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Maintain the duplicates in the sum of array integers

I wrote a program to:
Print the new array of elements
Print the sum of all elements (or integers)
Actually, I got it right, however, the little problem is, I want to maintain all the duplicates (still within the range of four largest elements). Here's what I mean:
Take an array of numbers: [4,5,-2,3,1,2,6,6]
The four largest numbers are 4,5,6,6. And their sum is 4+5+6+6=21
What the code is doing (not good):
Instead of getting "6,6,5,4" as (described above), the code is printing "6,5,4,3" with the sum as 18.
ALSO, when there are only four elements [with or without duplicates] as in [1,1,1,-5],
let it just add ALL elements. You guessed it, the sum of all elements is -2
How do I order the program to print the necessary duplicate(s) to make the four largest integers?
Here's my code...
var arr = new Array(4,5,-2,3,1,2,6,6);
// var arr = new Array(1,1,1,-5);
// var largArr = new Array();
function largest() {
largArr = Array(0, 0, 0, 0)
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[0]) {
largArr[0] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[1] && arr[i] < largArr[0]) {
largArr[1] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[2] && arr[i] < largArr[1]) {
largArr[2] = arr[i];
}
}
for (i = 0; i < arr.length; i++) {
if (arr[i] > largArr[3] && arr[i] < largArr[2]) {
largArr[3] = arr[i];
}
}
console.log(largArr[0], largArr[1], largArr[2], largArr[3]);
console.log(largArr[0] + largArr[1] + largArr[2] + largArr[3]);
}
largest();
I believe there is a genius out there who can help me solve this :)

You could get the top four and filter the original array.
function get4Largest(array) {
const top4 = [...array].sort((a, b) => b - a).slice(0, 4);
return array.filter(v => {
if (top4.includes(v)) {
top4.splice(top4.indexOf(v), 1);
return true;
}
});
}
console.log(get4Largest([4, 5, -2, 3, 1, 2, 6, 6]));
A different approach by keeping indices.
function get4Largest(array) {
return array
.map((v, i) => [v, i])
.sort(([a], [b]) => b - a)
.slice(0, 4)
.sort(([, a], [, b]) => a - b)
.map(([v]) => v);
}
console.log(get4Largest([4, 5, -2, 3, 1, 2, 6, 6]));

If you want sum of largest four numbers then you can easily do with sort, slice and reduce:
numbers.sort((a, b) => b - a).slice(0, 4).reduce((acc, curr) => acc + curr, 0)
const numbers = [4, 5, -2, 3, 1, 2, 6, 6];
const result = numbers
.sort((a, b) => b - a)
.slice(0, 4)
.reduce((acc, curr) => acc + curr, 0);
console.log(result);

You can use a reduce, sort and slice methods of an array like so:
function sumMaxValues (maxLength, values) {
return values
.sort((v1, v2) => v1 > v2 ? -1 : 1)
.slice(0, maxLength)
.reduce((sum, v) => sum + v, 0)
}
console.log(
sumMaxValues(4, [4, 5, -2, 3, 1, 2, 6, 6, 10]),
)
Edit: I fixed, a bug that #gog pointed out. The root cause of a problem was that sort when invoked without a compareFn then "the array elements are converted to strings, then sorted according to each character's Unicode code point value."(sort docs)

If for some reason you want to have a classical type of solution, that avoids modern javascript methods, here's one
const arr = Array(4, 5, -2, 3, 1, 2, 6, 6);
//const arr = Array(1, 1, 1, -5);
function largest(){
const largArr = Array(-1/0, -1/0, -1/0, -1/0);
for(let i = 0; i < arr.length; i++){
for(let j = 0; j < largArr.length; j++){
if(arr[i] > largArr[j]){
for(let k = largArr.length - 2; k >= j; k--){
largArr[k + 1] = largArr[k];
}
largArr[j] = arr[i];
break;
}
}
}
let sum = 0;
for(let j = 0; j < largArr.length; j++){
if(largArr[j] === -1/0){
largArr[j] = 0;
}
sum += largArr[j];
}
console.log(largArr, sum);
}
largest();
-1/0 stands for minus infinity (there can be no smaller number); you may also use Number.NEGATIVE_INFINITY for it. If it's too exotic for your needs, replace -1/0 with any number you are certain is less than any possible number in the array (that however, cannot be zero, since you allow negative numbers also).

Can someone explain me how does this javascript code works please?

Given a JavaScript function that takes in an array of numbers as the first and the only argument.
The function then removes one element from the array, upon removal, the sum of elements at odd indices is equal to the sum of elements at even indices. The function should count all the possible unique ways in which we can remove one element at a time to achieve balance between odd sum and even sum.
Example var arr = [2, 6, 4, 2];
Then the output should be 2 because, there are two elements 6 and 2 at indices 1 and 3 respectively that makes the combinations table.
When we remove 6 from the array
[2, 4, 2] the sum at odd indexes = sum at even indexes = 4
if we remove 2
[2, 6, 4] the sum at odd indices = sum at even indices = 6
The code below works perfectly. There might be other solutions but I want to understand this one, because I feel there is a concept I have to learn here. Can someone explain the logic of this algorithm please?
const arr = [2, 6, 4, 2];
const check = (arr = []) => {
var oddTotal = 0;
var evenTotal = 0;
var result = 0;
const arraySum = []
for (var i = 0; i < arr.length; ++i) {
if (i % 2 === 0) {
evenTotal += arr[i];
arraySum[i] = evenTotal
}
else {
oddTotal += arr[i];
arraySum[i] = oddTotal
}
}
for (var i = 0; i < arr.length; ++i) {
if (i % 2 === 0) {
if (arraySum[i]*2 - arr[i] + oddTotal === (arraySum[i - 1] || 0)*2 + evenTotal) {
result = result +1
};
} else if (arraySum[i]*2 - arr[i] + evenTotal === (arraySum[i - 1] || 0)*2 + oddTotal) {
result = result +1
}
}
return result;
};

LeetCode 768: Constructing a variation on a solution for Max Chunks To Make Sorted

LeetCode's Max Chunks To Make Sorted II challenge is:
Given an array arr of integers (not necessarily distinct), we split
the array into some number of "chunks" (partitions), and individually
sort each chunk. After concatenating them, the result equals the
sorted array.
What is the most number of chunks we could have made?
Example:
Input: arr = [2, 1, 3, 4, 4]
Output: 4
Explanation:
We can split into two chunks, such as [2, 1], [3, 4, 4].
However, splitting into [2, 1], [3], [4], [4] is the highest number of chunks possible.
The algorithm underlying the following solution is (the algorithm and the solution were posted as a comment on the solution page by a user named #benevolent. Unfortunately, I can't link to its comment):
If the largest number from arr[0] to (including) arr[k] is less than or equal to the smallest
number from arr[k+1] to the end, then we can split into two valid
chunks.
To illustrate:
left right
[......max] [min......]
To know the minimum element from k to arr.length-1, we can just
precompute from right to left.
The solution:
function maxChunksToSorted(arr) {
var minRight = Array(arr.length).fill(Number.MAX_SAFE_INTEGER);
for (var i = arr.length-2; i >= 0; --i) {
minRight[i] = Math.min(minRight[i+1], arr[i+1]);
}
var maxLeft = Number.MIN_SAFE_INTEGER;
var ans = 0;
for (var i = 0; i < arr.length; ++i) {
maxLeft = Math.max(maxLeft, arr[i]);
if (maxLeft <= minRight[i]) {
ans += 1
}
}
return ans;
};
console.log("expects: 1", "got:", maxChunksToSorted([5, 4, 3, 2, 1]));
console.log("expects: 4", "got:", maxChunksToSorted([2, 1, 3, 4, 4]));
My question:
I was trying to make a "mirror image" of the above solution, by "flipping" every action (e.g, the use of min becomes max, <= becomes >, and so on).
My maxArr indeed mirrors minRight (e.g., for [2, 1, 3, 4, 4], my maxArr is [MIN_SAFE_INTEGER, 1, 3, 4, 4], while the original minRight is [1, 3, 4, 4, MAX_SAFE_INTEGER]), but it clearly doesn't work, and I can't put my finger on the reason for that.
What's my fundamental problem?
Let me stress that I'm not looking for some other working solution. I'd like to understand what went wrong with my mirror solution, if it's even possible to make this mirror, and if not - what's the fundamental reason for that.
function maxChunksToSorted(arr) {
var maxArr = Array(arr.length).fill(Number.MIN_SAFE_INTEGER);
for (var i = 1; i <= arr.length; ++i) {
maxArr[i] = Math.max(maxArr[i-1], arr[i]);
}
var minLeft = Number.MAX_SAFE_INTEGER;
var ans = 0;
for (var i = 0; i < arr.length; ++i) {
minLeft = Math.min(minLeft, arr[i]);
if (minLeft > maxArr[i]) {
ans += 1
}
}
return ans;
};
console.log("expects: 1", "got:", maxChunksToSorted([5, 4, 3, 2, 1]));
console.log("expects: 4", "got:", maxChunksToSorted([2, 1, 3, 4, 4]));

This should do the job:
function chunk(list){
let sortedList = list.slice();
sortedList.sort();
var beginIndex = -1; var biggestFound;
var foundFirst = false; var foundLast = false;
for(var i = 0; i < list.length; i++){
if(beginIndex == -1) {
if(list[i] == sortedList[i]) print(list[i]);
else {beginIndex = i; biggestFound = list[i];}
}
else{
if(list[i] == sortedList[beginIndex]) foundFirst = true;
if(list[i] > biggestFound) biggestFound = list[i];
if(biggestFound == sortedList[i]) foundLast = true;
if(foundFirst && foundLast){
print(list.slice(beginIndex, i - beginIndex + 1));
foundFirst = false; foundLast = false; beginIndex = -1;
}
}
}
}
chunk([2,1,3,4,4]);
As I commented, if a chunk starts at position i, it must contain the element that corresponds to the position i in the sorted array and if it ends in position j, it must contain the element in the index j of the sorted array.
When both of these conditions are satisfied, you close the chunk and start a new one.
The complexity is O(n lg(n)), where n is the size of the array.

finding all missing elements in an array/range javascript

I am trying to to write a function to find all missing elements in an array. The series goes from 1...n. the input is an unsorted array and the output is the missing numbers.
below is what I have so far:
function findMissingElements(arr) {
arr = arr.sort();
var missing = [];
if (arr[0] !== 1) {
missing.unshift(1);
}
// Find the missing array items
for (var i = 0; i < arr.length; i++) {
if ((arr[i + 1] - arr[i]) > 1) {
missing.push(arr[i + 1] - 1);
}
}
return missing;
}
var numbers = [1, 3, 4, 5, 7, 8]; // Missing 2,6
var numbers2 = [5, 2, 3]; //missing 1, 4
var numbers3 = [1, 3, 4, 5, 7]; // Missing 2,6
console.log(findMissingElements(numbers)); // returns 2,6 correct
console.log(findMissingElements(numbers2)); // returns 1,4
console.log(findMissingElements(numbers3)); // returns 2, 6
I "manually" checked for the first element with an if block, is there any way to handle the case of the first element inside the for loop?

You can produce that by tracking which number should appear next and adding it to a list of missing numbers while it is less than the next number.
function findMissingElements(arr) {
// Make sure the numbers are in order
arr = arr.slice(0).sort(function(a, b) { return a - b; });
let next = 1; // The next number in the sequence
let missing = [];
for (let i = 0; i < arr.length; i++) {
// While the expected element is less than
// the current element
while (next < arr[i]) {
// Add it to the missing list and
// increment to the next expected number
missing.push(next);
next++;
}
next++;
}
return missing;
}

A not so efficient but more intuitive solution:
var n = Math.max.apply(null, arr); // get the maximum
var result = [];
for (var i=1 ; i<n ; i++) {
if (arr.indexOf(i) < 0) result.push(i)
}

Aside from the fact that your tests are not consistent, this feels a bit neater to me:
function findMissingElements (arr, fromFirstElement) {
arr.sort();
var missing = [];
var next = fromFirstElement ? arr[0] : 1;
while(arr.length) {
var n = arr.shift();
while (n != next) {
missing.push(next++);
}
next++;
}
return missing;
}
var numbers = [1, 3, 4, 5, 7, 8]; // Missing 2,6
var numbers2 = [5, 2, 3]; // Missing 1, 4
var numbers3 = [1, 3, 4, 5, 7]; // Missing 2, 6
console.log(findMissingElements(numbers)); // returns 2, 6
console.log(findMissingElements(numbers2)); // returns 1, 4
console.log(findMissingElements(numbers3)); // returns 2, 6
I've added an argument fromFirstElement which, if passed true, will enable you to start from a number defined by the first element in the array you pass.

Find the smallest sum of all indexes of unique number pairs summing to a given number

I want to loop through an array and then add each value to each other (except itself + itself) and if the sum of the two values that were looped through equals the second argument in my function, and the pair of values hasn't been encountered before, then remember their indices and, at the end, return the full sum of all remembered indices.
In other words, the problem statement is: given an array A of integers and a second value s that is a desired sum, find all pairs of values from array A at indexes i, j such that i < j and A[i] + A[j] = s, and return the sum of all indexes of these pairs, with the following restriction:
don't reuse value pairs, i.e. if two index pairs i, j and k, l satisfying the above conditions are found and if A[i] == A[k] and A[j] == A[l] or A[i] == A[l] and A[j] == A[k], then ignore the pair with the higher index sum.
Example
For example, functionName([1, 4, 2, 3, 0, 5], 7) should return 11 because values 4, 2, 3 and 5 can be paired with each other to equal 7 and the 11 comes from adding the indices of them to get to 11 where:
4 + 3 = 7
5 + 2 = 7
4 [index: 1]
2 [index: 2]
3 [index: 3]
5 [index: 5]
1 + 2 + 3 + 5 = 11
Example #2
functionName([1, 3, 2, 4], 4) would only equal 1, because only the first two elements can be paired to equal 4, and the first element has an index of 0 and the second 1
1 + 3 = 4
1 [index: 0]
3 [index: 1]
0 + 1 = 1
This is what I have so far:
function functionName(arr, arg) {
var newArr = [];
for(var i = 0; i < arr.length; i++){
for(var j = i + 1; j < arr.length; j++) {
if((arr[i] + arr[j]) === arg ) {
newArr.push(i , j);
}
}
}
if(newArr.length === 0) {
return console.log(0);
}
return console.log(newArr.reduce(function(a,b){return a + b}));
}
functionName([1, 4, 2, 3, 0, 5], 7);
The problem I have is that it all works but I have the issue that once it finds a pair that equals the second argument, then it's not supposed to use the same value pairs again but mine does, e.g.:
if the array is [1,1,1] and the second argument is 2, the loop will go through and find the answer but it continues to search after it finds the sum and I only want it to use the pair [1, 1] once, so if it finds a pair like this at indexes [0, 1] then it should not include any other pair that contains the value 1.
I was thinking that i could remove the rest of the values that are the same if more than 2 are found using filter leaving me with only 2 of the same value if there is in an array thus not having to worry about the loop finding a 1 + 1 twice but is this the best way to go about doing it?
I'm still new to this but looking forward to your comments
PS I'm planning on doing this using pure JavaScript and no libraries
Link to a JS fiddle that might make things easier to see what I have.
https://jsfiddle.net/ToreanJoel/xmumv3qt/

This is more complicated than it initially looks. In fact, making a loop inside a loop causes the algorithm to have quadratic time complexity with regard to the size of the array. In other words, for large arrays of numbers, it will take a very long time to complete.
Another way to handle this problem is to notice that you actually have to use each unique value in the array only once (or twice, if s is even and you have two s/2 values somewhere in the array). Otherwise, you would have non-unique pairs. This works because if you need pairs of numbers x and y such that x + y = s, if you know x, then y is determined -- it must be equal s - x.
So you can actually solve the problem in linear time complexity (to be fair, it's sometimes n*log(n) if all values in A are unique, because we have to sort them once).
The steps of the algorithm are as follows:
Make a map whose keys are values in array A, and values are sorted lists of indexes these values appear at in A.
Move through all unique values in A (you collected them when you solved step 1) in ascending order. For each such value:
Assume it's the lower value of the searched pair of values.
Calculate the higher value (it's equal to s - lower)
Check if the higher value also existed in A (you're doing it in constant time thanks to the map created in step 1).
If it does, add the lowest indexes of both the lower and the higher value to the result.
Return the result.
Here's the full code:
function findSumOfUniquePairs(numbers, sum) {
// First, make a map from values to lists of indexes with this value:
var indexesByValue = {},
values = [];
numbers.forEach(function (value, index) {
var indexes = indexesByValue[value];
if (!indexes) {
indexes = indexesByValue[value] = [];
values.push(value);
}
indexes.push(index);
});
values.sort();
var result = 0;
for (var i = 0, maxI = values.length; i < maxI; ++i) {
var lowerValue = values[i],
higherValue = sum - lowerValue;
if (lowerValue > higherValue) {
// We don't have to check symmetrical situations, so let's quit early:
break;
}
var lowerValueIndexes = indexesByValue[lowerValue];
if (lowerValue === higherValue) {
if (lowerValueIndexes.length >= 2) {
result += lowerValueIndexes[0] + lowerValueIndexes[1];
}
} else {
var higherValueIndexes = indexesByValue[higherValue];
if (higherValueIndexes) {
result += lowerValueIndexes[0] + higherValueIndexes[0];
}
}
}
return result;
}
document.write(findSumOfUniquePairs([1, 4, 2, 3, 0, 5], 7) + '<br>'); // 11;
document.write(findSumOfUniquePairs([1, 3, 2, 4], 4) + '<br>'); // 1
document.write(findSumOfUniquePairs([1, 1, 1], 2) + '<br>'); // 1
document.write(findSumOfUniquePairs([1, 1, 1, 1], 2) + '<br>'); // 1
document.write(findSumOfUniquePairs([1, 2, 3, 1, 2, 3, 1], 4) + '<br>'); // 7
document.write(findSumOfUniquePairs([5, 5, 1, 1, 1], 6) + '<br>'); // 2
document.write(findSumOfUniquePairs([0, 5, 0, 5, 1, 1, 1], 6) + '<br>'); // 5

This works, but it mucks up the initial array.
function functionName(arr, arg) {
var newArr = [];
for(var i = 0; i < arr.length; i++){
for(var j = i + 1; j < arr.length; j++) {
if((arr[i] + arr[j]) === arg ) {
newArr.push(i , j);
arr[i] = null;
arr[j] = null;
}
}
}
if(newArr.length === 0) {
return console.log(0);
}
return console.log(newArr.reduce(function(a,b){return a + b}));
}

Solution with loops with restart, if a sum is found. the found summands are stored in usedNumbers and later sorted and used to get the index for summing the index.
The sorting and the last index provides the correct start position for the Array.prototype.indexOf.
Edit:
what about [1,1,1,1], 2 ... should that be 6 or 1? – Jaromanda X 21
#JaromandaX that should be 1, after the pair is found with the values then it shouldn't look for a pair with the same values again – Torean
This version takes care of the requirement.
function f(array, sum) {
var arrayCopy = array.slice(0),
usedNumbers = [],
index = 0,
indexA = 0,
indexB,
a, b;
while (indexA < arrayCopy.length) {
indexB = indexA + 1;
while (indexB < arrayCopy.length) {
a = arrayCopy[indexA];
b = arrayCopy[indexB];
if (a + b === sum) {
usedNumbers.push(a, b);
arrayCopy = arrayCopy.filter(function (i) { return a !== i && b !== i; });
indexA--; // correction to keep the index
break;
}
indexB++;
}
indexA++;
}
return usedNumbers.sort().reduce(function (r, a, i) {
index = array.indexOf(a, i === 0 || a !== usedNumbers[i - 1] ? 0 : index + 1);
return r + index;
}, 0);
}
document.write(f([1, 4, 2, 3, 0, 5], 7) + '<br>'); // 11
document.write(f([1, 1, 1], 2) + '<br>'); // 1
document.write(f([5, 5, 1, 1, 1], 6) + '<br>'); // 2
document.write(f([0, 5, 0, 5, 1, 1, 1], 6) + '<br>'); // 5
document.write(f([1, 1, 1, 1], 2) + '<br>'); // 1

The solution below is very compact. It avoids unnecessary checks and loops only through the relevant elements. You can check the working codepen here:
http://codepen.io/PiotrBerebecki/pen/RRGaBZ.
function pairwise(arr, arg) {
var sum = 0;
for (var i=0; i<arr.length-1; i++) {
for (var j=i+1; j<arr.length; j++) {
if (arr[i] <= arg && arr[j] <= arg && arr[i] + arr[j] == arg) {
sum += i+j;
arr[i] = arr[j] = NaN;
}
}
}
return sum;
}
console.log( pairwise([1, 1, 0, 2], 2) ) // should return 6
Under the hood:
Start looping from the element with index (i) = 0.
Add a second loop only for the elements which are later in the array. Their index j is always higher than i as we are adding 1 to i.
If both elements (numbers) are less than or equal to to the arg, check if their sum equals to the arg. This avoids checking the sum if either of the numbers are greater than the arg.
If the pair has been found then change their values to NaN to avoid further checks and duplication.

This solution should have a time complexity of 0(n) or linear
Much faster than two nested for-loops. This function will give you the two indices that add up to the target number. It can easily be modified to solve any other configuration of this problem.
var twoSum = function(nums, target) {
const hash = {}
for(let i = 0; i < nums.length; i++) {
hash[nums[i]] = i
}
for(let j = 0; j < nums.length; j++) {
let numToFind = target - nums[j]
if(numToFind in hash && hash[numToFind] !== j) {
return [hash[numToFind], j]
}
}
return false
};
console.log(twoSum([1,2,3,5,7], 5))
In Python:
def twoSum(self, nums: List[int], target: int) -> List[int]:
myMap = {}
for i in range(len(nums)):
myMap[nums[i]] = i
for j in range(len(nums)):
numToFind = target - nums[j]
if numToFind in myMap and myMap[numToFind] != j:
return [myMap[numToFind], j]
print(twoSum([1,2,3,5,7], 5))
In Java:
import java.util.*;
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for(Integer i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for(Integer j = 0; j < nums.length; j++) {
Integer numToFind = target - nums[j];
Integer myInt = map.get(numToFind);
if(map.containsKey(numToFind) && myInt != j) {
return new int[] {myInt , j};
}
}
return new int[] {0, 0};
}
}
System.out.println(twoSum([1,2,3,5,7], 5))