)

Comparing Multiple Arrays Using Reduce

pretty new to Javascript and I've tried this question about 4 times now in a span of about a month and I am still unable to solve it.
So here is the question:
Construct a function intersection that compares input arrays and returns a new array with elements found in all of the inputs. BONUS: Use reduce!
The format is:
function intersection(arrays) {
// Your Code Goes Here
}
Test Case: Should log [15, 5]
console.log('Extensions 3 Test: ' + intersection([5, 10, 15, 20], [15, 88, 1, 5, 7]/*, [1, 10, 15, 5, 20]*/));
My current solution: Works for the case of only have two items to compare, but not for the third one, I could make it so that I would loop through and compare the obtained values with the next array but I don't think I am on the right path... Also, I am not using reduce to implement it... And I am not sure if I am supposed to be using 'arguments.' Any help is appreciated! Thank you so much.
function intersection(arrays) {
array = [];
for (var i = 0; i < arguments.length; i++)
array.push(arguments[i]);
var result = [];
for(var i = 0; i < array.length - 1; i++) {
for(var j = 0; j < array[i].length; j++) {
if (array[i+1].includes(array[i][j]))
result.push(array[i][j]);
}
}
return result;
}

Although, as several suggestions said, you could use underscore, lodash, or my personal favorite, Ramda (disclaimer: I'm one of the authors), this function should be straightforward enough that you wouldn't even consider a library for it. Here's a simple version:
const intersection = (xs, ys) => xs.filter(x => ys.indexOf(x) > -1);
intersection([5, 10, 15, 20, 3], [15, 88, 3, 1, 5, 7]); //=> [5, 15, 3]
const intersectAll = (...xss) => xss.reduce(intersection);
intersectAll([5, 10, 15, 20, 3], [15, 88, 3, 1, 5, 7], [1, 10, 15, 5, 20]); //=> [5, 15]
I would think that this is all you need, at least so long as you're worried only about reference/primitive equality and don't need to consider cases where you want to know that {x: 1} and {x: 1} are the same, even though they aren't the same reference. If you do need that, you might look to Ramda's intersection function.
Note that if includes were better supported, I would recommend this version instead, as it reads better:
const intersection = (xs, ys) => xs.filter(x => ys.includes(x));
Also, if you have no need for the binary function, you can make just a variadic version of it by combining the two above:
const intersection = (...xss) => xss.reduce((xs, ys) => xs.filter(x => ys.indexOf(x) > -1));

Maybe someone will finds it useful.
As an argument to the function you can give any number of arrays of any length and the function is compact, I think ;)
const findSimilar = (...arrays) => {
return arrays.reduce((includ, current) =>
Array.from(new Set(includ.filter((a) => current.includes(a))))
);
};
console.log(
findSimilar([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20])
);
And how it works:
Ok, first u take rest parameters(...arrays) as parameter of function, so u have
arrays = [ [5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20] ]
then in first iteration of reduce we have
includ = [5, 10, 15, 20] and current = [15, 88, 1, 5, 7]
on this two we use filter, what give us [5,15], i use Set to make shure there is no repetition and make array back (Array.from()), which is passed to the next iteration of reduce as "includ", at the next iteration we have
incude = [5,15] and current = [1, 10, 15, 5, 20] and so on ...
We can even use it like this
let result = [
[5, 10, 15, 20],
[15, 88, 1, 5, 7],
[1, 10, 15, 5, 20]
].reduce((includ, current) =>
Array.from(new Set(includ.filter((a) => current.includes(a))))
);
console.log(result);

Although not solving your problem directly, you can do what you're trying to do using the opensource library underscore.js.
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
You may be able to derive inspiration from the way that's been implemented. The above is the function call to their own _.intersection function which is also dependent on other underscore.js functions as you see below:
// Produce an array that contains every item shared between all the
// passed-in arrays.
_.intersection = function(array) {
if (array == null) return [];
var result = [];
var argsLength = arguments.length;
for (var i = 0, length = array.length; i < length; i++) {
var item = array[i];
if (_.contains(result, item)) continue;
for (var j = 1; j < argsLength; j++) {
if (!_.contains(arguments[j], item)) break;
}
if (j === argsLength) result.push(item);
}
return result;
};

Here is a solution using reduce, with the empty array passed in as intersection as the initial value.
Iterate the numbers and check if each one appears in one of the subarrays.
If it doesn't, set the Boolean isPresentInAll to false.
If it does appear in all three and it's not already present in the
intersection array, then push to the intersection array.
function intersection(arrayOfArrays) {
return arrayOfArrays.reduce(function(intersection, subArray) {
subArray.forEach(function(number) {
var isPresentInAll = true;
for (var i = 0; i < arrayOfArrays.length; i++) {
if (arrayOfArrays[i].indexOf(number) === -1) {
isPresentInAll = false;
}
}
if (isPresentInAll === true && intersection.indexOf(number) === -1) {
intersection.push(number);
}
});
return intersection;
}, []);
}

I think i got the right function for you.
(Note: results are not sorted!)
var intersection = function() {
// merge deduped arrays from arguments
var arrays = Array.prototype.reduce.call(arguments, function(carry, array) {
return [].concat(carry, array.filter(function(item, index, origin) {
return origin.indexOf(item) === index;
}));
}, []);
var results = arrays.reduce(function(carry, item, index, arr) {
if(
// just select items, which have more then 1 occurance
arr.filter(function(fItem) {
return fItem === item;
}).length > 1 &&
// ... and which are not already in results
!~carry.indexOf(item)
) {
carry = [].concat(carry,item);
}
return carry;
}, []);
return results;
};

Here's a version that uses 2 reduces.
The first iterates the arrays only once to create a hashmap object to track instance counts, the second to return values where counts match number of arguments
function intersection(){
// convert arguments to array of arrays
var arrays = [].slice.call(arguments);
// create an object that tracks counts of instances and is type specific
// so numbers and strings would not be counted as same
var counts= arrays.reduce(function(a,c){
// iterate sub array and count element instances
c.forEach(function(val){
var propName = typeof val + '|' + val;
// if array value not previously encountered add a new property
a[propName] = a[propName] || {count:0, value: val};
// increment count for that property
a[propName].count++;
});
return a;
},{});
// iterate above object to return array of values where count matches total arrays length
return Object.keys(counts).reduce(function(resArr, propName){
if(counts[propName].count === arrays.length){
resArr.push(counts[propName].value);
}
return resArr;
},[]);
}
console.log(intersection([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]))
Could use some fine tuning to make sure there are enough arguments and that they are all arrays

Here's what I came up with using vanilla javascript and one call to reduce.
function intersection(){
var arrays = [].slice.call(arguments);
var first = arrays[0];
var rest = arrays.slice(1);
return first.reduce(function(all, item, index){
var push = rest.every(function(subArray){
return subArray.indexOf(item) > -1;
});
if(push){
all.push(item);
}
return all;
},[])
}
console.log(intersection([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]));

function intersection(arrays) {
let common = arrays.reduce(function(accumulator, currentValue) {
return accumulator.filter(function(x){
return currentValue.indexOf(x) > -1;
})
})
return common;
}

To optimize your answer that couldn't work on more than 2 subarrays and didn't use reduce, here's the code that works for however many subarrays you pass in.
function intersection(arr1, arr2, arr3){
let ans = arr1[0]; // ans = [5,10,15,20]
for(let i = 0; i < ans.length; i++){ // i = 0...3
for(let j = 1; j < arr1.length; j++){ // j = 1...2
if(!(arr1[j].includes(ans[i]))){ // if the new subarray doesn't include an element in the ans
ans.splice(i, 1); // delete the element from ans
}
}
}
return ans;
}
const arr1 = [5, 10, 15, 20];
const arr2 = [15, 88, 1, 5, 7];
const arr3 = [1, 10, 15, 5, 20];
console.log(intersection([arr1, arr2, arr3])); // should log: [5, 15]