I am currently working on codeacademy's 'Credit Card Checker' using Luhn algorithm, however my code is returning false on valid arrays(which should return). Could you help me resolve the issue?
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = arr => {
let totalSum = 0;
let revList = arr.reverse();
for (let i = 0; i < revList.length; i++) {
let calcAmount = revList[i];
if (i !== 0 && i%2 === 0) {
calcAmount = revList[i] * 2;
if (calcAmount > 9 ) {
calcAmount -= 9;
totalSum += calcAmount;
} else {
totalSum += calcAmount;
}
} else {
totalSum += revList[i];
}
}
return (totalSum%10 === 0 ? true : false);
};
console.log(validateCred(valid3))
console.log(validateCred(valid4))
console.log(validateCred(valid5))
A simple mistake, check if (i%2 === 1) {, algorithm needs to multiply 2 for every second elements
arr.reverse() is also bad idea, original array is changed.
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = arr => {
let totalSum = 0;
let revList = arr.reverse();
for (let i = 0; i < revList.length; i++) {
let calcAmount = revList[i];
if (i%2 === 1) {
calcAmount = revList[i] * 2;
if (calcAmount > 9 ) {
calcAmount -= 9;
totalSum += calcAmount;
} else {
totalSum += calcAmount;
}
} else {
totalSum += revList[i];
}
}
return(totalSum%10 === 0 ? true : false);
};
batch.forEach(c => console.log(validateCred(c)));
Rewrite and fixed.
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
function luhnCheck(num) {
let digit, j, len, odd = true, sum = 0
const digits = (num + '').split('').reverse()
for (j = 0, len = digits.length; j < len; j++) {
digit = parseInt(digits[j], 10)
if ((odd = !odd)) digit *= 2
if (digit > 9) digit -= 9
sum += digit
}
return sum % 10 === 0
}
batch.forEach((item) => {
let number = item.join('')
console.log(number, luhnCheck(number))
})
Ideally, you should pass it in as a string and let the function do the splitting (separation of concerns) as in a real app your want to do various things on the entered number before running luhn like.. cardTypeFromNumber (visa, maestro, dankort, mastercard, amex, dinersclub, discover, chinaunionpay (doesn't use luhn), jcb mostly have all different lengths you wouldn't want to run luhn until you validate the length and format), formatCardNumber, validateCardExpiry, validateCardCVC to name a few.
The Luhn algorithm is explained on Wikipedia. Others have answered the question, this is just a different implementation.
function luhnCheck(cardNumber) {
let nums = cardNumber.split('').reverse();
let checkValue = nums.shift();
let luhnSum = nums.reduce((sum, n, i) => {
let val = n*(i % 2? 1 : 2);
sum += val > 9? val - 9 : val;
return sum;
}, 0);
return checkValue == 10 - (luhnSum % 10);
}
['4539677908016808', // valid1
'5535766768751439', // valid2
'4532778771091795', // invalid1
'5795593392134643', // invalid2
'344801968305414', // mystery1
'5466100861620239', // mystery2
].forEach(cardNum =>
console.log(luhnCheck(cardNum))
);
Related
I am learning Javascript and currently found myself with the following problem. I need to get each individual 3x3 block of a 9x9 2d array as a string, separated by commas.
What I mean is, for example, let's say I have the following array:
var 2dArray = [
[5, 3, 5, 6, 7, 8, 9, 1, 2],
[6, 7, 1, 1, 9, 5, 3, 4, 8],
[1, 9, 2, 3, 4, 2, 5, 6, 7],
[8, 5, 9, 7, 6, 1, 4, 2, 3],
[4, 2, 6, 5, 5, 3, 7, 9, 1],
[7, 1, 3, 1, 2, 4, 8, 5, 6],
[9, 6, 1, 5, 3, 7, 2, 8, 4],
[2, 8, 7, 5, 1, 9, 6, 3, 5],
[3, 4, 5, 6, 8, 6, 1, 7, 9] ]
The result should be something like 535671192,678195342, 912348657, ... and so on, until the string is made of all the 3x3 blocks.
I thought that making nested for loops would be the best approach, but got confused along the way and I would appreciate your help.
Thank you.
You can use two loops to iterate over the positions of all the possible top left corners and another two loops to get all the elements in that square.
var arr = [
[5, 3, 5, 6, 7, 8, 9, 1, 2],
[6, 7, 1, 1, 9, 5, 3, 4, 8],
[1, 9, 2, 3, 4, 2, 5, 6, 7],
[8, 5, 9, 7, 6, 1, 4, 2, 3],
[4, 2, 6, 5, 5, 3, 7, 9, 1],
[7, 1, 3, 1, 2, 4, 8, 5, 6],
[9, 6, 1, 5, 3, 7, 2, 8, 4],
[2, 8, 7, 5, 1, 9, 6, 3, 5],
[3, 4, 5, 6, 8, 6, 1, 7, 9]
];
let res = [];
for (let i = 0; i < 9; i += 3) {
for (let j = 0; j < 9; j += 3) {
let curr = "";
for (let k = 0; k < 3; k++) {
for (let l = 0; l < 3; l++) {
curr += arr[i + k][j + l];
}
}
res.push(curr);
}
}
console.log(res);
The issue can also be solved by loops but it's good practice to solve it using the Array.map() function.
let arrayRow = "";
const resultArray = Array2d.map((element, index) => {
return arrayRow + element.map((element, i) => {
return element.toString();
});
});
you should try like this too.
var arr = [
[5, 3, 5, 6, 7, 8, 9, 1, 2],
[6, 7, 1, 1, 9, 5, 3, 4, 8],
[1, 9, 2, 3, 4, 2, 5, 6, 7],
[8, 5, 9, 7, 6, 1, 4, 2, 3],
[4, 2, 6, 5, 5, 3, 7, 9, 1],
[7, 1, 3, 1, 2, 4, 8, 5, 6],
[9, 6, 1, 5, 3, 7, 2, 8, 4],
[2, 8, 7, 5, 1, 9, 6, 3, 5],
[3, 4, 5, 6, 8, 6, 1, 7, 9]
];
let res = [];
let i = 0,j = 0;
let justOneStr = "";
while(i < 9 && j < 9){
justOneStr += arr[i][j];
if(i % 3 == 2 && j % 3 == 2){
res.push(justOneStr);
justOneStr="";
}
if((j % 9 == 8 || j % 3 == 2) && i % 3 != 2){
j -= 3;
i += 1;
}
else if(j % 9 == 8 && i % 3 == 2){
j -= 9;
i += 1;
}
else if(i % 3 == 2 && j % 3 == 2){
i -= 2;
}
j++;
}
console.log(res);
Here's a fairly straightforward nested reduce() that returns an array of matrices of the specified size. It touches each element once, and uses the coordinates in the passed matrix to determine which sub-matrix and sub-matrix row to accumulate into.
const
matrix = [
[5, 3, 5, 6, 7, 8, 9, 1, 2],
[6, 7, 1, 1, 9, 5, 3, 4, 8],
[1, 9, 2, 3, 4, 2, 5, 6, 7],
[8, 5, 9, 7, 6, 1, 4, 2, 3],
[4, 2, 6, 5, 5, 3, 7, 9, 1],
[7, 1, 3, 1, 2, 4, 8, 5, 6],
[9, 6, 1, 5, 3, 7, 2, 8, 4],
[2, 8, 7, 5, 1, 9, 6, 3, 5],
[3, 4, 5, 6, 8, 6, 1, 7, 9]],
subdivide = (matrix, width, height) => {
return matrix.reduce((acc, row, i) => {
row.reduce((_acc, x, j, row) => {
const grid = Math.floor(j / width) + Math.floor(i / height) * (Math.ceil(row.length / width));
const gridRow = i % height;
_acc[grid] ??= [];
(_acc[grid][gridRow] ??= []).push(x);
return _acc;
}, acc)
return acc;
}, [])
},
subdividedmatrix = subdivide(matrix, 3, 3);
// log submatrices
subdividedmatrix.forEach(m => console.log(JSON.stringify(m)));
// map to strings
console.log(subdividedmatrix.map(submatrix => submatrix.flat().join('')))
.as-console-wrapper { max-height: 100% !important; top: 0; }
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I need this for a codewars challenge to make my code DRY
I have an array of arrays, lets say a 9x9 matrix.
const sudokuTest2= [
[5, 3, 4, 6, 7, 8, 9, 1, 2],
[6, 7, 2, 1, 9, 0, 3, 4, 9],
[1, 0, 0, 3, 4, 2, 5, 6, 0],
[8, 5, 9, 7, 6, 1, 0, 2, 0],
[4, 2, 6, 8, 5, 3, 7, 9, 1],
[7, 1, 3, 9, 2, 4, 8, 5, 6],
[9, 0, 1, 5, 3, 7, 2, 1, 4],
[2, 8, 7, 4, 1, 9, 6, 3, 5],
[3, 0, 0, 4, 8, 1, 1, 7, 9],
];
How can I divide it automatically into 9 equal areas, something which results from the following sample code:
const areas = [];
areas[0] = sudokuTest2[0]
.slice(0, 3)
.concat(sudokuTest2[1].slice(0, 3))
.concat(sudokuTest2[2].slice(0, 3));
areas[1] = sudokuTest2[3]
.slice(0, 3)
.concat(sudokuTest2[4].slice(0, 3))
.concat(sudokuTest2[5].slice(0, 3));
areas[2] = sudokuTest2[6]
.slice(0, 3)
.concat(sudokuTest2[7].slice(0, 3))
.concat(sudokuTest2[8].slice(0, 3));
areas[3] = sudokuTest2[0]
.slice(3, 6)
.concat(sudokuTest2[1].slice(3, 6))
.concat(sudokuTest2[2].slice(3, 6));
areas[4] = sudokuTest2[3]
.slice(3, 6)
.concat(sudokuTest2[4].slice(3, 6))
.concat(sudokuTest2[5].slice(3, 6));
areas[5] = sudokuTest2[6]
.slice(3, 6)
.concat(sudokuTest2[7].slice(3, 6))
.concat(sudokuTest2[8].slice(3, 6));
areas[6] = sudokuTest2[0]
.slice(6, 9)
.concat(sudokuTest2[1].slice(6, 9))
.concat(sudokuTest2[2].slice(6, 9));
areas[7] = sudokuTest2[3]
.slice(6, 9)
.concat(sudokuTest2[4].slice(6, 9))
.concat(sudokuTest2[5].slice(6, 9));
areas[8] = sudokuTest2[6]
.slice(6, 9)
.concat(sudokuTest2[7].slice(6, 9))
.concat(sudokuTest2[8].slice(6, 9));
For sure I could use some nested loops, but I'm curious if there is any solution with array methods.
What will be the best solution, what do you think?
You could take generate the wanted 3x3 parts by using a nested mapping.
const
sudoku = [[5, 3, 4, 6, 7, 8, 9, 1, 2], [6, 7, 2, 1, 9, 0, 3, 4, 9], [1, 0, 0, 3, 4, 2, 5, 6, 0], [8, 5, 9, 7, 6, 1, 0, 2, 0], [4, 2, 6, 8, 5, 3, 7, 9, 1], [7, 1, 3, 9, 2, 4, 8, 5, 6], [9, 0, 1, 5, 3, 7, 2, 1, 4], [2, 8, 7, 4, 1, 9, 6, 3, 5], [3, 0, 0, 4, 8, 1, 1, 7, 9]],
areas = Array
.from({ length: 3 })
.flatMap((_, i) => Array
.from(
{ length: 3 },
(__, j) => Array
.from({ length: 3 })
.flatMap((___, k) => sudoku[j * 3 + k].slice(i * 3, (i + 1) * 3)
)
));
areas.forEach(a => console.log(...a));
This can be done in with single forEach method. The idea behind this is dynamic indexing.
const matrix = [ [5, 3, 4, 6, 7, 8, 9, 1, 2], [6, 7, 2, 1, 9, 0, 3, 4, 9], [1, 0, 0, 3, 4, 2, 5, 6, 0], [8, 5, 9, 7, 6, 1, 0, 2, 0], [4, 2, 6, 8, 5, 3, 7, 9, 1], [7, 1, 3, 9, 2, 4, 8, 5, 6], [9, 0, 1, 5, 3, 7, 2, 1, 4], [2, 8, 7, 4, 1, 9, 6, 3, 5], [3, 0, 0, 4, 8, 1, 1, 7, 9],];
let areas = new Array(9).fill([]);
matrix.forEach((numbers, idx)=> {
let segment = Math.floor(idx / 3);
areas[segment] = [...areas[segment], ...matrix[idx].slice(0, 3)];
areas[segment + 3] = [...areas[segment + 3], ...matrix[idx].slice(3, 6)];
areas[segment + 6] = [...areas[segment + 6], ...matrix[idx].slice(6, 9)];
})
areas.forEach(area => console.log(...area));
This is the Python code which I want to convert to Javascript!
y[i] = x[i][:j] + x[i][j] + x[i][j + 1:]
I tried using the slice function but I cannot correctly implement it for a 2d array.
So that's just some python fancy work for list comprehension but I believe you could accomplish such a thing with:
const x = [
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9]
];
var y = new Array(81);
for (i = 0; i < x.length; i++) {
var row = new Array(20);
for (j = 0; j < x.length; j++) {
row.push(x[i].slice(j).concat(x[i][j]).concat(x[i].slice(0, j+1)))
}
y.push(row);
}
for(k = 0; k < y.length; k++) {
console.log(y[k]);
}
I'm not exceptionally good with javascript, but let me know if that helps.
-Or-
provide some more input/output so I can test with, any additional code, what you've tried or what you're expected output may look like.
I got stuck while practicing credit card checker practice.
My code:
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8]
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9]
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6]
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5]
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6]
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5]
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3]
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4]
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5]
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4]
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4]
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9]
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3]
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3]
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3]
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5]
// Add your functions below:
const validateCred = (array)=> {
let tempArrSub = []; //Holds values of the 9 subtracted from doubled elements bigger than 9
let tempArr = array; //Copies the values of the array passed into parameters
tempArr.pop();
tempArr.reverse();
for (let i = tempArr.length-1; i >=0; i-=2){ //Doubles every two elements from right to left
tempArr[i] *= 2;
}
for (let k = 0; k < tempArr.length; k++) { //Subtract 9 from every second element (right to left) if bigger than 9
if (tempArr[k] > 9){
tempArrSub.push(tempArr[k] - 9);
}
else {
tempArrSub.push(tempArr[k]);
}
}
let tempArrSum = 0;
for (let m = 0; m < tempArrSub.length; m++){ //Calculates the sum of all elements in the array
tempArrSum += tempArrSub[m];
}
tempArrSum += array.pop(); //Adds the last digit of initial array to the sum
if (tempArrSum % 10 === 0) { //Returns true if the sum is divisible by 10
return true;
}
else {
return false;
}
} //End Of Function
const findInvalidCards = (nestedArray) => {
let invalidCards = [];
let validCards = [];
for (let a = 0; a < nestedArray.length; a++){
if ( validateCred(nestedArray[a]) == true ) {
validCards.push(nestedArray[a]);
}
else {
invalidCards.push(nestedArray[a]);
}
}
console.log("Invalid cards: \n" + invalidCards);
console.log("Valid cards: \n" + validCards);
}
I checked all arrays one by one passing as an argument to validateCred() function. It's working and returning true or false for each credit card.
In order to automate process I wanted findInvalidCards() to find valid or invalid cards by calling validateCred() function inside itself and returning boolean value for each card.
Here I got stuck. Because it is returning true for the 1st card and false for the remaining cards. I've been playing with the code for the whole day, but I could not move forward. I rely on your help. Thanks in advance
I don't know if is it helpful to you or not. But I rewrite your code with an optimal way.
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8]
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9]
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6]
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5]
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6]
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5]
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3]
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4]
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5]
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4]
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4]
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9]
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3]
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3]
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3]
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
isValidCard = (array) => {
let copy = [...array];
const last = copy.pop();
copy.reverse();
copy = copy.map((dig, i) => ((i % 2 === 0 ? dig * 2 : dig)));
copy = copy.map(dig => (dig > 9 ? dig - 9 : dig));
const sum = copy.reduce((acc, curr) => acc + curr);
return ((sum + last) % 10 === 0);
}
findInvalidCards = (array) => {
const validList = [];
const invalidList = [];
array.forEach(card => {
if (isValidCard(card)) {
validList.push(card);
} else {
invalidList.push(card);
}
});
console.log('valid cards', validList);
console.log('invalid cards', invalidList);
}
findInvalidCards(batch);
I think your issue is this line:
let tempArr = array; //Copies the values of the array passed into parameters
According to your comment you want to copy the array, but just assigning it to another variable does not copies it. In order to copy just so something like this:
let copy = array.slice(0);
/* or (will not work for large arrays) */
let copy = [...array];
// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
const invalidCredit = [invalid1, invalid2, invalid3, invalid4, invalid5]
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = (arr) => {
let reversedArr = arr.reverse();
let oddOccurrences = [];
reversedArr.forEach((value, i) => i % 2 ? oddOccurrences.push(value *= 2) : oddOccurrences.push(value));
const checking = oddOccurrences.map((greater) => (greater > 9 ? (greater -= 9) : greater));
const sum = checking.reduce((acc, num) => {
return acc + num
}, 0)
// Conditional check
if (sum % 10 === 0)
return 'valid'
else return 'invalid'
}
console.log(validateCred([4, 5, 3, 9, 6, 8, 9, 8, 8, 7, 7, 0, 5, 7, 9, 8]));
findInvalidCards = (newArr) => {
const invalidList = [];
const validList = [];
for (let i = 0; i < newArr.length; i += 15) {
// inner loop
for (let j = 0; j < newArr.length; j++) {
validateCred(newArr[j]) === 'invalid' ? invalidList.push(newArr[j]) : validList.push(newArr[j]);
}
};
console.log(invalidList)
}
findInvalidCards(batch);
In javascript I have an array as follows:
var foo = [2, 2, 4, 4, 128, 2, 2, 1, 4, 18, 27, 16, 2, 1, 18, 21, 5, 1, 128, 1, 2, 2, 1, 18, 12, 60, 2, 28, 1, 17, 2, 3, 4, 2, 2, 2, 1, 27, 2, 17, 7, 2, 2, 2, 5, 1, 2, 4, 7, 1, 2, 1, 1, 1, 2, 1, 5, 7, 2, 7, 6, 1, 7, 1, 5, 8, 4];
And I am interested in finding a way (within one loop, not multiple) to derive a subset array of the highest 10 values, where the previous position of the value is the 'key' (so simulating a Map object):
eg:
var fooTopTen = [[4, 128], [18, 128], [25, 60], [27, 28], [10, 27], [37, 27], [15, 21], [9, 18], [14, 18], [23, 18]];
My previous answer used a reverse index table, but contained some bugs - which are now fixed - and was harder to understand than the following code.
This is actually the slowest of all solutions given in the answers - for maximum performance, check my other answer.
var foo = [2, 2, 4, 4, 128, 2, 2, 1, 4, 18, 27, 16, 2, 1, 18, 21, 5, 1, 128, 1, 2, 2, 1, 18, 12, 60, 2, 28, 1, 17, 2, 3, 4, 2, 2, 2, 1, 27, 2, 17, 7, 2, 2, 2, 5, 1, 2, 4, 7, 1, 2, 1, 1, 1, 2, 1, 5, 7, 2, 7, 6, 1, 7, 1, 5, 8, 4];
var fooTopTen = [];
// add index to values
for(var i = 0, len = foo.length; i < len; ++i)
fooTopTen.push([i, foo[i]]);
// sort first by value (descending order), then by index (ascending order)
fooTopTen.sort(function(t1, t2) {
return t2[1] - t1[1] || t1[0] - t2[0];
});
// shorten array to correct size
fooTopTen.length = 10;
// output top ten to check result
document.writeln('[[' + fooTopTen.join('], [') + ']]');
The second part of the comparison function (the one comparing the indices) is not needed, as sort() is stable in most implementations (this isn't required by ECMA according to MDC). I'll leave it in as an example to how sorting with multiple requirements can be done...
This runs once through the main array it searches, inserting items at the appropriate place in the results array:
function top10(arr) {
var results = [[0,Number.MAX_VALUE],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0]];
for (var i=0; i<arr.length; i++) {
// search from back to front
for (var j=9; j>=0; j--) {
if (arr[i] <= results[j][1]) {
if (j==9)
break;
results.splice(j+1, 0, [i, arr[i]]);
results.pop();
break;
}
}
}
return results.slice(1);
}
For large arrays this should even be rather fast, since most times the inner loop should only do one iteration.
Here's the de-bugged version of my previous answer using an index table. I did a little benchmarking and for the input given in the question, this solition will be faster than anything else which has been suggested in this thread till now:
var foo = [2, 2, 4, 4, 128, 2, 2, 1, 4, 18, 27, 16, 2, 1, 18, 21, 5, 1, 128, 1, 2, 2, 1, 18, 12, 60, 2, 28, 1, 17, 2, 3, 4, 2, 2, 2, 1, 27, 2, 17, 7, 2, 2, 2, 5, 1, 2, 4, 7, 1, 2, 1, 1, 1, 2, 1, 5, 7, 2, 7, 6, 1, 7, 1, 5, 8, 4];
var indexTable = {}, uniqueValues = [];
// --- build reverse index table, find unique values
for(var i = foo.length; i--; ) {
var value = foo[i];
if(indexTable.hasOwnProperty(value))
indexTable[value].push(i);
else {
indexTable[value] = [i];
uniqueValues.push(value);
}
}
// --- sort unique values in ascending order
uniqueValues.sort(function(i1, i2) {
return i1 - i2;
});
// --- find ten greatest values
var fooTopTen = [], k = 0;
for(var i = uniqueValues.length; k < 10 && i--; ) {
var value = uniqueValues[i],
indices = indexTable[value];
for(var j = indices.length; k < 10 && j--; )
fooTopTen[k++] = [indices[j], value];
}
// --- output result
document.writeln('[[' + fooTopTen.join('], [') + ']]');
var foo = [2, 2, 4, 4, 128, 2, 2, 1, 4, 18, 27, 16, 2, 1, 18, 21, 5, 1, 128, 1, 2, 2, 1, 18, 12, 60, 2, 28, 1, 17, 2, 3, 4, 2, 2, 2, 1, 27, 2, 17, 7, 2, 2, 2, 5, 1, 2, 4, 7, 1, 2, 1, 1, 1, 2, 1, 5, 7, 2, 7, 6, 1, 7, 1, 5, 8, 4];
var index = 0;
var result = foo.map( function(a){ return [index++, a]; } )
.sort( function(a,b){ return (a[1] < b[1]); } )
.splice( 0, 10 );
document.write(result.join( ' ' ));
If foo is very large compared to the size of result required, it may be quicker to iterate over foo insertion-sorting each element into result as we come across it.
// Sorting method
function sortNumber(a, b) {
return a - b;
}
// Find the offset of an element in array
function findOffset(element, array) {
for (var i = 0; i < array.length; i++) {
if (array[i] == element) {
// Make sure we don't find it again
array[i] = null;
return i;
}
}
}
var foo = [2, 2, 4, 4, 128, 2, 2, 1, 4, 18, 27, 16, 2, 1, 18, 21, 5, 1, 128, 1, 2, 2, 1, 18, 12, 60, 2, 28, 1, 17, 2, 3, 4, 2, 2, 2, 1, 27, 2, 17, 7, 2, 2, 2, 5, 1, 2, 4, 7, 1, 2, 1, 1, 1, 2, 1, 5, 7, 2, 7, 6, 1, 7, 1, 5, 8, 4];
// Copies
var bar = foo.slice();
var baz = foo.slice();
var fooTopTen = new Array(10);
// Sort
bar.sort(sortNumber).reverse();
// Create the results
for (var i = 0; i < 10; i++) {
fooTopTen[i] = new Array(2);
fooTopTen[i][0] = findOffset(bar[i], baz);
fooTopTen[i][1] = bar[i];
}
computer-science-y answer:
problem statement: Given a large array X of length N, and a small number m < N (here m=10), produce an array Y of length m where each element of Y contains the pair {i,X[i]} such that the set of X{i} are the m largest elements of X.
If m is much smaller than N, then loop over elements of X and sort them into Y, discarding pairs to maintain at most m elements. (i.e. as moonshadow mentioned)
Sorting X will cost you O(N log N) elements. Iterating over X and sorting into Y should cost you only O(N log m) elements.