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How to solve codility minMaxDivision

I have been trying to wrap my head around this codility question for 1H30,and how to solve with binary search. I found the answer but I cant understand the logic behind it. Can someone who gets it kindly walk me through this answer.
This is the question
Task description
You are given integers K, M and a non-empty zero-indexed array A
consisting of N integers. Every element of the array is not greater
than M.
You should divide this array into K blocks of consecutive elements.
The size of the block is any integer between 0 and N. Every element of
the array should belong to some block.
The sum of the block from X to Y equals A[X] + A[X + 1] + ... + A[Y].
The sum of empty block equals 0.
The large sum is the maximal sum of any block.
For example, you are given integers K = 3, M = 5 and array A such
that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2
A[6] = 2
The array can be divided, for example, into the following blocks:
[2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15; [2], [1, 5, 1,
2], [2, 2] with a large sum of 9; [2, 1, 5], [], [1, 2, 2, 2] with a
large sum of 8; [2, 1], [5, 1], [2, 2, 2] with a large sum of 6.
The goal is to minimize the large sum. In the above example, 6 is the
minimal large sum.
Write a function:
function solution(K, M, A);
that, given integers K, M and a non-empty zero-indexed array A
consisting of N integers, returns the minimal large sum.
For example, given K = 3, M = 5 and array A such that:
A[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2
A[6] = 2
the function should return 6, as explained above.
Assume that:
N and K are integers within the range [1..100,000];
M is an integer within the range [0..10,000];
each element of array A is an integer within the range [0..M].
This is the answer I could get my hands on
function solution(K, M, A) {
var begin = A.reduce((a, v) => (a + v), 0)
begin = parseInt((begin+K-1)/K, 10);
var maxA = Math.max(A);
if (maxA > begin) begin = maxA;
var end = begin + M + 1;
var res = 0;
while(begin <= end) {
var mid = (begin + end) / 2;
var sum = 0;
var block = 1;
for (var ind in A) {
var a = A[ind];
sum += a;
if (sum > mid) {
++block;
if (block > K) break;
sum = a;
}
}
if (block > K) {
begin = mid + 1;
} else {
res = mid;
end = mid - 1;
}
}
return res;
}

I would like to give the more detailed explanation of the algorithm that I have implemented and then one correct implementation in C++.
Find the maximum element in the input array. We could also use M, but M does not necessarily occur. A smaller number could be the maximum, so it is slight optimisation.
Calculate the sum of the input array. This would be the maximum largest sum.
Apply binary search, where the start is the maximum element and the end is the sum. The minimum largest sum would be in this range.
For each trial, check whether we can squeeze the elements into fewer blocks than the block number requested. If it is fewer, it is okay because we can use empty blocks. If it is equal, that is also acceptable. However, it is greater, then we can conclude that the tried minimum largest sum needs to be higher to allow individual blocks to be larger to reduce the block count.
One general principle can be observed above that the more fairly we distribute the sums of the blocks, the largest will become the minimum possible. For this, we need to squeeze as many elements into an individual block as possible.
If the number of blocks for a tried minimum largest sum is smaller than the expected block count, then we can try a slightly smaller minimum largest sum, otherwise we have to try a bit greater until we eventually find the best number.
As far as runtime complexity goes, the solution is O(n * log(N * M)) because the binary search is logarithmic. The sum can be N number of times the maximum element M at worst, which results in an N * M range to bisect with binary search. The inner iteration will go through all the elements, so that is N times. Therefore, it is O(N * log(N * M)) which is equivalent to O(N * log(N + M).
int check(vector<int>& A, int largest_sum)
{
int sum = 0;
int count = 0;
for (size_t i = 0; i < A.size(); ++i) {
const int e = A[i];
if ((sum + e) > largest_sum) { sum = 0; ++count; }
sum += e;
}
return count;
}
int solution(int K, int /*M*/, vector<int> &A)
{
int start = *max_element(A.begin(), A.end());
int end = accumulate(A.begin(), A.end(), 0);
while (start <= end) {
int mid = (start + end) / 2;
if (check(A, mid) < K) end = mid - 1;
else start = mid + 1;
}
return start;
}

This is a binary search on the solution. For each candidate solution, we iterate over the whole array once, filling array blocks to the maximum sum the block can be before exceeding the candidate. If the sum is not achievable, there is no point in trying a smaller sum so we search the space of higher possible candidates. And if the sum is achievable, we try the space of lower candidates, while we can.

I have changed a little bit the code so it's more clear, but here is my explanation:
/*
K = numberOfBlocks
M = maxNumber
A = array
*/
function solution(numberOfBlocks, maxNumber, array) {
let begin = array.reduce((a, b) => (a + b), 0); // Calculate total sum of A
begin = Math.ceil(begin / numberOfBlocks); // Calculate the mean of each theorethical block
begin = Math.max(begin, Math.max(...array)); // Set begin to the highest number in array if > than the mean
// In short: begin is now the smallest possible block sum
// Calculate largest possible block sum
let end = begin + maxNumber + 1;
var result = 0;
while (begin <= end) {
// Calculate the midpoint, which is our result guess
const midpoint = (begin + end) / 2;
let currentBlockSum = 0;
let block = 1;
for (let number of array) {
currentBlockSum += number;
// If currentBlockSum > midpoint means that we are
// in a different block...
if (currentBlockSum > midpoint) {
++block;
// ...so we reset sum with the current number
currentBlockSum = number;
// but if we are out of blocks, our guess (midpoint) is incorrect
// and we will have to adjust it
if (block > numberOfBlocks)
break;
}
}
// If we are out of blocks
// it means that our guess (midpoint) is bigger than we thought
if (block > numberOfBlocks) {
begin = midpoint + 1;
// else, it's smaller
} else {
result = midpoint;
end = midpoint - 1;
}
}
return result;
}

Looking and testing at the solutions, none of them actually work.
I decided to spend some time on it, here is my solution (working for any use case with maximum performance).
using System;
using System.Linq;
class Solution
{
public int solution(int K, int M, int[] A)
{
int start = Math.Max((int)Math.Ceiling((decimal)A.Sum()/(decimal)K), A.Max());
int end = A.Sum();
int result = 0;
while(start <= end)
{
int dicotomie = (end + start) / 2;
if(calculateNbBlocks(dicotomie, A) <= K)
{
result = dicotomie;
end = dicotomie - 1;
}
else
start = dicotomie + 1;
}
return result;
}
public int calculateNbBlocks(int dicotomie, int[] A)
{
int nbBlocks = 1;
int sum = 0;
for(int i=0; i<A.Length; i++)
{
sum += A[i];
if(sum > dicotomie)
{
sum = A[i];
nbBlocks++;
}
}
return nbBlocks;
}
}

Scored 100% on Codility (https://app.codility.com/demo/results/trainingQYJ68K-KJR/) using const midpoint = Math.floor((begin + end) / 2);
instead of
const midpoint = (begin + end) / 2;
after copying Javascript code in answer by pytness (Nov 5, 2020 at 8:40).
/*
K = numberOfBlocks
M = maxNumber
A = array
*/
function solution(numberOfBlocks, maxNumber, array) {
let begin = array.reduce((a, b) => (a + b), 0); // Calculate total sum of A
// console.log("total sum of A: ", begin);
begin = Math.ceil(begin / numberOfBlocks); // Calculate the mean of each theoretical block
// console.log('Math.ceil(begin / numberOfBlocks): ', begin);
begin = Math.max(begin, Math.max(...array)); // Set begin to the highest number in array if > than the mean
// console.log('Math.max(begin, Math.max(...array)): ', begin);
// In short: begin is now the smallest possible block sum
// Calculate largest possible block sum
let end = begin + maxNumber + 1;
// console.log("end: ", end);
var result = 0;
while (begin <= end) {
// Calculate the midpoint, which is our result guess
const midpoint = Math.floor((begin + end) / 2);
// console.log("midpoint: ", midpoint);
let currentBlockSum = 0;
let block = 1;
for (let number of array) {
currentBlockSum += number;
// console.log("currentBlockSum: ", currentBlockSum);
// If currentBlockSum > midpoint means that we are
// in a different block...
if (currentBlockSum > midpoint) {
// console.log("currentBlockSum > midpoint");
++block;
// console.log("block: ", block);
// ...so we reset sum with the current number
currentBlockSum = number;
// console.log("currentBlockSum: ", currentBlockSum);
// but if we are out of blocks, our guess (midpoint) is incorrect
// and we will have to adjust it
if (block > numberOfBlocks) {
// console.log("block > numberOfBlocks before break");
// console.log("block: ", block);
// console.log("break");
break;
}
}
}
// If we are out of blocks, it means that our guess for midpoint is too small.
if (block > numberOfBlocks) {
// console.log("block > numberOfBlocks before begin");
begin = midpoint + 1;
// console.log("begin: ", begin);
}
// Else, it's too big.
else {
// console.log("block <= numberOfBlocks");
result = midpoint;
// console.log("result: ", result);
end = midpoint - 1;
// console.log("end: ", end);
}
}
// console.log("result: ", result);
return result;
}

that returns the sum of all numbers between two chosen numbers

I'm totally stuck at one excersise question. Can someone help me out with this question?
Create a function sumRangeNumbers() that returns the sum of all numbers
between two chosen numbers. The function should take two arguments, one
representing the lowest boundary and one that represents the highest
boundary. For example, the arguments 10 and 20 should return the sum of
10+11+12+13...+20.
for (var i = 0; i < 82; i++) {
document.write(i + i + '+');
}
How do I write the loop that sums all the numbers with an function?

The answer of DCR already provides a nice implementation and is probably what you were looking for. However, with a little mathematical knowledge you can make the function a little easier.
We know that the sum of 1 to n is n(n+1)/2 by looking at this wikipedia page.
The sum of a to b, is simply the sum of 1 to b minus the sum of 1 to a - 1 (we also want to include a itself).
The sum between a and b is then b(b + 1)/2 - (a - 1)(a)/2 and therefore your function becomes:
const a = 10
const b = 20
function sumRangeNumbers(a, b) {
const high = Math.max(a, b);
const low = Math.min(a, b);
return high * (high + 1) / 2 - (low - 1) * (low) / 2;
}
console.log(sumRangeNumbers(a, b)); // 165
console.log(sumRangeNumbers(b, a)); // 165
console.log(sumRangeNumbers(5, 7)); // 18

function sumRangeNumber (num1, num2) {
let total = 0
for (let i = num1; i <= num2; i++) {
total += i
}
return total
}
You are on the right track with a for loop. What we did here was in place of declaring i as zero we passed the low value and in the comparison we pass the high value. This creates the range ie 10-20. From there each loop we add I too total which is declared outside fo the loop so as to not have it reset and we add to it.
As a previous comment mentioned, this is kinda doing your HW for you, so give the above function a shot and play around with it and change things to make sure you understand whats happening.

you need to first create a function and then you need to call it.
function sum(x,y){
var sum = 0;
for(let i = x;i<=y;i++){
sum = sum + i;
}
console.log(sum)
}
sum(1,10);

const sumRange = (num1, num2) => (
min = Math.min(num1, num2),
Array(Math.abs(num1 - num2) + 1)
.fill().map((_, i) => i + min)
.reduce((sum, el) => sum + el, 0)
);
console.log(sumRange(20, 10));
console.log(sumRange(10, 20));

function sumRangeNumbers(lower, upper) {
let total = 0;
for (let index=lower; index<=upper; index++) {
total = total + index;
}
return total;
}
document.write(sumRangeNumbers(10,20));

Simple:
function sumRangeNumbers(from, to) {
let result = 0;
for (let i = from; i <= to; i++) {
result += i;
}
return result;
}

If the numbers belong to range of natural numbers, then why do you need a loop. Just use the fact that sum from low to high=
((high-low+1) * (low + high)) / 2

Give this a shot:
function getSum(x,y) {
sum += x
console.log(sum, x)
if (x == y) { return sum }
else { return getSum(x+1,y) }
}
sum = 0

Here's a simple example using your current attempt. Keep in mind, you'll want to some error handling for cases where they give you an invalid high/low number.
IE:
if (lowNum >= highNum) { console.error('invalid range'); }
and maybe this too
if (typeof lowNum !== 'number' && typeof highNum !== 'number') { console.error('Invalid inputs'); }
function sumUp(lowNum, highNum) {
if (lowNum >= highNum) { throw new Error('Invalid Range'); }
// Initialize total at 0
let total = 0;
// Loop through numbers between lowNum and highNum.
// Do not include lowNum and highNum in the addition
for (let i = lowNum + 1; i < highNum; i++) {
// Increment the total with the 'in-between number (i)'
total += i;
}
// Return the result
return total;
}
// Test 1 (should be 44)
console.log(2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, sumUp(1, 10));
// Test 2 (should be 315)
console.log(50 + 51 + 52 + 53 + 54 + 55, sumUp(49, 56));
// If you really want to do document.write
document.write(sumUp(49, 56));
// Test 3 (should fail)
console.log(sumUp(15, 3));

Optimising Javascript

I was given a quiz and I had gotten the answer wrong and It's been bugging me ever since so I thought I'd ask for your thoughts
I needed to optimise the following function
function sumOfEvenNumbers(n) {
var sum = 0;
for(let i = 2; i < n;i++){
if(i % 2 == 0) sum += i;
}
return sum;
}
console.log(sumOfEvenNumbers(5));
I came up with
function sumOfEvenNumbers(n) {
var sum = 0;
while(--n >= 2) sum += n % 2 == 0? n : 0
return sum;
}
console.log(sumOfEvenNumbers(5));
What other ways were there?

It's a bit of a math question. The sum appears to be the sum of an arithmitic sequence with a common difference of 2. The sum is:
sum = N * (last + first) / 2;
where N is the number of the numbers in the sequence, last is the last number of those numbers, and first is the first.
Translated to javascript as:
function sumOfEvenNumbers(n) {
return Math.floor(n / 2) * (n - n % 2 + 2) / 2;
}
Because the number of even numbers between 2 and n is Math.floor(n / 2) and the last even number is n - n % 2 (7 would be 7 - 7 % 2 === 6 and 8 would be 8 - 8 % 2 === 8). and the first is 2.

Sum of n numbers:
var sum = (n * (n+1)) / 2;
Sum of n even numbers:
var m = Math.floor(n/2);
var sum = 2 * (m * (m+1) /2);

You can compute these sums using an arithmetic sum formula in constant time:
// Return sum of positive even numbers < n:
function sumOfEvenNumbers(n) {
n = (n - 1) >> 1;
return n * (n + 1);
}
// Example:
console.log(sumOfEvenNumbers(5));
Above computation avoids modulo and division operators which consume more CPU cycles than multiplication, addition and bit-shifting. Pay attention to the limited range of the bit-shifting operator >> though.
See e.g. http://oeis.org/A002378 for this and other formulas leading to the same result.

First thing is to eliminate the test in the loop:
function sumOfEvenNumbers(n) {
var sum = 0;
var halfN= Math.floor(n/2);
for(let i = 1; i < n/2;i++) {
sum += i;
}
return sum * 2;
}
Then we can observe that is just calculating the sum of all the integers less than a limit - and there is a formula for that (but actually formula is for less-equal a limit).
function sumOfEvenNumbers(n) {
var halfNm1= Math.floor(n/2)-1;
var sum = halfNm1 * (halfNm1+1) / 2;
return sum * 2;
}
And then eliminate the division and multiplication and the unnecessary addition and subtraction:
function sumOfEvenNumbers(n) {
var halfN= Math.floor(n/2);
return (halfN-1) * halfN;
}

Your solution computes in linear (O(N)) time.
If you use a mathematical solution, you can compute it in O(1) time:
function sum(n) {
let half = Math.ceil(n/2)
return half * (half + 1)
}

Because the question is tagged ecmascript-6 :)
const sumEven = x => [...Array(x + 1).keys()].reduce((a, b) => b % 2 === 0 ? a + b : a, 0);
// set max number
console.log(sumEven(10));

How can we get the number of all possible non-negative integers similar to a given non-negative integer by re-arranging it using javascript

Given a non-negative number say 1213, it should return 12 because there are 12 possible integers similar to 1213 i.e., 1123,1132,1213,1231,1312,1321,2113,2131,2311,312,3121 and 3211. Same with 10, it should return 1 and 12 should return 2 and if the number is 120 it should return 4 as combinations are 120,102,210,201.

You can use this formula to get the total number of unique permutations excluding permutations with leading zero.
Lets define some symbols:
n = Total Number of digits
z = Number of zeros
r1, r2, ..., rn = repetition count of digits with count > 1
fact(p) = factorial of number of p
Total permutations = (n - z) * fact(n - 1) / fact(r1) * fact(r2) * .... * fact(rn)
For example, for 1213,
n = 4, z = 0, r1 (digit 1) = 2
permutations = (4 - 0) * fact(4 - 1) / fact(2) = 4 * 6 / 2 = 12
You can easily convert this to program.
function factorial(n) {
if (n <=1)
return 1;
return n * factorial(n-1);
}
function getPermutations(number) {
var n = number.toString().split('').length;
var r = {};
number.toString().split('').forEach(function(digit){
r[digit] = r[digit] || 0;
r[digit] += 1;
});
var z = number.toString().split('').reduce(function(count, digit) {
return (digit === '0') ? count + 1 : count;
}, 0);
var denominator = Object.keys(r).map(function (key) { return r[key]; }).reduce(function(result, curr) {
return result * factorial(curr);
}, 1);
//console.log(n, r, z);
return (n - z) * factorial(n - 1) / denominator;
}
var result = getPermutations(1216);
console.log(result);
Note : This is basic implementation and would not be the most optimum. Also, factorial calculation involves large numbers and would probably fail for large inputs.

You are looking for an anagram algorithm :
This script find every anagram of a string then delete every number starting with zero :
var allAnagrams = function(arr) {
var anagrams = {};
arr.forEach(function(str) {
var recurse = function(ana, str) {
if (str === '')
anagrams[ana] = 1;
for (var i = 0; i < str.length; i++)
recurse(ana + str[i], str.slice(0, i) + str.slice(i + 1));
};
recurse('', str);
});
return Object.keys(anagrams);
}
var arr = ['120']; //declare your number
var anag = allAnagrams(arr); //send it to the function
for (var i in anag) { //delete leading 0
if((anag[i].charAt(0)) === '0' ) {
anag.splice(i);
}
}
console.log(anag); //print array
console.log(anag.length); // print length
Here the output will be :
["102", "120", "201", "210"]
4

Weighted average from an array

I have to write the following weighted average formula in JavaScript:
Average = (p1*p2*x1 + p3*p4*x2 + ... +p(n-2)*p(n-1)*xn) / (p1*p2 + p3*p4 + ... + p(n-2)p(n-1) )
The formula gives the average of x values.
I also have an array populated with n elements in JavaScript:
Array = (p1,p2,x1,p3,p4,x2....)
...where pi are the weights and xi the values I want to find the average for.
How can I write the formula using this array?

I would probably use the following strategy:
Create two new arrays (probably weights and values).
Iterate over the original array in steps of 3; multiplying the pn's and pushing the result into weights and pushing the xn into values.
Iterate over the new arrays, creating the weighted total (the left hand of the division) and the total weight (right hand of the division).
Divide one by the other. Done.
In other words, something like this:
function weighted_average(input) {
var weights = [];
var values = [];
var weighted_total = 0;
var total_weight = 0;;
if (input.length % 3 !== 0) {
throw new Error("Input array length is not a multiple of 3.");
}
for (var i = 0; i < input.length; i += 3) {
weights.push(input[i] * input[i + 1]);
values.push(input[i + 2]);
}
for (var i = 0; i < weights.length; i += 1) {
weighted_total += weights[i] * values[i];
total_weight += weights[i];
}
return weighted_total / total_weight;
}
You'll have to verify whether this does exactly what you're after, though. No guarantees. ;)
JSFiddle demo: jsfiddle.net/Z8seZ
Of course, you could skip the intermediary arrays to make it a bit faster. But the above is more explicit and more readable, and therefore more maintainable (e.g. you could easily split out the actual algorithm and create different "wrapper" functions for different forms of input). I would only optimize it if working with (really) large data sets.

Here's a functional approach, requiring ES5:
var w = a.unzip(3).map(function(v, i, a) {
var weight = v[0] * v[1];
var sum = weight * v[2];
return [sum, weight];
}).reduce(function(p, c, i, a) {
return [p[0] + c[0], p[1] + c[1]];
}, [0, 0]);
var aw = w[0] / w[1];
which in pseudo-code is:
split the array into chunks of three
convert each three [p1, p2, x ] into a pair [ p1 * p2 * x , p1 * p2 ]
sum the pairs (along the array, not within each pair)
divide one by the other
and where the (non-standard) unzip function which chunks the array is:
Object.defineProperty(Array.prototype, 'unzip', {
value: function(n) {
n = n || 2;
return this.reduce(function(p, c, i, a) {
if (i % n === 0) {
p.push(a.slice(i, i + n));
}
return p;
}, []);
}
});

ES6 one-liner for an array of objects xs containing the keys w as weight and v as value:
((_w, _v) => _v / _w)(...xs.reduce((r, o) => [r[0] + o[w], r[1] + o[w] * o[v]], [0, 0]))