given a matrix like this one:
1 2 3
4 5 6
7 8 9
which can be represented as a 2 dimensional array:
arr = [[1,2,3], [4,5,6], [7,8,9]];
rotate the array so that it is read diagonally at a 45 degree angle and prints out this:
1
4 2
7 5 3
8 6
9
I spent a while coming up with a solution that I don't even fully intuitively understand, but it works, at least for 3x3 and 4x4 matrices. I was hoping to see more logical and clean implementations.
Here's my solution:
arr = [[1,2,3,0],[4,5,6,0],[7,8,9,0], [0,0,0,0]];
// arr[i][j];
transform(arr);
function transform(ar) {
// the number of lines in our diagonal matrix will always be rows + columns - 1
var lines = ar.length + ar[0].length - 1;
// the length of the longest line...
var maxLen = ~~(ar.length + ar[0].length)/2;
var start = 1;
var lengths = [];
// this for loop creates an array of the lengths of each line, [1,2,3,2,1] in our case
for (i=0;i<lines; i++) {
lengths.push(start);
if (i+1 < maxLen) {
start++;
} else {
start--;
}
}
// after we make each line, we're going to append it to str
var str = "";
// for every line
for(j=0; j<lengths.length; j++) {
// make a new line
var line = "";
// i tried to do it all in one for loop but wasn't able to (idk if it's possible) so here we use a particular for loop while lengths of the lines are increasing
if (j < maxLen) {
// lengths[j] is equal to the elements in this line, so the for loop will run that many times and create that many elements
for(c=0; c<lengths[j]; c++) {
// if ar[r][c], the pattern here is that r increases along rows (as we add new lines), and decreases along columns. c stays the same as we add rows, and increases across columns
line += ar[lengths[j]-1-c][c] + " ";
// when we've added all the elements we need for this line, add it to str with a line break
if (c == lengths[j]-1) {
line += "\n"; str += line;
}
}
} else {
// when we're headed down or decreasing the length of each line
for (r=0; r<lengths[j]; r++) {
// the pattern here tripped me up, and I had to introduce another changing variable j-maxLen (or the distance from the center). r stays the same as rows increase and decreases across columns. c increases along rows and decreases across columns
line += ar[lengths[j]-r+j-maxLen][j-maxLen+r +1] + " ";
// that's all our elements, add the line to str;
if (r == lengths[j] -1) {
line += "\n"; str += line;
}
}
}
}
console.log(str);
}
The main idea is to partition the original matrix indexed by (i,j) according to i+j.
This is expressed in the code snippet rotated[i+j].push(arr[i][j]) below:
arr = [[1,2,3], [4,5,6], [7,8,9]];
var summax = arr.length + arr[0].length - 1; // max index of diagonal matrix
var rotated = []; // initialize to an empty matrix of the right size
for( var i=0 ; i<summax ; ++i ) rotated.push([]);
// Fill it up by partitioning the original matrix.
for( var j=0 ; j<arr[0].length ; ++j )
for( var i=0 ; i<arr.length ; ++i ) rotated[i+j].push(arr[i][j]);
// Print it out.
for( var i=0 ; i<summax ; ++i ) console.log(rotated[i].join(' '))
Output:
1
4 2
7 5 3
8 6
9
In Ruby
Produces same output:
puts arr.transpose.flatten.group_by.with_index { |_,k|
k.divmod(arr.size).inject(:+) }.values.map { |a| a.join ' ' }
function transform(ar) {
var result = [],
i, x, y, row;
for (i = 0; i < ar.length; i++) {
row = [];
for (x = 0, y = i; y >= 0; x++, y--) {
row.push(ar[y][x]);
}
result.push(row);
}
for (i = 1; i < ar[0].length; i++) {
row = [];
for (x = i, y = ar[0].length - 1; x < ar[0].length; x++, y--) {
row.push(ar[y][x]);
}
result.push(row);
}
return result;
}
This returns the rotated array, to print it out as you go just replace each result.push(row); line with console.log(row.join(" "));.
Here's my approach:
var origMatrix = [[1,2,3,4,5], [4,5,6,7,8], [9,10,11,12,13], [14,15,16,17,18], [19,20,21,22,23]];
var maxSize = origMatrix.length;//Presuming all internal are equal!
var rotatedMatrix = [];
var internalArray;
var keyX,keyY,keyArray;
for(var y=0;y<((maxSize * 2)-1);y++){
internalArray = [];
for(var x=0;x<maxSize;x++){
keyX = x;
keyY = y - x;
if(keyY > -1){
keyArray = origMatrix[keyY];
if(typeof(keyArray) != 'undefined' && typeof(keyArray[keyX]) != 'undefined'){
internalArray.push(keyArray[keyX]);
}
}
}
rotatedMatrix.push(internalArray);
}
//log results
for(var i=0;i<rotatedMatrix.length;i++){
console.log(rotatedMatrix[i]);
}
Here's a JSFiddle of it in action (open the Console to see the results)
The Idea: Walk the Diagonals and Clip
You could use the diagonal enumeration from Cantor, see Cantor pairing function,
which is used to show that the set N x N has the same cardinality as the set N (i.e. natural numbers can be mapped one to one to pairs of natural numbers) and combine it with a condition that one skips those values which lie outside the rectangular matrix.
The Cantor pairing function pi takes two natural numbers i and j, i.e. the pair (i, j) and maps it to a natural number k
pi : |N x |N -> |N : pi(i, j) = k
Use the reverse mapping to get this
pi^-1 : |N -> |N x |N : pi^-1(k) = (i, j)
i.e. one enumerates the cells of the "infinite Matrix" N x N diagonally.
So counting up k and applying the inverse function will give the proper pair of indices (i, j) for printing the rotated matrix.
Example:
0->(0, 0) 2->(0, 1) | 5->(0, 2) 9->(0, 3) . .
1->(1, 0) 4->(1, 1) | 8->(1, 2)
3->(2, 0) 7->(2, 2) |
---------------------+ <- clipping for 3 x 2 matrix
6->(3, 0)
.
.
Calculation of the Inverse Cantor Pair Function
For input k these formulas give the pair (i, j):
w = floor((sqrt(8*k + 1) - 1) / 2)
t = (w*w + w) / 2
j = k - t
i = w - j
See the link given above for a derivation.
Resulting Algorithm
Given a m x n matrix A: i from [0, .., m - 1] enumerates the rows, and j from [0, .., n - 1] enumerates the columns
Start at k = 0
calculate the corresponding index pair (i, j)
print the matrix value A[i, j] if the indices i and j lie within your matrix dimensions m and n
print a new line, once your i hit the top of the matrix, i.e. if i == 0
increment k
continue with step 2 until you arrived at the index pair (i, j) = (n - 1, n - 1)
Sample Implementation in JavaScript
Note: I tried this out in the MongoDB shell, using its print() function.
Helper functions
function sprint(k) {
var s = '' + k;
while (s.length < 3) {
s = ' ' + s;
}
return s;
}
function print_matrix(a) {
var m = a.row_size;
var n = a.column_size;
for (var i = 0; i < m; i++) {
var s = '';
for (var j = 0; j < n; j++) {
s += sprint(a.value[i][j]);
}
print(s);
}
}
The Inverse of the Cantor pairing function
// inverse of the Cantor pair function
function pi_inv(k) {
var w = Math.floor((Math.sqrt(8*k + 1) - 1) / 2);
var t = (w*w + w) /2;
var j = k - t;
var i = w -j;
return [i, j];
}
The algorithm
// "rotate" matrix a
function rot(a) {
var m = a.row_size;
var n = a.column_size;
var i_max = m - 1;
var j_max = n - 1;
var k = 0;
var s = '';
do {
var ij = pi_inv(k);
var i = ij[0];
var j = ij[1];
if ((i <= i_max) && (j <= j_max)) {
s += sprint(a.value[i][j]);
}
if (i == 0) {
print(s);
s = '';
}
k += 1
} while ((i != i_max) || (j != j_max));
print(s);
}
Example usage
// example
var a = {
row_size: 4,
column_size: 4,
value: [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]
};
print('in:');
print_matrix(a);
print('out:');
rot(a);
Output for 4x4 Matrix
in:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
out:
1
5 2
9 6 3
13 10 7 4
14 11 8
15 12
16
This method works for any m x n Matrix, e.g. 4 x 5:
in:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
out:
1
6 2
11 7 3
16 12 8 4
17 13 9 5
18 14 10
19 15
20
or 4 x 3:
in:
1 2 3
4 5 6
7 8 9
10 11 12
out:
1
4 2
7 5 3
10 8 6
11 9
12
Related
Can someone explain to me why my fibonacciGenerator function doesn't work with this code? I understand why it works with the second code tho but I just don't get why the first one doesn't.
function fibonacciGenerator(n) {
if (n > 0) {
var fArray = [];
fArray.push(0);
if (n >= 2) {
fArray.push(1);
}
for (var i = 0; i < n; i++) {
fArray.push(fArray[i] + fArray[i + 1]);
}
console.log(fArray);
}
}
fibonacciGenerator(1);
fibonacciGenerator(2);
Second code working :
function fibonacciGenerator(n) {
if (n > 0) {
var fArray = [];
fArray.push(0);
if (n >= 2) {
fArray.push(1);
}
for (var i = 2; i < n; i++) {
fArray.push(fArray[i - 1] + fArray[i - 2]);
}
console.log(fArray);
}
}
fibonacciGenerator(1);
fibonacciGenerator(2);
The first code is printing 2 extra Fibonacci number this is because:
you are first pushing 0 and 1 into the array as:
var fArray = [];
fArray.push(0);
if (n >=2 ){
fArray.push(1);
}
and then you loop over again till n times. Because of this reason it prints two extra Fibonacci numbers.
the solution is to either loop over n-2 time or to use the second code.
var fArray = [];
fArray.push(0);
if (n >= 2) {
fArray.push(1);
}
The initial condition is to cover n=1: [0] and n=2: [0,1]
The 2nd code is working because the loop only starts when n is greater than i, so means it skips the loop with n < 2.
For your problem, you don't skip the loop when n < 2.
for (var i = 0; i < n; i++) {
fArray.push(fArray[i] + fArray[i + 1]);
}
You can imagine the result will be like below when n < 2 with your loop.
Note that the inital value is fArray = [0]
fArray.push(fArray[0] + fArray[1]); //fArray[1] is undefined because you only have 1 item in your array
In this case fArray[0] + fArray[1] ==> 0 + undefined = NaN
So that's why your logic does not work when n < 2
To correct it, you need to avoid the loop if n < 2
//if n=1 or n=2, it won't trigger the loop due to `i < n-2`
for (var i = 0; i < n-2; i++) {
fArray.push(fArray[i] + fArray[i + 1]);
}
The idea to have i start with 0 instead of 2, and to adjust the body of the loop accordingly, is fine, but there is one thing that the first version didn't adjust: the stop condition of the loop.
By setting i=0, the first version loops 2 times more than the second version. You should also alter the end condition in the same way: instead of i < n, it should have i < n - 2, so to ensure the number of iterations is the same as in the second version.
Not related to your question, but the console.log should better be placed outside of the function. The job of the function should be to return the array, not to print it. So also, when n > 0 is false, it should return an empty array.
function fibonacciGenerator(n) {
var fArray = [];
if (n > 0) {
fArray.push(0);
if (n >= 2) {
fArray.push(1);
}
for (var i = 0; i < n - 2; i++) {
fArray.push(fArray[i] + fArray[i + 1]);
}
}
return fArray;
}
console.log(fibonacciGenerator(1));
console.log(fibonacciGenerator(2));
First you have to identify the pattern.
Fibonacci series -> 0 1 1 2 3 5 8 13 21
Term -> 0 1 2 3 4 5 6 7 8
0th term =0, 1st term =1
From the second term,
2nd = 1st term + 0th term = 1+0 = 1
3rd = 2nd term + 1st term = 1+1 = 2
4th = 3rd term + 2nd term = 2+1 = 3
5th = 4th term + 3rd term = 3+2 = 5
nth = (n-1) + (n-2)
since the first 2 terms are fixed, you have to start for loop from i= 2.
Also, according to the above shown pattern, you have to use following code inside the for loop.
fArray.push(fArray[i - 1] + fArray[i-2]);
given matrix / 2D array
1 2 3
4 5 6
7 8 9
we have to print
7 4 1 5 9 6 3 in javascript without using inbuilt functions
Here's a hint.
Assuming your matrix is presented as a 1-d array of integers:
m = [1,2,3,4,5,6,7,8,9];
which is logically assumed to be a "square" size (1x1, 2x2, 3x3, 4x4, etc...)
So the above array is logically assumed to be:
1 2 3
4 5 6
7 8 9
Let's use a helper function to help us obtain any that value at any given (x,y) position. With (0,0), being the top left value of the matrix
let getVal = (m, col, row) => {
let rowlen = Math.round(Math.sqrt(m.length));
let index = row * rowlen + col;
return m[index];
}
Hence, getVal(m, 0, 0) returns 1 for the top left corner and getVal(m,2,2) returns 9 for the value in the bottom right corner.
Now with that little helper function provided, do you think you can implement the three for loops for "going up from the bottom left", "diagonal from top left to bottom right", and "going up from the bottom right"?
var output= "";
var length = matrix.length;
// console.log(length)
for(var i = length - 1; i >= 0; i--){
output= output+ matrix[i][0] + " ";
}
// console.log(bag);
for(var j = 1; j < length; j++){
output= output+ matrix[j][j] + " ";
}
// console.log(bag);
for(var k = length - 2; k >= 0; k--){
output= output+ matrix[k][length - 1] + " ";
}
console.log(output);
}
The question is as follows:
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
I have written the following code, but I am stuck in terms of how to flag the row as one time to be downward moving, where I increment the start row, but when it's zigzagging back to the top, it should be decremented. I am unable to figure out the logic to make this work without affecting the downward movement. Any help would be appreciated.
const convert = (s, numRows) => {
let startRow = 0
let endRow = numRows - 1
let startColumn = 0
let endColumn = Math.floor((s.length / 2) - 1)
s = s.split('')
let results = []
// to setup the columns
for (let i = 0; i < numRows; i++) {
results.push([])
}
while (startRow <= endRow && startColumn <= endColumn && s.length) {
for (let i = startRow; i <= endRow; i++) {
results[i][startColumn] = s.shift()
}
for (let i = endRow - 1; i >= startRow; i--) {
results[i][startColumn + 1] = s.shift()
startColumn++
}
//this line seems to be the issue
startRow++
}
return results
}
console.log(convert('PAYPALISHIRING', 4))
I rewrote your while loop as follows where I simply walk a "zigzag" pattern! Hopefully, it is simple enough to understand.
let c=0, row=0,col=0, down=0;
while(c<s.length) {
results[row][col]=s[c];
if(down==0) { // moving down
row++;
if(row==numRows) {
down = 1;
col++;
row-=2;
}
} else { // moving up
row--;
col++;
if(row==0) {
down=0;
}
}
c++;
}
Ps. Above code does not handle numRows < 3 so you have to manage them before this loop.
My precalculus is a little rusty, but the logic behind this problem seems like a sine wave. I made a math error somewhere in creating the sin equation that prevents this from working (r never equals c with the current paramaters), but hopefully this will help if this is the direction you choose to go in.
/*If x-axis is position in string, and y-axis is row number...
n=number of rows
Equation for a sin curve: y = A sin(B(x + C)) + D
D=vertical shift (y value of mid point)
D=median of 1 and n
n: median:
1 1
2 1.5
3 2
4 2.5
5 3
6 3.5
7 4
median=(n+1)/2
D=(n+1)/2
A=amplitude (from the mid-point, how high does the curve go)
median + amplitude = number of rows
amplitude = number of rows - median
A=n-D
C=phase shift
Phase shift for a sin curve starting at its lowest point: 3π/2
(so at time 1, row number is 1, and curve goes up from there)
C=3π/2
Period is 2π/B
n p
3 4
4 6
5 8
6 10
period=2(n-1)
2(n-1)=2π/B
B(2(n-1)=2π
B=2π/2(n-1)
B=π/(n-1)
Variables:
s = string
n = number of rows
c = current row number being evaluated
p = position in string
r = row number
*/
var output='';
function convert(s,n) {
D=(n+1)/2
A=n-D
C=(3*Math.PI)/2
B=Math.PI/(n-1)
for (c=1;c<=n;c++) { //loop from 1st row to number of rows
for (p=1;p<=s.length;p++) { //loop from 1st to last character in string
r=A*Math.sin(B*(p+C))+D //calculate the row this character belongs in
if (r==c) { output+= s.charAt(r) } //if the character belongs in this row, add it to the output variable. (minus one because character number 1 is at position 0)
}}
//do something with output here
}
I wrote a code with javascript for this problem :
"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000."
but the result is false and i don't know why? can you help me guys
my code is :
function multipleSum(n){
var sum = 0;
for(var i = 1; i<n; i++){
var m3 = 3 * i;
var m5 = 5 * i;
if(m3 < n ){
sum=sum+m3
}
if(m5 < n ){
sum=sum+m5;
}
//if(m3 > n && m5 > n) {console.log(m3,m5,sum);break;}
}
return sum
}
console.log(multipleSum(1000)) //266333 but correct one is 233168 why?
Your logic is flawed. You should be iterating on each number (specified in range), and see if the modulus of the number with 3 or 5 is 0 or not. If modulus is zero, it means the number is divisible.
function multipleSum(n){
var sum = 0;
for(var i = 1; i<n; i++){
if(i % 3 == 0 || i % 5 ==0){ // gives reminder of 0, divisible by either 3 or 5
sum += i; // add in sum if that's the case.
}
}
return sum
}
console.log(multipleSum(1000))
Edit: tried some time understanding why you went with multiply approach, I think you are gathering factors and want to break out early from the loop instead of iterating on entire collection. This should help you:
function multipleSum(n){
var sum = 0;
for(var i = 1; i<n; i++){
var m3 = i * 3;
var m5 = i * 5;
if(m3 > n) break; // breaks early!!
if(m3 < n) sum += m3
if(m5 < n && m5 % 3 != 0) sum += m5; // make sure number is not divisible by 3, say m5 = 15, it will be captured as multiple of 3 anyway, we don't want duplicates.
}
return sum
}
console.log(multipleSum(1000))
Your logic is flawed in the way that, all the multiplications of 3 * 5 is doubled. Remember, you have:
3 * 1
5 * 1
3 * 2
3 * 3
5 * 2
3 * 4
3 * 5
5 * 3 // Here comes the dupe.
I would do this in a different way.
Get all the multiples of 3 in an array.
Get all the multiples of 5 in an array.
Break the loop when both the multiplications are greater than n.
Merge both the arrays.
Remove the duplicates.
Add everything using the .reduce() function.
var num = 1000;
var m3 = [];
var m5 = [];
for (i = 0; i < num; i++) {
if (i * 3 < num)
m3.push(i * 3);
if (i * 5 < num)
m5.push(i * 5);
if (i * 3 > num)
break;
}
m35 = m3.concat(m5);
m35u = m35.filter(function(item, pos) {
return m35.indexOf(item) == pos;
});
console.log(m35u.reduce((a, b) => a + b, 0));
I get 233168 as answer.
You can try this one liner (your home work: explain how this works ;):
console.log(
Array.from({length: 1000})
.reduce( (p, n, i) => p + (i % 3 === 0 || i % 5 === 0 ? i : 0), 0 )
);
Try this, maybe answer you.Thank
const solution = (numb) => {
const collectedNumb = [];
const maxDividing = parseInt(numb / 3);
for (let idx = 1; idx <= maxDividing; idx++) {
const multipled3 = idx * 3;
const multipled5 = idx * 5;
multipled3 < numb && collectedNumb.push(multipled3);
multipled5 < numb && collectedNumb.push(multipled5);
}
const uniqCollected = [...new Set(collectedNumb)].sort((a, b)=> a-b);
console.log(uniqCollected);
const reduced = uniqCollected.reduce((acc, numb) => acc + numb, 0);
return reduced;
};
Given the triangle of consecutive odd numbers:
1
3 5
7 9 11
13 15 17 19
21 23 25 27 29
// Calculate the row sums of this triangle from the row index (starting at index 1) e.g.:
rowSumOddNumbers(1); // 1
rowSumOddNumbers(2); // 3 + 5 = 8
I tried to solve this using for loops:
function rowSumOddNumbers(n){
let result = [];
// generate the arrays of odd numbers
for(let i = 0; i < 30; i++){
// generate sub arrays by using another for loop
// and only pushing if the length is equal to current j
let sub = [];
for(let j = 1; j <= n; j++){
// if length === j (from 1 - n) keep pushing
if(sub[j - 1].length <= j){
// and if i is odd
if(i % 2 !== 0){
// push the i to sub (per length)
sub.push(i);
}
}
}
// push everything to the main array
result.push(sub);
}
// return sum of n
return result[n + 1].reduce(function(total, item){
return total += item;
});
}
My code above is not working. Basically I was planning to 1st generate an array of odd numbers less than 30. Next I need to create a sub array base on the length of iteration (j) that would from 1 - n (passed). Then finally push it to the main array. And then use reduce to get the sum of all the values in that index + 1 (since the index starts at 1).
Any idea what am I missing and how to make this work?
Most code problems involve some analysis first in order to spot patterns which you can then convert into code. Looking at the triangle, you'll see the sum of each row follows a pattern:
1: 1 === 1 ^ 3
2: 3 + 5 = 8 === 2 ^ 3
3: 7 + 9 + 11 = 27 === 3 ^ 3
... etc
So from the analysis above you can see that your code could probably be simplified slightly - I won't post an answer, but think about using Math.pow.
No need for any loops.
function rowSumOddNumbers(n) {
// how many numbers are there in the rows above n?
// sum of arithmetic sequence...
let numbers_before_n_count = (n - 1) * n / 2;
let first_number_in_nth_row = numbers_before_n_count * 2 + 1;
let last_number_in_nth_row = first_number_in_nth_row + 2 * (n - 1);
// sum of arithmetic sequence again...
return n * (first_number_in_nth_row + last_number_in_nth_row) / 2;
}