Develop Reference

JavaScript Recursive function return undefined instead of an array

I have the next function:
function solveSudoku(prev_tab, fila, columna) {
let tab = _.cloneDeep(prev_tab);
let sig_fila = fila;
let sig_col = columna;
if (fila === 8 && columna === 8) {
//console.log(tab);
return tab;
}
if (columna === 8) {
sig_col = 0;
sig_fila = sig_fila + 1
} else {
sig_col = sig_col + 1;
}
if ((tab[fila][columna]) !== '') {
solveSudoku(tab, sig_fila, sig_col)
} else {
for (let num = 1; num <= 9; num++) {
if (numeroValido(tab, num, fila, columna)) {
tab[fila][columna] = num;
//tab.toString();
solveSudoku(tab, sig_fila, sig_col)
}
}
}
}
it returns undefined instead of a 2D array, i already try to add return in every recursive call =>
return solveSudoku( tab, sig_fila, sig_col )
but now that doesn't work either

I'm not really familiar with algorithms for solving sudoku, so I don't know if the algorithm below is correct.
But you need to ensure that the result of the recursion is returned. In my update below, I return the first recursive call. In the loop, I only return it if the recursion successfully found a solution, otherwise the loop continues trying other numbers in the column.
function solveSudoku(prev_tab, fila, columna) {
let tab = _.cloneDeep(prev_tab);
let sig_fila = fila;
let sig_col = columna;
if (fila === 8 && columna === 8) {
//console.log(tab);
return tab;
}
if (columna === 8) {
sig_col = 0;
sig_fila = sig_fila + 1
} else {
sig_col = sig_col + 1;
}
if ((tab[fila][columna]) !== '') {
return solveSudoku(tab, sig_fila, sig_col)
} else {
for (let num = 1; num <= 9; num++) {
if (numeroValido(tab, num, fila, columna)) {
tab[fila][columna] = num;
//tab.toString();
let result = solveSudoku(tab, sig_fila, sig_col);
if (result) { // continue searching if the recursion failed
return result;
}
}
}
}
}

What the meaning of return [''] in JavaScript?

Please see this leetcode solution. In the function it returns [''] which actually return an array of answer. Could someone tell me what's going on there?
[The problem is solved. Actually it will return in the middle of the code.]
https://leetcode.com/problems/remove-invalid-parentheses/discuss/154272/JavaScript-BFS-solution
function removeInvalidParentheses(s) {
let queue = new Set([s]);
while (queue.size) {
const next = new Set();
for (let v of queue) {
if (isValid(v)) {
return [...queue].filter(isValid);
}
for (let i = 0; i < v.length; i++) {
next.add(v.slice(0, i) + v.slice(i+1));
}
}
queue = next;
}
return [''];
}
function isValid(str) {
let bal = 0;
for (let ch of str) {
if (ch === '(') {
bal++;
} else if (ch === ')') {
bal--;
}
if (bal < 0) {
return false;
}
}
return bal === 0;
}

The function returns an array with a single empty string if the prior code (line 7) does not return a result. It is simply a default value so that calling code sees some result from the method.
function removeInvalidParentheses(s) {
let queue = new Set([s]);
while (queue.size) {
const next = new Set();
for (let v of queue) {
if (isValid(v)) {
return [...queue].filter(isValid);
}
for (let i = 0; i < v.length; i++) {
next.add(v.slice(0, i) + v.slice(i+1));
}
}
queue = next;
}
return [''];
}
function isValid(str) {
let bal = 0;
for (let ch of str) {
if (ch === '(') {
bal++;
} else if (ch === ')') {
bal--;
}
if (bal < 0) {
return false;
}
}
return bal === 0;
}

making custom validation for password field in react

I am making a custom registration page with only 2 values Email and Password, later I will add confirm password as well, for my password field I have some restrictions and I am using some regex and also some custom made code to make the validation.
this is my validateField:
validateField(fieldName, value) {
let fieldValidationErrors = this.state.formErrors;
let emailValid = this.state.emailValid;
let passwordValid = this.state.passwordValid;
//let passwordValidConfirm = this.state.passwordConfirmValid;
switch(fieldName) {
case 'email':
emailValid = value.match(/^([\w.%+-]+)#([\w-]+\.)+([\w]{2,})$/i);
fieldValidationErrors.email = emailValid ? '' : ' is invalid';
break;
case 'password':
passwordValid = (value.length >= 5 && value.length <= 32) && (value.match(/[i,o,l]/) === null) && /^[a-z]+$/.test(value) && this.check4pairs(value) && this.check3InRow(value);
fieldValidationErrors.password = passwordValid ? '': ' is not valid';
break;
default:
break;
}
this.setState({formErrors: fieldValidationErrors,
emailValid: emailValid,
passwordValid: passwordValid,
//passwordValidConfirm: passwordValidConfirm
}, this.validateForm);
}
as you can see for
passwordValid
I have made some methods, this one
check3InRow
doesnt work the way I want it to work, this one makes sure, you have at least 3 letters in your string that are in a row so like "abc" or "bce" or "xyz".
check3InRow(value){
var counter3 = 0;
var lastC = 0;
for (var i = 0; i < value.length; i++) {
if((lastC + 1) === value.charCodeAt(i)){
counter3++;
if(counter3 >= 3){
alert(value);
return true;
}
}
else{
counter3 = 0;
}
lastC = value.charCodeAt(i);
}
return false;
}
this doesnt work correctly so it should accept this:
aabcc
as a password but not:
aabbc

You are starting your counter from 0 and looking for greater than equal to 3 which will never be 3 for 3 consecutive characters. Rest everything is fine with your code.
check3InRow(value) {
var counter3 = 1;
var lastC = 0;
for (var i = 0; i < value.length; i++) {
if ((lastC + 1) === value.charCodeAt(i)) {
counter3++;
if (counter3 >= 3) {
alert(value);
return true;
}
} else {
counter3 = 1;
}
lastC = value.charCodeAt(i);
}
return false;
}

Can we not do a simple version of that function? Like
function check3InRow2(value){
for (var i = 0; i < value.length-2; i++) {
const first = value.charCodeAt(i);
const second = value.charCodeAt(i+1);
const third = value.charCodeAt(i+2);
if(Math.abs(second - first) === 1 && Math.abs(third-second) === 1){
return true;
}
}
return false;
}
I mean complexity wise it is O(N) so maybe we can give this a try
Also adding the your function. When you are AT a char then you should consider counter with 1. Because if another one matches it will be 2 consecutive values.
function check3InRow(value) {
var counter3 = 1;
var lastC = value.charCodeAt(0);
for (var i = 1; i < value.length; i++) {
if ((lastC + 1) === value.charCodeAt(i)) {
counter3++;
if (counter3 >= 3) {
return true;
}
} else {
counter3 = 1;
}
lastC = value.charCodeAt(i);
}
return false;
}

Stacks with Javascript

I am working with stacks on the infamous 'Braces' problem and I got to a halt. This should be an easy fix but my eyes are not much of a help at the moment.
The first call to the function is working like a charm but the second one is running an extra time and I can't see why.
The first call returns 01110 which is correct but the second returns 011110 which is not...
If you can't read the code here, got to the fiddle
//constructor function for our Stack class
function Stack() {
this.dataStore = [];
this.top = 0;
this.push = push;
this.pop = pop;
this.peek = peek;
this.length = length;
this.clear = clear;
}
function push(element) {
this.dataStore[this.top++] = element;
}
function pop() {
return this.dataStore[--this.top];
}
function peek() {
return this.dataStore[this.top - 1];
}
function length() {
return this.top;
}
function clear() {
this.top = 0;
}
function braces(expression) {
for (var i = 0; i < expression.length; i++) {
//if the number of elements in the expression is odd, it is guaranteed not to have a matching expression
//therefore we print 0
if (expression[i].length%2 !== 0) {
console.log(0);
} else {
var s = new Stack();
var startPoint = expression[i].charAt(0);
//if the expression starts with an open brace it means we will not have a matching expression so we print 0
if (startPoint == '(' || startPoint == '{' || startPoint == '[') {
for (var j = 0; j < expression[i].length; j++) {
var char = expression[i].charAt(j);
var h = '';
if (char == '(' || char == '{' || char == '[') {
s.push(char);
} else {
h = s.peek();
if (h == "(" && char == ")") {
s.pop();
} else if (h == "{" && char == "}") {
s.pop();
} else if (h == "[" && char == "]") {
s.pop();
}
}
}
} else {
console.log(0);
}
if (s.length() == 0) {
console.log(1)
} else {
console.log(0);
}
}
}
}
var expr = [ "}()()", "[]({})", "([])", "{()[]}", "([)]" ]; //working
var expr2 = [ "}()(){", "[]({})", "([])", "{()[]}", "([)]" ]; //running an extra time
braces(expr);

Change this:
else {
console.log(0);
continue; //this is new
}
if (s.length() == 0) {
Your function would log both 0 and 1/0 if the startpoint is not { or ( or [ and the length of s was 0

Your stack functions are all outside of the scope of Stack() and therefore the data probably won't be what you expect. You can start fixing this by putting functions inside the Stack() function:
function Stack() {
this.dataStore = [];
this.top = 0;
this.push = push;
this.pop = pop;
this.peek = peek;
this.length = length;
this.clear = clear;
this.pop = function () {
// pop
}
this.push = function () {
// code
}
this.peek = function () {
// code
}
}
That way, the methods all have access to the same data.

called function refuses to return true

Even when sum === largest in subset(), I don't get a "true" in my console. The true is sent back only to the calling function? Then how do I get the "return true" to behave as a newbie would expect?
function ArrayAdditionI(arr)
{
arr.sort();
var largest = arr.pop()
subset([], arr, largest);
}
function subset(soFar, rest, largest)
{
var sum = 0;
if (rest.length === 0)
{
for(var i=0; i<soFar.length; i++)
{
sum+= soFar[i];
}
if (sum === largest)
{
return true;
}
}
else
{
var soFar2 = soFar.slice(0);
soFar2.push(rest[0]);
subset(soFar,rest.slice(1),largest);
subset(soFar2, rest.slice(1),largest);
}
}
ArrayAdditionI([85,3,88,2])

function ArrayAdditionI(arr)
{
arr.sort();
var largest = arr.pop()
var ret = subset([], arr, largest);
// do something with ret
}
function subset(soFar, rest, largest)
{
var sum = 0;
if (rest.length === 0)
{
for(var i=0; i<soFar.length; i++)
{
sum+= soFar[i];
}
if (sum === largest)
{
return true;
}
}
else
{
var soFar2 = soFar.slice(0);
soFar2.push(rest[0]);
subset(soFar,rest.slice(1),largest);
subset(soFar2, rest.slice(1),largest);
}
return false;
}
ArrayAdditionI([85,3,88,2])