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How to split array based on length and removing first item and adding new item to the end

I have an array [1,2,3,4,5,6,7,8,9].
I want it to split as below using JavaScript:
[[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8],[5,6,7,8,9]]

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const len = 5;
const result = [...Array(arr.length - len + 1)].map((_, i) => arr.slice(i, len + i));
console.log(result);

You can accomplish this using multiple slices.
const getSlices = (arr, len) => {
const res = [];
for(let i = 0; i <= arr.length - len; i++){
res.push(arr.slice(i, i + len));
}
return res;
};
console.log(getSlices([1,2,3,4,5,6,7,8,9], 5));

const split = (arr, l) => arr.reduce((res, el, i) => {
const slice = arr.slice(i, i + l);
if(slice.length == l) res.push(slice);
return res;
}, []);
console.log(split([1,2,3,4,5,6,7,8,9], 5))

let arr = [1,2,3,4,5,6,7,8,9];
function mySplit(arr, n) {
let result = [];
for(let i=0; i<=arr.length-n; i++) {
result.push(arr.slice(i, i+n));
}
return result;
}
console.log( mySplit(arr, 5) );

Find duplicate pairs in an array

I have an array a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7]
I want to fetch all the duplicate pair in this array list.
Since there are pairs of 2 and 7 the output should be -
Output: [2, 7]
I tried writing my own logic but I am very weak in that area. Can somebody help?
function getDuplicateArrayElements(arr){
let sorted_arr = arr.slice().sort();
let results = [];
for (let i = 0; i < sorted_arr.length; i++) {
let matchingElementCount = 1;
for (let j = i + 1; j < sorted_arr.length - i; j++) {
if (sorted_arr[j] === sorted_arr[i]) {
++matchingElementCount;
} else {
if(matchingElementCount % 2 === 0) {
results.push(sorted_arr[i]);
}
i = j - 1;
break;
}
}
}
return results; } var a = [1,1,1,2,2,3,3,3,4,6,6,6,7,7]; var duplicateValues= getDuplicateArrayElements(a);

You can achieve your result by using reduce and forEach.
const arr = [1,1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
// Generate a hashmap from the given array for counting the frequency.
const hashMap = arr.reduce((a, c) => {
a[c] = (a[c] || 0) + 1;
return a;
}, {});
const pair = [];
// If the frequency is divided by 2 then push the key of the hashMap into pair array.
Object.entries(hashMap).forEach(([k, v]) => {
if (v % 2 === 0) {
[...Array(Math.floor(v / 2))].forEach(_ => pair.push(k));
}
})
console.log(pair);

You can grab the frequency of each number, and then filter out any which have an odd frequency. You can then .flatMap() the frequencies to an array containing your number for each pair you found like so:
const a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
const freq = a.reduce((m, n) => m.set(n, (m.get(n) || 0) + 1), new Map);
const res = [...freq].filter(([n, count]) => count % 2 == 0).flatMap(([n, c]) => Array(c/2).fill(n));
console.log(res);
This way, if you have four 1s (ie: two pairs of 1s), the filter will pick up on that, allowing you to flat-map the [1, 4] array to an array of [1, 1], which is merged into the larger resulting array.

You could create a helper map and keep the counts of each number as the values and the numbers itself as the keys. After iterating through the array, you only need to find the ones with a count divisible by 2:
var a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 7, 7]
function findDuplicates(arr) {
const map = {};
for (const curr of arr) {
if (!map[curr]) {
map[curr] = 0;
}
map[curr]++;
}
const res = [];
for (const key in map) {
if (map.hasOwnProperty(key) && map[key] % 2 === 0) {
res.push(Number.parseInt(key));
}
}
return res;
}
console.log(findDuplicates(a));

You can first count the occurrence of each numbers and if it is greater than 0 and divisible by 2 then add these to final result else don't
function getDuplicateArrayElements(arr) {
let map = {}
let results = [];
for (let num of arr) {
map[num] = map[num] || 0
map[num]++
}
return Object.keys(map)
.filter(v => map[v] && map[v] % 2 === 0)
.map(v => new Array(map[v]/2).fill(+v))
.flat()
.sort()
}
var a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 7, 7,8,8,8,8];
var duplicateValues = getDuplicateArrayElements(a);
console.log(duplicateValues)

const a = {};
[1,1,1,2,2,3,3,3,4,6,6,6,7,7].forEach(v => {a[v] = a[v] ? a[v] + 1 : 1});
const l = [];
Object.keys(a).forEach(k => !(a[k] % 2) && l.push(k));

Here you go:
function getDuplicateArrayElements(arr){
var dupilcates=arr.filter(x => arr.filter(y=>y==x).length==2);
var found=[];
for(var i=0;i<dupilcates.length;i=i+2)
found.push(dupilcates[i]);
return found;
}

This will give you the desired pairs. with [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 6, 7, 7] input it will return [1,1,2,6,6,7]:
function getDuplicateArrayElements(arr){
let sorted_arr = arr.slice().sort();
let results = [];
let i = 0;
while (i < sorted_arr.length) {
let counter = 1;
let j = i;
while (sorted_arr[j] === sorted_arr[j+1]) {
counter++;
j++;
}
if (counter%2 == 0) {
results.push(...Array(counter/2).fill(sorted_arr[i]))
}
i += counter;
}
return results;
}
var a = [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 6, 7, 7];
console.log(getDuplicateArrayElements(a));

Another rather concise solution:
a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
uniques = new Set(a); //filter out duplicates
res = [];
uniques.forEach((key)=>{
if(a.filter(elem => elem === key).length === 2){res.push(key)};
//filter out only the elements which match the key being tested
//if there are 2, push to result
})
Edit: even more concise, but perhaps less efficient:
a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
res = Array.from(new Set(a.filter(elem => a.filter(el => el === elem).length === 2)));

Javascript has awesome JSON object, in my opinion, you can use json as a dictionary;
{ key: _value }.
Loop throw array one times, no sort, no slice
key is array's element value, _value is frequency
var frequencies = {};
for (let i = 0; i < a.length; a++) {
if (result[a[i]] == 'over') continue;
if (result[a[i]] == undefined) { // First times
result[a[i]] = 1
} else if (result[a[i]] == 1) { // Second times
result[a[i]] = 2
} else { // Ignore if > 2
result[a[i]] = 'over'
}
}
// result: {1: "over", 2: 2, 3: "over", 4: "over", 6: "over", 7: 2}
so now pick keys have value equal 2

function getDuplicateArrayElements(numbers: number[]): number[] {
const occurences = new Map<number, number>();
for (let number of numbers) {
if (occurences.has(number)) {
const current = occurences.get(number)!;
occurences.set(number, current + 1);
} else
occurences.set(number, 1)
}
return (
Array
.from(occurences.entries())
.reduce<number[]>(
(accumulator, [key, value]) => {
if (value === 2) {
return accumulator.concat(key)
}
return accumulator
},
[]
)
)
}
const a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 6, 6, 6, 7, 7];
getDuplicateArrayElements(a); // [2, 7]

Sort Array to 1st min, 1st max, 2nd min, 2nd max, etc

Write a JS program to return an array in such a way that the first element is the first minimum and the second element is the first maximum and so on.
This program contains a function which takes one argument: an array. This function returns the array according to the requirement.
Sample Input: array=[2,4,7,1,3,8,9]. Expected Output: [1,9,2,8,3,7,4].
const arrsort=(arr)=>{
return arr.sort(function(a, b){return a - b});
}
const test=(arr)=>{
arr=arrsort(arr);
var arr2=[];
var j=0;
var k=arr.length-1;
for (var i=0;i<arr.length-1;i++){
if(i%2===0){
arr2.push(arr[j]);
j++;
}
else{
arr2.push(arr[k]);
k--;
}
}
return arr2;
}

Instead of using two indices, you could shift and pop the values of a copy of the sorted array.
var array = [2, 4, 7, 1, 3, 8, 9]
const arrsort = arr => arr.sort((a, b) => a - b);
const test = (arr) => {
var copy = arrsort(arr.slice()),
result = [],
fn = 'pop';
while (copy.length) {
fn = { pop: 'shift', shift: 'pop' }[fn];
result.push(copy[fn]());
}
return result;
}
console.log(test(array));

You can first sort() the array in ascending order and then loop through half of the array. And push() the values at corresponding indexes.
let arr = [2,4,7,1,3,8,9];
function order(arr){
let res = [];
arr = arr.slice().sort((a,b) => a-b);
for(let i = 0; i < Math.floor(arr.length/2); i++){
res.push(arr[i],arr[arr.length - 1 - i]);
}
return arr.length % 2 ? res.concat(arr[Math.floor((arr.length - 1)/2)]) : res;
}
console.log(order(arr))

You could sort the array, then copy and reverse and push to another array
const a = [2,4,7,1,3,8,9];
a.sort();
const b = a.slice().reverse();
const res = [];
for (let i = 0; i < a.length; i++) {
if (res.length < a.length) res.push(a[i]);
if (res.length < a.length) res.push(b[i]);
}
console.log(res);
Or use a Set
const a = [2,4,7,1,3,8,9];
a.sort();
const b = a.slice().reverse();
const res = new Set();
a.forEach((e, i) => (res.add(e), res.add(b[i])));
console.log(Array.from(res));

There are many ways are available to do this. And my solution is one of themm i hope.
Find max and min value, push them into another array. And delete max, min from actual array.
let array=[2,4,7,1,3,8,9];
let finalArray = [];
let max, min;
for(let i = 0; i < array.length; i++) {
max = Math.max(...array);
min = Math.min(...array);
finalArray.push(min);
finalArray.push(max);
array = array.filter(function(el) {
return el != max && el != min;
})
}
console.log(finalArray);

After sorting array this would work
myarr = [1,2,3,4,5,6];
let lastindex = myarr.length-1;
for(let i = 1 ; i <= myarr.length/ 2; i = i+2) {
ele = myarr[i];
myarr[i] = myarr[lastindex];
myarr[lastindex] = ele;
lastindex--;
}
Final Output will be: [1, 6, 3, 5, 4, 2]

You can use two iterators after sorting your array, one goes ascending and the other goes descending, until they cross each other.
Here's the code:
const array = [2, 4, 7, 1, 3, 8, 9];
const test = arr => {
const result = [];
const sortedArr = array.sort((a, b) => a - b);
for (let i = 0, j = sortedArr.length - 1; i <= j; i++, j--) {
result.push(sortedArr[i]);
i == j || result.push(sortedArr[j]);
}
return result;
};
console.log(test(array));

You can easily achieve the result using two pointer algorithm
function getValue(arr) {
const result = [];
let start = 0,
end = arr.length - 1;
while (start < end) result.push(arr[start++], arr[end--]);
if (start === end) result.push(arr[start]);
return result;
}
const array = [2, 4, 7, 1, 3, 8, 9];
const sorted = array.sort();
console.log(getValue(sorted));

Group array elements into set of n

I have an array
let arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
I want to group it into a set of n arrays such that first n elements in result[0] next n elements in result[1] and if any element is remaining it is discarded.
let sampleOutput = [[0, 1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12, 13]] for n = 7;
Here is my code:
function group5(arr, len) {
let result = [];
let loop=parseInt(arr.length/len)
for (let i=0; i<arr.length; i+=len) {
let x = []; let limitReached = false;
for (let j=0; j<len; j++) {
if (arr[i+j]) {
x.push(arr[i+j]);
} else {
limitReached = true;
break;
}
}
if (!limitReached) {
result.push(x);
} else {
break;
}
}
return result;
}
But I am unable to get expected result. I have tried following things.
Map function
Running i loop to arr.len
Checking arr.len % 7
Creating an array for every third element.
This question is not duplicate of Split array into chunks because I have to discard extra elements that can not be grouped into sets of n.
I have to keep the original array Immutable because I am using this on props in a child component. I need a function that does not modify the original array.

It's pretty straigthforward using Array.from
const list = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14];
function chunkMaxLength(arr, chunkSize, maxLength) {
return Array.from({length: maxLength}, () => arr.splice(0,chunkSize));
}
console.log(chunkMaxLength(list, 7, 2));

What about :
function group5(arr, len) {
let chunks = [];
let copy = arr.splice(); // Use a copy to not modifiy the original array
while(copy.length > len) {
chunks.push(copy.splice(0, len));
}
return chunks;
}

You could use a combination of reduce and filter to achieve the expected result. This example gives you a third control over length which makes the code a bit more reuseable.
let arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
const groupNumber = 7;
const groupCount = 2;
const groupArray = (group, size, length) => group.reduce((accumulator, current, index, original) =>
((index % size) == 0)
? accumulator.concat([original.slice(index, index + size)])
: accumulator, []
).filter((single, index) => index < length)
const test = groupArray(arr, groupNumber, groupCount);
console.log(test);
Step by Step
const groupArray = (group, size, length) => {
// if (index modulus size) equals 0 then concat a group of
// length 'size' as a new entry to the accumulator array and
// return it, else return the accumulator
const reducerFunc = (accumulator, current, index, original) =>
((index % size) == 0)
? accumulator.concat([original.slice(index, index + size)])
: accumulator
// if the current index is greater than the supplied length filter it out
const filterFunc = (single, index) => index < length;
// reduce and filter original group
const result = group.reduce(reducerFunc, []).filter(filterFunc)
return result;
}

Also (apart from the existing approaches) you can have a recursive approach like this
function chunks(a, size, r = [], i = 0) {
let e = i + size;
return e <= a.length ? chunks(a, size, [...r, a.slice(i, e)], e) : r;
}
function chunks(a, size, r = [], i = 0) {
let e = i + size;
return e <= a.length ? chunks(a, size, [...r, a.slice(i, e)], e) : r;
}
var arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
console.log('Chunk with 3: ', chunks(arr, 3));
console.log('Chunk with 4: ', chunks(arr, 4));
console.log('Chunk with 5: ', chunks(arr, 5));
console.log('Chunk with 6: ', chunks(arr, 6));
console.log('Chunk with 7: ', chunks(arr, 7));

I able to solve the problem with this code
function groupN(n, arr) {
const res = [];
let limit = 0;
while (limit+n <= arr.length) {
res.push(arr.slice(limit, n + limit));
limit += n
}
return res
}

I usually prefer declarative solutions (map, reduce, etc), but in this case I think a for is more understandable:
function groupArray(array, num) {
const group = [];
for (let i = 0; i < array.length; i += num) {
group.push(array.slice(i, i + num));
}
return group;
}

How to push elements of a 1d array into a 9x9 array in javascript? [duplicate]

Imagine I have an array:
A = Array(1, 2, 3, 4, 5, 6, 7, 8, 9);
And I want it to convert into 2-dimensional array (matrix of N x M), for instance like this:
A = Array(Array(1, 2, 3), Array(4, 5, 6), Array(7, 8, 9));
Note, that rows and columns of the matrix is changeable.

Something like this?
function listToMatrix(list, elementsPerSubArray) {
var matrix = [], i, k;
for (i = 0, k = -1; i < list.length; i++) {
if (i % elementsPerSubArray === 0) {
k++;
matrix[k] = [];
}
matrix[k].push(list[i]);
}
return matrix;
}
Usage:
var matrix = listToMatrix([1, 2, 3, 4, 4, 5, 6, 7, 8, 9], 3);
// result: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

You can use the Array.prototype.reduce function to do this in one line.
ECMAScript 6 style:
myArr.reduce((rows, key, index) => (index % 3 == 0 ? rows.push([key])
: rows[rows.length-1].push(key)) && rows, []);
"Normal" JavaScript:
myArr.reduce(function (rows, key, index) {
return (index % 3 == 0 ? rows.push([key])
: rows[rows.length-1].push(key)) && rows;
}, []);
You can change the 3 to whatever you want the number of columns to be, or better yet, put it in a reusable function:
ECMAScript 6 style:
const toMatrix = (arr, width) =>
arr.reduce((rows, key, index) => (index % width == 0 ? rows.push([key])
: rows[rows.length-1].push(key)) && rows, []);
"Normal" JavaScript:
function toMatrix(arr, width) {
return arr.reduce(function (rows, key, index) {
return (index % width == 0 ? rows.push([key])
: rows[rows.length-1].push(key)) && rows;
}, []);
}

This code is generic no need to worry about size and array, works universally
function TwoDimensional(arr, size)
{
var res = [];
for(var i=0;i < arr.length;i = i+size)
res.push(arr.slice(i,i+size));
return res;
}
Defining empty array.
Iterate according to the size so we will get specified chunk.That's why I am incrementing i with size, because size can be 2,3,4,5,6......
Here, first I am slicing from i to (i+size) and then I am pushing it to empty array res.
Return the two-dimensional array.

The cleanest way I could come up with when stumbling across this myself was the following:
const arrayToMatrix = (array, columns) => Array(Math.ceil(array.length / columns)).fill('').reduce((acc, cur, index) => {
return [...acc, [...array].splice(index * columns, columns)]
}, [])
where usage would be something like
const things = [
'item 1', 'item 2',
'item 1', 'item 2',
'item 1', 'item 2'
]
const result = arrayToMatrix(things, 2)
where result ends up being
[
['item 1', 'item 2'],
['item 1', 'item 2'],
['item 1', 'item 2']
]

How about something like:
var matrixify = function(arr, rows, cols) {
var matrix = [];
if (rows * cols === arr.length) {
for(var i = 0; i < arr.length; i+= cols) {
matrix.push(arr.slice(i, cols + i));
}
}
return matrix;
};
var a = [0, 1, 2, 3, 4, 5, 6, 7];
matrixify(a, 2, 4);
http://jsfiddle.net/andrewwhitaker/ERAUs/

Simply use two for loops:
var rowNum = 3;
var colNum = 3;
var k = 0;
var dest = new Array(rowNum);
for (i=0; i<rowNum; ++i) {
var tmp = new Array(colNum);
for (j=0; j<colNum; ++j) {
tmp[j] = src[k];
k++;
}
dest[i] = tmp;
}

function matrixify( source, count )
{
var matrixified = [];
var tmp;
// iterate through the source array
for( var i = 0; i < source.length; i++ )
{
// use modulous to make sure you have the correct length.
if( i % count == 0 )
{
// if tmp exists, push it to the return array
if( tmp && tmp.length ) matrixified.push(tmp);
// reset the temporary array
tmp = [];
}
// add the current source value to the temp array.
tmp.push(source[i])
}
// return the result
return matrixified;
}
If you want to actually replace an array's internal values, I believe you can call the following:
source.splice(0, source.length, matrixify(source,3));

This a simple way to convert an array to a two-dimensional array.
function twoDarray(arr, totalPerArray) {
let i = 0;
let twoDimension = []; // Store the generated two D array
let tempArr = [...arr]; // Avoid modifying original array
while (i < arr.length) {
let subArray = []; // Store 2D subArray
for (var j = 0; j < totalPerArray; j++) {
if (tempArr.length) subArray.push(tempArr.shift());
}
twoDimension[twoDimension.length] = subArray;
i += totalPerArray;
}
return twoDimension;
}
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
twoDarray(arr, 3); // [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]

function changeDimension(arr, size) {
var arrLen = arr.length;
var newArr = [];
var count=0;
var tempArr = [];
for(var i=0; i<arrLen; i++) {
count++;
tempArr.push(arr[i]);
if (count == size || i == arrLen-1) {
newArr.push(tempArr);
tempArr = [];
count = 0;
}
}
return newArr;
}
changeDimension([0, 1, 2, 3, 4, 5], 4);

function matrixify(array, n, m) {
var result = [];
for (var i = 0; i < n; i++) {
result[i] = array.splice(0, m);
}
return result;
}
a = matrixify(a, 3, 3);

function chunkArrToMultiDimArr(arr, size) {
var newArray = [];
while(arr.length > 0)
{
newArray.push(arr.slice(0, size));
arr = arr.slice(size);
}
return newArray;
}
//example - call function
chunkArrToMultiDimArr(["a", "b", "c", "d"], 2);

you can use push and slice like this
var array = [1,2,3,4,5,6,7,8,9] ;
var newarray = [[],[]] ;
newarray[0].push(array) ;
console.log(newarray[0]) ;
output will be
[[1, 2, 3, 4, 5, 6, 7, 8, 9]]
if you want divide array into 3 array
var array = [1,2,3,4,5,6,7,8,9] ;
var newarray = [[],[]] ;
newarray[0].push(array.slice(0,2)) ;
newarray[1].push(array.slice(3,5)) ;
newarray[2].push(array.slice(6,8)) ;
instead of three lines you can use splice
while(array.length) newarray.push(array.splice(0,3));

const x: any[] = ['abc', 'def', '532', '4ad', 'qwe', 'hf', 'fjgfj'];
// number of columns
const COL = 3;
const matrix = array.reduce((matrix, item, index) => {
if (index % COL === 0) {
matrix.push([]);
}
matrix[matrix.length - 1].push(item);
return matrix;
}, [])
console.log(matrix);

Using the Array grouping proposal (currently stage 3), you can now also do something like the following:
function chunkArray(array, perChunk) {
return Object.values(array.group((_, i) => i / perChunk | 0));
}
See also the MDN documentation for Array.prototype.group().

Simplest way with ES6 using Array.from()
const matrixify = (arr, size) =>
Array.from({ length: Math.ceil(arr.length / size) }, (v, i) =>
arr.slice(i * size, i * size + size));
const list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] ;
console.log(matrixify(list, 3));

Another stab at it,
Creating an empty matrix (Array of row arrays)
Iterating arr and assigning to matching rows
function arrayToMatrix(arr, wantedRows) {
// create a empty matrix (wantedRows Array of Arrays]
// with arr in scope
return new Array(wantedRows).fill(arr)
// replace with the next row from arr
.map(() => arr.splice(0, wantedRows))
}
// Initialize arr
arr = new Array(16).fill(0).map((val, i) => i)
// call!!
console.log(arrayToMatrix(arr, 4));
// Trying to make it nice
const arrToMat = (arr, wantedRows) => new Array(wantedRows).fill(arr)
.map(() => arr.splice(0, wantedRows))
(like in: this one)
(and: this one from other thread)
MatArray Class?
Extending an Array to add to a prototype, seems useful, it does need some features to complement the Array methods, maybe there is a case for a kind of MatArray Class? also for multidimensional mats and flattening them, maybe, maybe not..

1D Array convert 2D array via rows number:
function twoDimensional(array, row) {
let newArray = [];
let arraySize = Math.floor(array.length / row);
let extraArraySize = array.length % row;
while (array.length) {
if (!!extraArraySize) {
newArray.push(array.splice(0, arraySize + 1));
extraArraySize--;
} else {
newArray.push(array.splice(0, arraySize));
}
}
return newArray;
}
function twoDimensional(array, row) {
let newArray = [];
let arraySize = Math.floor(array.length / row);
let extraArraySize = array.length % row;
while (array.length) {
if (!!extraArraySize) {
newArray.push(array.splice(0, arraySize + 1));
extraArraySize--;
} else {
newArray.push(array.splice(0, arraySize));
}
}
return newArray;
}
console.log(twoDimensional([1,2,3,4,5,6,7,8,9,10,11,12,13,14], 3))

Short answer use:
const gridArray=(a,b)=>{const d=[];return a.forEach((e,f)=>{const
h=Math.floor(f/b);d[h]=d[h]||[],d[h][f%b]=a[f]}),d};
Where:
a: is the array
b: is the number of columns

An awesome repository here .
api : masfufa.js
sample : masfufa.html
According to that sample , the following snippet resolve the issue :
jsdk.getAPI('my');
var A=[1, 2, 3, 4, 5, 6, 7, 8, 9];
var MX=myAPI.getInstance('masfufa',{data:A,dim:'3x3'});
then :
MX.get[0][0] // -> 1 (first)
MX.get[2][2] // ->9 (last)