Develop Reference

For Loop that calculates sum of squred numbers in array

I am working on a challenge and in need of some help:
Write a for loop that calculates sum of squares of items in an array of numbers. Example: For array [ 1, 2, 3, 4 ] it calculates the sum of squares as 30 (i.e. 1 + 4 + 9 + 16). I have a fiddle set up if anyone wants to have a closer look. Thanks for your help!
https://jsfiddle.net/jamie_shearman/2drt56e3/23/
var aFewNumbers = [ 1, 2, 3, 7 ];
var squareOfAFewNumbers = 0;
for( var i = 0; i <= aFewNumbers; i++ ) {
squareOfAFewNumbers = squareOfAFewNumbers * aFewNumbers[i] ;
}
console.log( squareOfAFewNumbers );

Your math is wrong. As an obvious issue, the variable starts at 0 and then is multiplied by each array element; 0 times anything is 0, so it will always remain 0 no matter what the values are. Not to mention your loop isn't looking at the length of the array as a stopping condition, which it should be, since you want to iterate from the beginning to the end of the array.
You need to iterate through to the array's length, square each array element, and add that square to the variable:
for( var i = 0; i < aFewNumbers.length; i++ ) {
squareOfAFewNumbers += aFewNumbers[i] * aFewNumbers[i];
}
If you can use ES6, you can even use higher-order array functions to simplify this more:
var squareOfAFewNumbers = aFewNumbers.reduce((result, entry) => result + entry * entry, 0);

There are multiple approaches you can take for reaching the desired result, but as you've mentioned that you must write a for loop; therefore, I sorted the answers by having that in mind.
Using the for loop
let numbers = [1, 2, 3, 7],
sum = 0;
for(let i = 0; i < numbers.length; i++) {
sum += Math.pow(numbers[i], 2)
}
// Sum is now 63
Using forEach method of the array object
let numbers = [1, 2, 3, 7],
sum = 0;
numbers.forEach(number => sum += Math.pow(number, 2))
// Sum is now 63
Oneliner
let sum = [1, 2, 3, 7].reduce((a, c) => a + Math.pow(c, 2))
The reduce method uses an accumulator to store temporary results, the accumulator is passed to the callback as the first argument and the second argument is the element's value, you can read more about the reduce method here.

You can use JavaScript pow() Method to create the square and sum it to the sumSquareOfAFewNumbers.
var aFewNumbers = [ 1, 2, 3, 7 ];
var sumSquareOfAFewNumbers = 0;
aFewNumbers.forEach(function(element) {
sumSquareOfAFewNumbers += Math.pow(element, 2);
});
console.log(sumSquareOfAFewNumbers)

How do I get an Array item's opposite index and value?

I have an array that I created using Array(...) and Array.prototype.map, like this:
var array_name = Array(255).map(function(undef, i) {
return i + 1;
});
The values of this array are:
[1, 2, 3, ..., 253, 254, 255]
This is an array that won't get modified, so the first value of this array will always be 1 and the last value of this array will always be 255.
I already know the index of each value is {value} - 1, so 200 would be 199, 199 would be 198, so on and so forth.
Let's say I want 255's opposite value, which would be 0, I could get that using array_name[0], but what if I wanted 200's opposite value, how would I know what the opposite index of 199 is so I could get it's value?

Do:
opposite_index = arr.length - index - 1
For example:
a = [1,2,3,4,5,6,7,8,9,10]
index = 3
a[index]
4
It's opposite is 7 so:
opposite_index = a.length - index - 1
a[opposite_index]
7
With reverse as per #Maheer Ali suggestion:
a.reverse()[index]
7

Your Array(255).map() create undefined array value.So do with Array#from length object.And pass your value.get index of the value and match with reverse array you get opposite value
let check = (val) => {
var array_name = Array.from({length:255},(a,b)=>b+1);
var nor_ind = array_name.indexOf(val);
var re_in = array_name.reverse(array_name).indexOf(val)
return ({nor_val:val,nor_ind:nor_ind,opp_val:re_in})
}
console.log(check(254))

First of all the code you provided doesn't create array [1,2,3...255]. It will create it will 255 empty items first you need to fill().
var arr = Array(255).fill().map((a,i) => i+1);
//Create an array which will have revese items.
let revarr= arr.reverse()
console.log(revarr[0]) //255
console.log(revarr[254]) // 1
If you don't want to create a reverse arr. You can create a function
var arr = Array(255).fill().map((a,i) => i+1);
const opp = (arr,num) => arr[arr.length - num - 1];
console.log(opp(arr,0));
console.log(opp(arr,254));

First, you gotta understand that there is weird behavior concerning Array(n).map(f) (it won't create the array you're expecting), see this answer for explanation, second, do this to get the opposite values:
/* fill it first with .fill(), see the question I linked for more explanation */
var array = Array(255).fill(undefined).map(function(undef, i) {
return i + 1;
});
function opposite(array, n) {
return array[array.length - n];
}
console.log(opposite(array, 255));
console.log(opposite(array, 200));
console.log(opposite(array, 199));
console.log(opposite(array, 1));
Notice that length - n is used instead of length - n - 1, because we're dealing with values from 1 to n, not from 0 to n - 1.

Subtract the index from (length-1) -> max index of the array
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
function findOpp(index, length) {
maxIndex = length - 1;
if (index <= maxIndex && index >= 0) {
return maxIndex - index;
} else {
return 'You have enter a wrong index';
}
}
console.log(findOpp(-1, 10));
console.log(findOpp(0, 10));
console.log(findOpp(1, 10));
console.log(findOpp(2, 10));
console.log(findOpp(4, 10));
console.log(findOpp(5, 10));
console.log(findOpp(6, 10));
console.log(findOpp(7, 10));
console.log(findOpp(8, 10));
console.log(findOpp(9, 10));
console.log(findOpp(10, 10));

Using Maheer Ali's suggestion I managed to get the desired result by reversing the array and using indexOf to get the index of that number:
var numbers = Array(255).map(function(v, i) {
return i + 1;
});
var opposite_brightness = numbers.reverse()[numbers.indexOf(brightness)];

Javascript - Function for each which sums and multiplies every int in array

Javascript is something new for me, and we have to do homework.
I have created new array:
var numbers = [1,2,3,4,5,6];
And with function forEeach I should achieve result like in console.log:
console.log(numbers[0]*numbers[1]+numbers[0]+numbers[1]);
I've tested many things, but I don't have any idea how to pull out signle init...
I know it should be simple, but I've stucked.
Thanks for help!

From your question looks like your problem is interacting with the current element of the forEach loop.
var numbers = [1,2,3,4,5,6]
// this will print every number in the array
// note that index numbers are not needed to get elements from the array
numbers.forEach(function(num){
console.log(num)
})
Now, if what you're trying t achieve is sum and multiply every int (as stated in the question title), you could do it like this
var numbers = [1,2,3,4,5,6]
var sumResult = 0
var multiplicationResult = 1
// the function will be evaluated for every element of the array
numbers.forEach(function(num){
sumResult += num
multiplicationResult *= num
})
console.log('Sum', sumResult)
console.log('Multiplication', multiplicationResult)
However, a more appropiate approach could be obtained by using reduce like this:
var numbers = [1,2,3,4,5,6]
var sumResult = numbers.reduce(function(result, num){
return num+result
}, 0)
var multiplicationResult = numbers.reduce(function(result, num){
return num*result
}, 1)
console.log('Sum', sumResult)
console.log('Multiplication', multiplicationResult)
Hope this helps.
More info:
Reduce # MDN
ForEach # MDN

To pull out a single number for the provided array you using the indexer/bracket notation which is specifying a number (length of array - 1) in brackets, like below:
var numbers = [1, 2, 3, 4, 5, 6];
numbers[0]; // selects the first number in the array
numbers[1]; // selects second number etc.
To sum up the numbers using forEach, simply do:
var sum = 0;
numbers.forEach(function(number) {
sum += number; // add number to sum
});
forEach goes through all the numbers in the numbers array, passing in each number to the function defined and then adds the number to the sum variable.

If you want your results, use map(). Unlike forEach(), map() will always return results in a new array. It wasn't very clear as to what expression you are expected to use or what the result of said expression should be so this demo will do the following on each iteration:
A = current value * next value
B = current value + next value
C = A + B;
Demo
const num = [1, 2, 3, 4, 5, 6];
let arr = num.map(function(n, idx, num) {
let next = num[idx + 1];
if (!next > 0) {
next = idx + 2;
}
let subSUM = n + next;
let subPRD = n * next;
let subRES = subPRD + subSUM;
return subRES;
});
console.log(arr);

Giving an original array and a new array that each element's value indicates the count of elements on the left that are bigger

Giving an array, say [4,2,1,3,5], based on this array, we have a new array, which each number shows the count of elements on its left that are bigger than itself, which is [0,1,2,1,0]. Now write a function with given input of [0,1,2,1,0], return the original array. The range of array is 1 ~ n (n is the size of the array, you can assume all numbers in the original array are consecutive if sorted)
Now to recover the original array, I have tried a way to solve the problem by iterating through the range of array from the end to the front like this:
My approach:
say the range is 1 ~ 5, the original array would be [1, 2, 3, 4, 5] if sorted. Iterate from the end to the beg,
so first 5, there is no element can be bigger than 5, so its maximum count of bigger elements would be 0, then 4 would have 1 as its maximum count of bigger elements, 3 to 2, etc. Store the key-value pairs into an object.
Now iterating through the input from back to front,
0 -> 5
1 -> can be 4, 3, or 2
2 -> can be either 3, 2, or 1
1 -> any number bigger than the first one.
0 -> (can be anything, since 5 is taken, so it can be either 1, 2, 3, or 4)
Simply to map each element of the input as value to its key from the map is not enough. What would be an intuitive way to approach this with optimal performance? (avoiding O(n ^2) if possible.)

Initially make an AVL Tree from numbers 1 to n.
Start from rear i.e. at nth index (considering 1 based index).
Now the high level outline level of the algorithm should look like this:
1. At any ith index, say the number in array(not the originial array) is j
2. Search the number which is at (i-j)th position in your AVL tree(This can be done in O(logn) time. Comment if you need more explanation on this)
3. The element in the AVL tree is your required element. Delete that element from AVL tree.(O(logn))
So the total complexity would be O(nlogn).
Walkthrough
Initially the tree will contain all 5 elements.
You start at index 5(1-based indexing). Element is 0, i.e. i=5, j=0. So 5th largest element which is 5.
Now the tree contains four elements 1,2, 3, and 4. i=4, j=1. So 4-1 i..e 3rd largest element which is 3 in this case.
i=3, j=2. (3-2)rd largest element is 1 since the tree contains (1, 2, 4).
And so on.
Using Tree to find the ith largest number
We can do this by, storing the count of number of nodes in left subtree at the root node. So consider a tree, having elements 1, 2, 3,4 and 5 and tree structure as following:
4(3)
/ \
3(1) 5(0)
/ \
1(0) 2(0)
At root, number 4 is the value and the number in round bracket has the number of nodes in left subtree.
While constructing(insertion and deletion too) the tree, we can maintain the count.
Now, to find the ith node, say we want suppose 3rd nodes in the given tree. We start with the root, it says it has 3 elements smaller than it to the left so we move to left. Now the root i.e. 3 has 1 smaller left element which is less than 3(ith element) so we move to right of it. Subtract 1(the left count)+1(the root itself) out of 3. Now the root is 2 we want 1st element, the left count is 0. Hence the 1st element of the subtree rooted at 2 is 2.
Basic pseudocode is below:
while(true){
count = root->leftCount;
if((count+1)<i){
//move to right
i-=(count+1);
root = root->right;
}
else if(i==(count+1)){
//root is the ith node
break;
} else{
//move to the levft
root=root->left
}
}

You could use Array#reduceRight and use the value as negative index for generating the original array.
var array = [1, 2, 3, 4, 5],
countLeft = [0, 1, 2, 1, 0],
result = countLeft.reduceRight(function (r, a) {
return array.splice(array.length - 1 - a, 1).concat(r);
}, []);
console.log(result);
Shorter version with ES6 and reverse base array.
var array = [5, 4, 3, 2, 1],
countLeft = [0, 1, 2, 1, 0],
indices = array.map((_, i) => i),
result = [];
countLeft.forEach(a => {
result.unshift(array[indices[a]]);
indices = indices.filter((_, i) => i !== a);
});
console.log(result);
At last a proposal with complexity between O(n*(n-1)/2) and O(n).
This version uses a lazy array with progressive reduction of the length for every iteration. At the end, the offset array has zero elements.
var array = [5, 4, 3, 2, 1],
countLeft = [0, 1, 2, 1, 0],
result = [],
length = array.length;
countLeft.forEach((offset => (offset.length = countLeft.length, a => {
var i = offset[a] || 0;
result.unshift(array[i + a]);
offset.length--;
while (i < offset.length) {
offset[i] = (offset[i] || 0) + 1;
i++;
}
}))([]));
console.log(result);
A linear version, heavily inspired by the proposal of Oriol
var array = [1, 2, 3, 4, 5],
countLeft = [0, 1, 2, 1, 0],
swap = [],
i = 0,
l,
temp;
while (i < countLeft.length) {
l = countLeft[i];
while (l) {
swap.push(i + l - countLeft[i]);
l--;
}
i++;
}
i = swap.length;
while (i--) {
temp = array[swap[i]];
array[swap[i]] = array[swap[i] - 1];
array[swap[i] - 1] = temp;
}
console.log(array);

Here is a possible solution. See inline comments for a brief description of this method.
var a = [0,1,2,1,0],
n, b = [], res = [];
// build b = [5,4,3,2,1]
// we use this array to keep track of values to be pushed in res[],
// sorted in descending order
for(n = a.length; n > 0; n--) {
b.push(n);
}
// for each element of a, starting from the end:
// find correct value in b and remove it from b
while(a.length) {
res.push(b.splice(a.pop(), 1)[0]);
}
res = res.reverse();
console.log(res);
Output:
[4, 2, 1, 3, 5]

I propose an approach based on a custom sort, based on mergesort:
Split the array of inversions into two halves
Sort each part recursively, from greatest to lowest, maintaining stability
Merge the two parts
The difference with mergesort is the merge part. If we choose the j-th element of right part instead of the i-th of the left one, it will advance some elements, and therefore its number of inversions must be reduced by that amount.
Like mergesort, the complexity is O(n log n)
function undoInversions(inversions) {
function reorder(arr, from=0, to=arr.length) {
// Based on a stable decreasing mergesort
if(from >= to) return []; // Unusual base case
if(to === from + 1) return [arr[from]]; // Base case
var m = Math.floor((from + to)/2);
var l = reorder(arr, from, m), // Left recursive call
r = reorder(arr, m, to), // Right recursive call
ret = [], i=0, j=0;
while(i < l.length && j < r.length) { // Merge
if(r[j].value - l.length + i >= l[i].value) {
r[j].value -= l.length - i; // Reduce number of inversions
ret.push(r[j++]);
} else {
ret.push(l[i++]);
}
}
while(i < l.length) ret.push(l[i++]); // Merge remaining, if any
while(j < r.length) ret.push(r[j++]); // Merge remaining, if any
return ret;
}
var array = new Array(inversions.length);
reorder(inversions.map(function(inv, idx) {
return {value: inv, originalIndex: idx}; // Keep track of indices
})).forEach(function(obj, idx) {
if(obj.value !== 0) throw 'Invalid input';
array[obj.originalIndex] = idx + 1; // Invert the permutation
});
return array;
}
console.log(JSON.stringify(undoInversions([0,1,2,1,0])));
Here is an example to understand how it works:
[0,1,2,1,0] ~ [4,2,1,3,5]
⤩ ⤧
[0,0,2,1,0] ~ [2,4,1,3,5]
⤩ ⤧
[0,1,0,1,0] ~ [2,1,4,3,5]
⤩ ⤧
[0,0,0,1,0] ~ [1,2,4,3,5]
⤩ ⤧
[0,0,0,0,0] ——→ [1,2,3,4,5]
That is, each array of inversions corresponds to a permutation. We apply a permutation σ which transforms the input to the array of inversions [0,0,0,0,0], which corresponds to the permutation [1,2,3,4,5]. Since we kept track of the original indices, now we only need the to apply σ⁻¹ to [1,2,3,4,5] in order to get the permutation corresponding to the input.

Pushing unique numbers within a range to an array in Javascript

I am trying to insert numbers from 1 to 15 into an array.
And here is the code:
<html>
<head></head>
<body>
<button id="myBtn" value="HELLO">HELLO</button>
<script type="text/javascript">
var btn = document.getElementById("myBtn");
var num = null;
btn.addEventListener('click', function() {
var questions = new Array();
num = Math.floor(Math.random() * 14 + 2);
questions.push(num);
for (var i = 1; i <= 15; i++) {
num = Math.floor(Math.random() * 14 + 2);
if (questions.indexOf(num) !== -1) {
alert(num + " Exists in array. So not pushing it");
} else {
alert(num + " is not found. So pushing it");
questions.push(num);
}
console.log(questions);
}
alert(questions);
})
</script>
</body>
</html>
If you run this with the console open. You will notice that though a number is not in the array the in operator still discards the number without pushing. Can I know why and how to correct this?
And also is there any better way to insert x number of numbers in random order each time.

You should not use in operator with Arrays. What you should have done is
if (questions.indexOf(num) !== -1) {
When you use in with an Array, it will not check the values but the indices of the Array. That is why your code fails.
Please check this answer of mine to know more about, why you should not use in operator with Arrays.
The best way to generate N unique random numbers is to, generate the list of numbers and then shuffle them, like this
function getRandomNumbers() {
var rand, index = 0,
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
array.forEach(function(value) {
rand = Math.floor(Math.random() * ++index);
array[index - 1] = array[rand];
array[rand] = value;
});
return array;
}
console.log(getRandomNumbers());
This is adopted from the _.shuffle function of Underscore.js library, which shuffles the list of data with Fisher-Yates Shuffle algorithm.

the in operator works on objects, so you're really checking to see if your array has an index, not a value.