Exercise - find a passcode using dimensional arrays and function

user_17510308 0
Viewed 444

I´m stuck on this:

Write a function named getValidPassword that takes a two dimensional array as parameter.

Each entry in the first array represents a passcode. You need to find the passcode that has no odd digits and returns that passcode from your function. Here’s an example:

var loggedPasscodes =[
[1, 4, 4, 1],
[1, 2, 3, 1],
[2, 6, 0, 8],
[5, 5, 5, 5],
[4, 3, 4, 3]
];
getValidPassword(loggedPasscodes) // returns the array: [2, 6, 0, 8]

Tried this way, but error message shows: >>>>Code is incorrect Function getValidPassword is not working as requested." [2,6,0,8].

`var loggedPasscodes=[
    [1, 4, 4, 1],
[1, 2, 3, 1],
[2, 6, 0, 8],
[5, 5, 5, 5],
[4, 3, 4, 3],
];
var getValidPassword = function(getPassword){
var passcode = [];
var cache = [];
for (var i = 0; i < getPassword.length; i++){
for (var j = 0; j < getPassword[i].length; j++){
if(getPassword[i][j] % 2 === 0){           cache.push(getPassword[i][j]);
}
else {break;
}
if(getPassword[i].length === cache.length){
passcode= cache.slice();
}
}
}
return passcode;
};
console.log('[' + getValidPassword(loggedPasscodes) + ('') + ']');`

Also this:

function getValidPassword(loggedPasscodes) {
    return loggedPasscodes.filter(passcode => passcode.every(n => n % 2 === 0));
};

let loggedPasscodes =[
    [1, 4, 4, 1],
    [1, 2, 3, 1],
    [2, 6, 0, 8],
    [5, 5, 5, 5],
    [10, 2, 4, 42],
    [4, 3, 4, 3]
];

console.log(getValidPassword(loggedPasscodes));
and


function getValidPassword(loggedPasscodes) {
    for (let i = 0; loggedPasscodes.length > i; i++) {
        let passcode = loggedPasscodes[i];
        let temImpar = false;
        for (let j = 0; passcode.length > j; j++) {
            if (passcode[j] % 2 !== 0)  { // ímpar
                temImpar = true;
                 números)
                break;
            }
        }
        if (! temImpar) {
            return passcode; 
    }
};



 let loggedPasscodes =[
        [1, 4, 4, 1],
        [1, 2, 3, 1],
        [2, 6, 0, 8],
        [5, 5, 5, 5],
        [4, 3, 4, 3]
    ];
    
    console.log(getValidPassword(loggedPasscodes)); // [2, 6, 0, 8

]

Also this:

function getValidPassword(loggedPasscodes) {
let codes = [];
for (let i = 0; loggedPasscodes.length > i; i++) {
    let passcode = loggedPasscodes[i];
    let temImpar = false;
    for (let j = 0; passcode.length > j; j++) {
        if (passcode[j] % 2 !== 0)  { // ímpar
            temImpar = true;
           
            break;
        }
    }
    if (! temImpar) {
        codes.push(passcode); 
    }
}
return codes;
};

let loggedPasscodes =[
    [1, 4, 4, 1],
    [1, 2, 3, 1],
    [2, 6, 0, 8],
    [5, 5, 5, 5],
    [4, 3, 4, 3]
];

console.log(getValidPassword(loggedPasscodes))

This one:

function getValidPassword(loggedPasscodes) {
    return loggedPasscodes.find(passcode => passcode.every(n => n % 2 === 0));
};

let loggedPasscodes =[
    [1, 4, 4, 1],
    [1, 2, 3, 1],
    [2, 6, 0, 8],
    [5, 5, 5, 5],
    [4, 3, 4, 3]
];

console.log(getValidPassword(loggedPasscodes)); // [2, 6, 0, 8]

Also try to replicate something like this example:

function retornaNNumerosPares(n) {
let numerosPares = [];
for (let i = 0; numerosPares.length < n; i++) {
    if (i % 2 == 0) {
        numerosPares.push(i);
    }
}
return numerosPares;

}

console.log(retornaNNumerosPares(5));

But none of the above are working. Can you please help me, I´m almost finnished all the exercices.

Can you please help?

multidimensional-array arrays javascript
4 Answers

I'd start by checking the first value of each array to knock out the uninteresting values. Any matches do my second loop and upon any detection it's even cancel out.

If a complete match is made you'd then return the result as show in comment or push it and continue finding more codes.

let loggedPasscodes = [
  [1, 4, 4, 1],
  [1, 2, 3, 1],
  [2, 6, 0, 8],
  [5, 5, 5, 5],
  [4, 3, 4, 3]
];

function getValidPassword(loggedPasscodes) {
  let Codes = [];
  for (let i = 0; i < loggedPasscodes.length; i++) {
    console.log("Read Array:", i);
    if (loggedPasscodes[i][0] % 2 == 0) { // Check if first value is even
      let GoodCode = true;
      for (let x = 1; x < 4; x++) { // Let's fast check the array now
        if (loggedPasscodes[i][x] % 2 != 0) {
          GoodCode = false;
          break; // Break if it's a waste of time (odds)
        }
      }
      //if (GoodCode == true) Codes.push(loggedPasscodes[i]); // Build our array if there's more codes even.
      if (GoodCode == true) return loggedPasscodes[i];
    }
  }
  return Codes;
};
console.log(getValidPassword(loggedPasscodes)); // [2, 6, 0, 8]

user_6058500

A fast approach takes every chance to end an unnecessary loop.

This approach iterate the array of codes and iterates the code as well by check unwanted numbers, then perform a continue of the outer loop, otherwise return the found array at the end of the inner loop.

const
    getValidPassword = codes => {
        outer: for (const code of codes) {
            for (const value of code) if (value % 2) continue outer;
            return code;
        }
    },
    loggedPasscodes = [[1, 4, 4, 1], [1, 2, 3, 1], [2, 6, 0, 8], [5, 5, 5, 5], [4, 3, 4, 3]];

console.log(getValidPassword(loggedPasscodes));

user_14476750

It was tough but I finally got it accepted, even though I had the same solution with many other versions.

Nevertheless I like Nina Scholz version better for being really really short and elegant.

Here's my working code:

var getValidPassword = function(pass) {
    var buffer = [];
    var evenArrays = [];
    
    for (var i = 0; i < pass.length; i++) {
        for (var j = 0; j < pass[i].length; j++) {
            // Testing if cell is odd or even
            if (pass[i][j]%2 !== 0) {
                break;
            } else {
                buffer.push(pass[i][j]);
            }
        }
        // Before changing line pass buffer to the evenArray if buffer is complete
        if(pass[i].length === buffer.length) {
            evenArrays = buffer;
        }
        buffer=[];
    }
    return evenArrays;
};

var loggedPasscodes = [
    [1, 4, 4, 1],
    [1, 2, 3, 1],
    [2, 6, 0, 8],
    [5, 5, 5, 5],
    [4, 3, 4, 3]
];

console.log(getValidPassword(loggedPasscodes));

user_27897300

The problem is that you were expected to return the array, but you returned a copy of the array instead. When comparing two arrays directly, Java only compares the memory address.

int[] a=new int[] {2,6,0,8};
int[] b=new int[] {2,6,0,8};

if (a!=b) System.out.println("Error. I was expecting [2,6,0,8]").

This comparison would work because a and b reference the same memory address.

int[] a=new int[] {2,6,0,8};
int[] b=a;

if (a==b) System.out.println("Your answer is correct.");
user_10318870
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