We Keep Coding

Extract last ","

This is my code
function convertToCurr(value) {
var x = value.toString().length;
var z = x % 3;
var a = 0;
if (z == 0) {
a = (x / 3) - 1;
}
else {
a = (x / 3);
}
var last = 0;
var vals = [];
var i;
for (i = 1; i <= a; i++) {
steps = 3;
start = x - steps * i;
end = start + steps;
last = end - steps;
vals.unshift(value.toString().slice(start, end));
}
vals.unshift("R " + value.toString().slice(0, last));
return vals.join();
}
basicIO.write(convertToCurr(input));
context.log.INFO("log data");
}
These are my outputs
{"output":"R 1,000,000,.00","log":["log data"]}
{"output":"R 1,000,.00","log":["log data"]}
I need to exctract the last "," so that the amounts make sense

The most straightforward solution is to perform a String.prototype.replace() on the final join().
function convertToCurr(value) {
var x = value.toString().length;
var z = x % 3;
var a = 0;
if (z == 0) {
a = (x / 3) - 1;
}
else {
a = (x / 3);
}
var last = 0;
var vals = [];
var i;
for (i = 1; i <= a; i++) {
steps = 3;
start = x - steps * i;
end = start + steps;
last = end - steps;
vals.unshift(value.toString().slice(start, end));
}
vals.unshift("R " + value.toString().slice(0, last));
return vals.join().replace(',.', '.');
}
console.log(convertToCurr(10000.75));
console.log(convertToCurr(10.01));
console.log(convertToCurr(1000));
console.log(convertToCurr(7002344));
It should be noted that replace() only replaces a single instance of the substring you provide it in the input string, but this doesn't matter here since ,. only appears in your output string one time.

function convertToCurr(value) {
var x = value.toString().length;
var z = x % 3;
var a = 0;
var combinedString; // holds string value after join
if (z == 0) {
a = x / 3 - 1;
} else {
a = x / 3;
}
var last = 0;
var vals = [];
var i;
for (i = 1; i <= a; i++) {
steps = 3;
start = x - steps * i;
end = start + steps;
last = end - steps;
vals.unshift(value.toString().slice(start, end));
}
vals.unshift("R " + value.toString().slice(0, last));
combinedString = vals.join(); // join array into string
return combinedString.replace(",.", "."); // replace ,. with .
}
console.log(convertToCurr(50000000.15));

Browse is slowing down for a JavaScript code

I'm working on a JavaScript library for me. This library will help you to convert a binary number into a decimal number. The default JavaScript function can only convert before point (integers). But this code will convert floating numbers also (hopefully).
In order to test this when I ran this code on Firefox returned,
This page is slowed down your browser.
& Google Chrome returned nothing but just loading.
So I want to know what is the problem??
Here is my code
var x = "101.11";
var r = 0;
var ra = 0;
for (var i = 0; i < x.length; ++i) {
if (x.charAt(i) == ".") {
var Ap = i;
}
}
for (var j = 0; j < (x.length - Ap); ++j) {
var a = x.charAt(j);
r = r + a * Math.pow(2, ((x.length - Ap - 1) - j));
}
for (var k = Ap + 1;
(x.length - Ap) < k; ++k) {
var b = x.charAt(k);
ra = ra + b * Math.pow(2, (Ap - k));
}
document.write(r);
if (ra <= 0) {
document.write("." + ra);
}

Here is the solution. I'm very much thankful to you people.
/*
It's a dumb library for converting binary fraction number into a desimal number.
Created by : Adib Rahman
Last update : 12:36 30/01/2020
*/
function print(something)
{
document.write(something);
}
var x = "010101010.111";
var r = 0;
var ra = 0;
for(var i=0;i<x.length;++i)
{
if(x.charAt (i)== ".")
{
var Ap=i;
}
}
var beforePoint = (x.length-(x.length-Ap-1))-1;
var afterPoint = x.length-Ap-1;
for(var j=0;j<beforePoint;++j)
{
r = r+(x.charAt(j))*Math.pow(2,(beforePoint-j-1));
}
for(var k=Ap+1;k<x.length;++k)
{
var ra = ra+(x.charAt(k))*Math.pow(2,(Ap-k));
}
//Final Result
if(ra >=0)
{
print(r+ra);
}
else
{
print(r);
}

How can I make ASCII send character instead of code

How can I convert charCode to the actual letter/digit/symbol that was typed?
I mean if A is typed on a keyboard the result on PHP should be the letter A and not the charCode for A.
var ucode = function(s) {
var len = s.length;
var rs = "";
for ( var i = 0; i < len; i++) {
var k = s.substring(i, i + 1);
rs += "$" + (s.charCodeAt(i) + "1") + ";";
}
return rs;
};

if s is string, you can use String.prototype.slice and change your line of code like something below
var ucode = function(s) {
var len = s.length;
var rs = "";
for ( var i = 0; i < len; i++) {
rs += s.slice(i);
}
return rs;
};

Longest palindrome in a string

I wrote the following function to find the longest palindrome in a string. It works fine but it won't work for words like "noon" or "redder". I fiddled around and changed the first line in the for loop from:
var oddPal = centeredPalindrome(i, i);
to
var oddPal = centeredPalindrome(i-1, i);
and now it works, but I'm not clear on why. My intuition is that if you are checking an odd-length palindrome it will have one extra character in the beginning (I whiteboarded it out and that's the conclusion I came to). Am I on the right track with my reasoning?
var longestPalindrome = function(string) {
var length = string.length;
var result = "";
var centeredPalindrome = function(left, right) {
while (left >= 0 && right < length && string[left] === string[right]) {
//expand in each direction.
left--;
right++;
}
return string.slice(left + 1, right);
};
for (var i = 0; i < length - 1; i++) {
var oddPal = centeredPalindrome(i, i);
var evenPal = centeredPalindrome(i, i);
if (oddPal.length > result.length)
result = oddPal;
if (evenPal.length > result.length)
result = evenPal;
}
return "the palindrome is: " + result + " and its length is: " + result.length;
};
UPDATE:
After Paul's awesome answer, I think it makes sense to change both variables for clarity:
var oddPal = centeredPalindrome(i-1, i + 1);
var evenPal = centeredPalindrome(i, i+1);

You have it backwards - if you output the "odd" palindromes (with your fix) you'll find they're actually even-length.
Imagine "noon", starting at the first "o" (left and right). That matches, then you move them both - now you're comparing the first "n" to the second "o". No good. But with the fix, you start out comparing both "o"s, and then move to both "n"s.
Example (with the var oddPal = centeredPalindrome(i-1, i); fix):
var longestPalindrome = function(string) {
var length = string.length;
var result = "";
var centeredPalindrome = function(left, right) {
while (left >= 0 && right < length && string[left] === string[right]) {
//expand in each direction.
left--;
right++;
}
return string.slice(left + 1, right);
};
for (var i = 0; i < length - 1; i++) {
var oddPal = centeredPalindrome(i, i + 1);
var evenPal = centeredPalindrome(i, i);
if (oddPal.length > 1)
console.log("oddPal: " + oddPal);
if (evenPal.length > 1)
console.log("evenPal: " + evenPal);
if (oddPal.length > result.length)
result = oddPal;
if (evenPal.length > result.length)
result = evenPal;
}
return "the palindrome is: " + result + " and its length is: " + result.length;
};
console.log(
longestPalindrome("nan noon is redder")
);

This will be optimal if the largest palindrome is found earlier.
Once its found it will exit both loops.
function isPalindrome(s) {
//var rev = s.replace(/\s/g,"").split('').reverse().join(''); //to remove space
var rev = s.split('').reverse().join('');
return s == rev;
}
function longestPalind(s) {
var maxp_length = 0,
maxp = '';
for (var i = 0; i < s.length; i++) {
var subs = s.substr(i, s.length);
if (subs.length <= maxp_length) break; //Stop Loop for smaller strings
for (var j = subs.length; j >= 0; j--) {
var sub_subs = subs.substr(0, j);
if (sub_subs.length <= maxp_length) break; // Stop loop for smaller strings
if (isPalindrome(sub_subs)) {
maxp_length = sub_subs.length;
maxp = sub_subs;
}
}
}
return maxp;
}

Here is another take on the subject.
Checks to make sure the string provided is not a palindrome. If it is then we are done. ( Best Case )
Worst case 0(n^2)
link to
gist
Use of dynamic programming. Break each problem out into its own method, then take the solutions of each problem and add them together to get the answer.
class Palindrome {
constructor(chars){
this.palindrome = chars;
this.table = new Object();
this.longestPalindrome = null;
this.longestPalindromeLength = 0;
if(!this.isTheStringAPalindrome()){
this.initialSetupOfTableStructure();
}
}
isTheStringAPalindrome(){
const reverse = [...this.palindrome].reverse().join('');
if(this.palindrome === reverse){
this.longestPalindrome = this.palindrome;
this.longestPalindromeLength = this.palindrome.length;
console.log('pal is longest', );
return true;
}
}
initialSetupOfTableStructure(){
for(let i = 0; i < this.palindrome.length; i++){
for(let k = 0; k < this.palindrome.length; k++){
this.table[`${i},${k}`] = false;
}
}
this.setIndividualsAsPalindromes();
}
setIndividualsAsPalindromes(){
for(let i = 0; i < this.palindrome.length; i++){
this.table[`${i},${i}`] = true;
}
this.setDoubleLettersPlaindrome();
}
setDoubleLettersPlaindrome(){
for(let i = 0; i < this.palindrome.length; i++){
const firstSubstring = this.palindrome.substring(i, i + 1);
const secondSubstring = this.palindrome.substring(i+1, i + 2);
if(firstSubstring === secondSubstring){
this.table[`${i},${i + 1}`] = true;
if(this.longestPalindromeLength < 2){
this.longestPalindrome = firstSubstring + secondSubstring;
this.longestPalindromeLength = 2;
}
}
}
this.setAnyPalindromLengthGreaterThan2();
}
setAnyPalindromLengthGreaterThan2(){
for(let k = 3; k <= this.palindrome.length; k++){
for(let i = 0; i <= this.palindrome.length - k; i++){
const j = i + k - 1;
const tableAtIJ = this.table[`${i+1},${j-1}`];
const stringToCompare = this.palindrome.substring(i, j +1);
const firstLetterInstringToCompare = stringToCompare[0];
const lastLetterInstringToCompare = [...stringToCompare].reverse()[0];
if(tableAtIJ && firstLetterInstringToCompare === lastLetterInstringToCompare){
this.table[`${i},${j}`] = true;
if(this.longestPalindromeLength < stringToCompare.length){
this.longestPalindrome = stringToCompare;
this.longestPalindromeLength = stringToCompare.length;
}
}
}
}
}
printLongestPalindrome(){
console.log('Logest Palindrome', this.longestPalindrome);
console.log('from /n', this.palindrome );
}
toString(){
console.log('palindrome', this.palindrome);
console.log(this.table)
}
}
// const palindrome = new Palindrome('lollolkidding');
// const palindrome = new Palindrome('acbaabca');
const palindrome = new Palindrome('acbaabad');
palindrome.printLongestPalindrome();
//palindrome.toString();

function longestPalindrome(str){
var arr = str.split("");
var endArr = [];
for(var i = 0; i < arr.length; i++){
var temp = "";
temp = arr[i];
for(var j = i + 1; j < arr.length; j++){
temp += arr[j];
if(temp.length > 2 && temp === temp.split("").reverse().join("")){
endArr.push(temp);
}
}
}
var count = 0;
var longestPalindrome = "";
for(var i = 0; i < endArr.length; i++){
if(count >= endArr[i].length){
longestPalindrome = endArr[i-1];
}
else{
count = endArr[i].length;
}
}
console.log(endArr);
console.log(longestPalindrome);
return longestPalindrome;
}
longestPalindrome("abracadabra"));

let str = "HYTBCABADEFGHABCDEDCBAGHTFYW12345678987654321ZWETYGDE";
let rev = str.split("").reverse().join("").trim();
let len = str.length;
let a="";
let result = [];
for(let i = 0 ; i < len ; i++){
for(let j = len ; j > i ; j--){
a = rev.slice(i,j);
if(str.includes(a)){
result.push(a);
break;
}
}
}
result.sort((a,b) => { return b.length - a.length})
let logPol = result.find((value)=>{
return value === value.split('').reverse().join('') && value.length > 1
})
console.log(logPol);

function longest_palindrome(s) {
if (s === "") {
return "";
}
let arr = [];
let _s = s.split("");
for (let i = 0; i < _s.length; i++) {
for (let j = 0; j < _s.length; j++) {
let word = _s.slice(0, j + 1).join("");
let rev_word = _s
.slice(0, j + 1)
.reverse()
.join("");
if (word === rev_word) {
arr.push(word);
}
}
_s.splice(0, 1);
}
let _arr = arr.sort((a, b) => a.length - b.length);
for (let i = 0; i < _arr.length; i++) {
if (_arr[arr.length - 1].length === _arr[i].length) {
return _arr[i];
}
}
}
longest_palindrome('bbaaacc')
//This code will give you the first longest palindrome substring into the string

var longestPalindrome = function(string) {
var length = string.length;
var result = "";
var centeredPalindrome = function(left, right) {
while (left >= 0 && right < length && string[left] === string[right]) {
//expand in each direction.
left--;
right++;
}
return string.slice(left + 1, right);
};
for (var i = 0; i < length - 1; i++) {
var oddPal = centeredPalindrome(i, i + 1);
var evenPal = centeredPalindrome(i, i);
if (oddPal.length > 1)
console.log("oddPal: " + oddPal);
if (evenPal.length > 1)
console.log("evenPal: " + evenPal);
if (oddPal.length > result.length)
result = oddPal;
if (evenPal.length > result.length)
result = evenPal;
}
return "the palindrome is: " + result + " and its length is: " + result.length;
};
console.log(longestPalindrome("n"));
This will give wrong output so this condition need to be taken care where there is only one character.

public string LongestPalindrome(string s) {
return LongestPalindromeSol(s, 0, s.Length-1);
}
public static string LongestPalindromeSol(string s1, int start, int end)
{
if (start > end)
{
return string.Empty;
}
if (start == end)
{
char ch = s1[start];
string s = string.Empty;
var res = s.Insert(0, ch.ToString());
return res;
}
if (s1[start] == s1[end])
{
char ch = s1[start];
var res = LongestPalindromeSol(s1, start + 1, end - 1);
res = res.Insert(0, ch.ToString());
res = res.Insert(res.Length, ch.ToString());
return res;
}
else
{
var str1 = LongestPalindromeSol(s1, start, end - 1);
var str2 = LongestPalindromeSol(s1, start, end - 1);
if (str1.Length > str2.Length)
{
return str1;
}
else
{
return str2;
}
}
}

This is in JS ES6. much simpler and works for almost all words .. Ive tried radar, redder, noon etc.
const findPalindrome = (input) => {
let temp = input.split('')
let rev = temp.reverse().join('')
if(input == rev){
console.log('Palindrome', input.length)
}
}
//i/p : redder
// "Palindrome" 6

Got 90% of the JavaScript code - can't figure out the rest

So I am trying to model Gram-Schmidt for any size N×N matrix, and I have officially hit a roadblock I can't get past. I know it's a matter of looping this correctly, but I can't figure out what the problem is. Remember I do not want to just pass in a 3×3 matrix, but any size N×N.
The course notes QR Decomposition with Gram-Schmidt explains exactly what I want to do. Very simple calculation by the way. In the course notes ||u|| means that it is the sum of the square of the elements, so sqrt(x12 + x22 + x32 + .... + xn2).
The multiplication symbol is actually the dot product.
The code I wrote so far is listed below. What is wrong with it?
function qrProjection(arr) {
var qProjected = [];
var tempArray = [];
var aTemp = arr;
var uTemp = new Array(arr.length);
var uSquareSqrt = new Array(arr.length);
var eTemp = [];
var sum = 0;
var sumOfSquares = 0;
var breakCondition = 0;
var secondBreakCondition = 0;
var iterationCounter = 0;
//Build uTemp Array
for (i = 0; i < arr.length; i++) {
uTemp[i] = new Array(arr[i].length);
}
for (i = 0; i < arr.length; i++) {
eTemp[i] = new Array(arr[i].length);
}
uTemp[0] = aTemp[0];
for (j = 0; j <= arr.length; j++) {
for (l = 0; l < arr[j].length; l++) {
if (breakCondition == 1) break;
sumOfSquares = Math.pow(uTemp[j][l], 2) + sumOfSquares;
}
if (breakCondition == 0) {
uSquareSqrt[j] = Math.sqrt(sumOfSquares);
sumOfSquares = 0;
}
for (i = 0; i < arr[j].length; i++) {
if (breakCondition == 1) break;
eTemp[j][i] = (1 / (uSquareSqrt[j])) * (uTemp[j][i]);
}
breakCondition = 1;
if (iterationCounter == 0) {
for (m = 0; m < arr[j].length; m++) {
matrixDotProduct = aTemp[j + 1][m] * eTemp[j][m] + matrixDotProduct;
}
}
else {
for (m = 0; m < arr[j].length; m++) {
for (s = 0; s <= iterationCounter; s++) {
matrixDotProduct = aTemp[j + 1][s] * eTemp[m][s] + matrixDotProduct;
}
for (t = 0; t < arr[j].length; t++) {
uTemp[j + 1][t] = aTemp[j + 1][t] - eTemp[j][t] * matrixDotProduct;
}
}
}
if (iterationCounter == 0) {
for (m = 0; m < arr[j].length; m++) {
uTemp[j + 1][m] = aTemp[j + 1][m] - eTemp[j][m] * matrixDotProduct;
}
}
matrixDotProduct = 0;
for (l = 0; l < arr[j].length; l++) {
sumOfSquares = Math.pow(uTemp[j + 1][l], 2) + sumOfSquares;
}
uSquareSqrt[j + 1] = Math.sqrt(sumOfSquares);
sumOfSquares = 0;
for (i = 0; i < arr[j].length; i++) {
eTemp[j + 1][i] = (1 / (uSquareSqrt[j + 1])) * (uTemp[j + 1][i]);
}
iterationCounter++;
}
qProjected = eTemp;
return qProjected;
}

I must apologize that instead of tweaking your code, I wrote my own from scratch:
/* Main function of interest */
// Each entry of a matrix object represents a column
function gramSchmidt(matrixA, n) {
var totalVectors = matrixA.length;
for (var i = 0; i < totalVectors; i++) {
var tempVector = matrixA[i];
for (var j = 0; j < i; j++) {
var dotProd = dot(matrixA[i], matrixA[j], n);
var toSubtract = multiply(dotProd, matrixA[j], n);
tempVector = subtract(tempVector, toSubtract, n);
}
var nrm = norm(tempVector, n);
matrixA[i] = multiply(1 / nrm, tempVector, n);
}
}
/*
* Example usage:
* var myMatrix = [[1,0,0],[2,3,0],[5,4,7]];
* gramSchmidt(myMatrix, 3);
* ==> myMatrix now equals [[1,0,0],[0,1,0],[0,0,1]]
* 3 here equals the number of dimensions per vector
*/
/* Simple vector arithmetic */
function subtract(vectorX, vectorY, n) {
var result = new Array(n);
for (var i = 0; i < n; i++)
result[i] = vectorX[i] - vectorY[i];
return result;
}
function multiply(scalarC, vectorX, n) {
var result = new Array(n);
for (var i = 0; i < n; i++)
result[i] = scalarC * vectorX[i];
return result;
}
function dot(vectorX, vectorY, n) {
var sum = 0;
for (var i = 0; i < n; i++)
sum += vectorX[i] * vectorY[i];
return sum;
}
function norm(vectorX, n) {
return Math.sqrt(dot(vectorX, vectorX, n));
}
Note that the algorithm above computes the Gram-Schmidt orthogonalization, which is the matrix [e1 | e2 | ... | en], not the QR factorization!