Stagger, stack, sum

Clash Royale CLAN TAG#URR8PPP

up vote
7
down vote

favorite

Inspired by this Stack Overflow question.

The challenge

Input

An array of square matrices containing non-negative integers.

Output

A square matrix built from the input matrices as follows.

Let $N times N$ be the size of each input matrix, and $P$ the number of input matrices.

For clarity, consider the following example input matrices ($N=2$, $P=3$):

 3 5
 4 10

 6 8
12 11

 2 0
 9 1

1. Start with the first input matrix.


2. Shift the second input matrix N−1 steps down and N−1 steps right, so that its upper-left entry coincides with the lower-right entry of the previous one.


3. 
   

   Imagine the second, shifted matrix as if it were stacked on top of the first. Sum the two values at the coincident entry. Write the other values, and fill the remaining entries with 0 to get a $(2N-1)times(2N-1)$ matrix. With the example input, the result so far is

   
   
   

    3 5 0
    4 16 8
    0 12 11
   

   
   


4. 
   

   For each remaining input matrix, stagger it so that its upper-left coincides with the lower-right of the accumulated result matrix so far. In the example, including the third input matrix gives

   
   
   

    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   



5. The ouput is the $((N−1)P+1)times((N−1)P+1)$ matrix obtained after including the last input matrix.

Additional rules and clarifications

• 
  $N$ and $P$ are positive integers.


• You can optionally take $N$ and $P$ as additional inputs.


• Input and output can be taken by any reasonable means. Their format is flexible as usual.


• Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.


• Shortest code in bytes wins.

Test cases:

In each case, input matrices are shown first, then the output.

1. 
   

   $N=2$, $P=3$:

   
   
   

    3 5
    4 10
   
    6 8
   12 11
   
    2 0
    9 1
   
    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   


2. 
   

   $N=2$, $P=1$:

   
   
   

    3 5
    4 10
   
    3 5
    4 10
   

   
   


3. 
   

   $N=1$, $P=4$:

   
   
   

    4
   
    7
   
   23
   
    5
   
   39
   

   
   


4. 
   

   $N=3$, $P=2$:

   
   
   

   11 11 8
    6 8 12
   11 0 4
   
    4 1 13
    9 19 11
   13 4 2
   
   11 11 8 0 0
    6 8 12 0 0
   11 0 8 1 13
    0 0 9 19 11
    0 0 13 4 2
   

   
   


5. 
   

   $N=2$, $P=4$:

   
   
   

   14 13
   10 0
   
   13 20
   21 3
   
    9 22
    0 8
   
   17 3
   19 16
   
   14 13 0 0 0
   10 13 20 0 0
    0 21 12 22 0
    0 0 0 25 3
    0 0 0 19 16
   

   
   

code-golf number array-manipulation integer

share|improve this question

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How long is your MATL solution for this?
  – Giuseppe
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Thanks! Corrected
  – Luis Mendo
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
  – Luis Mendo
  28 mins ago
  
  

  
  
  
  

add a comment | 

up vote
7
down vote

favorite

Inspired by this Stack Overflow question.

The challenge

Input

An array of square matrices containing non-negative integers.

Output

A square matrix built from the input matrices as follows.

Let $N times N$ be the size of each input matrix, and $P$ the number of input matrices.

For clarity, consider the following example input matrices ($N=2$, $P=3$):

 3 5
 4 10

 6 8
12 11

 2 0
 9 1

1. Start with the first input matrix.


2. Shift the second input matrix N−1 steps down and N−1 steps right, so that its upper-left entry coincides with the lower-right entry of the previous one.


3. 
   

   Imagine the second, shifted matrix as if it were stacked on top of the first. Sum the two values at the coincident entry. Write the other values, and fill the remaining entries with 0 to get a $(2N-1)times(2N-1)$ matrix. With the example input, the result so far is

   
   
   

    3 5 0
    4 16 8
    0 12 11
   

   
   


4. 
   

   For each remaining input matrix, stagger it so that its upper-left coincides with the lower-right of the accumulated result matrix so far. In the example, including the third input matrix gives

   
   
   

    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   



5. The ouput is the $((N−1)P+1)times((N−1)P+1)$ matrix obtained after including the last input matrix.

Additional rules and clarifications

• 
  $N$ and $P$ are positive integers.


• You can optionally take $N$ and $P$ as additional inputs.


• Input and output can be taken by any reasonable means. Their format is flexible as usual.


• Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.


• Shortest code in bytes wins.

Test cases:

In each case, input matrices are shown first, then the output.

1. 
   

   $N=2$, $P=3$:

   
   
   

    3 5
    4 10
   
    6 8
   12 11
   
    2 0
    9 1
   
    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   


2. 
   

   $N=2$, $P=1$:

   
   
   

    3 5
    4 10
   
    3 5
    4 10
   

   
   


3. 
   

   $N=1$, $P=4$:

   
   
   

    4
   
    7
   
   23
   
    5
   
   39
   

   
   


4. 
   

   $N=3$, $P=2$:

   
   
   

   11 11 8
    6 8 12
   11 0 4
   
    4 1 13
    9 19 11
   13 4 2
   
   11 11 8 0 0
    6 8 12 0 0
   11 0 8 1 13
    0 0 9 19 11
    0 0 13 4 2
   

   
   


5. 
   

   $N=2$, $P=4$:

   
   
   

   14 13
   10 0
   
   13 20
   21 3
   
    9 22
    0 8
   
   17 3
   19 16
   
   14 13 0 0 0
   10 13 20 0 0
    0 21 12 22 0
    0 0 0 25 3
    0 0 0 19 16
   

   
   

code-golf number array-manipulation integer

share|improve this question

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How long is your MATL solution for this?
  – Giuseppe
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Thanks! Corrected
  – Luis Mendo
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
  – Luis Mendo
  28 mins ago
  
  

  
  
  
  

add a comment | 

up vote
7
down vote

favorite

up vote
7
down vote

favorite

Inspired by this Stack Overflow question.

The challenge

Input

An array of square matrices containing non-negative integers.

Output

A square matrix built from the input matrices as follows.

Let $N times N$ be the size of each input matrix, and $P$ the number of input matrices.

For clarity, consider the following example input matrices ($N=2$, $P=3$):

 3 5
 4 10

 6 8
12 11

 2 0
 9 1

1. Start with the first input matrix.


2. Shift the second input matrix N−1 steps down and N−1 steps right, so that its upper-left entry coincides with the lower-right entry of the previous one.


3. 
   

   Imagine the second, shifted matrix as if it were stacked on top of the first. Sum the two values at the coincident entry. Write the other values, and fill the remaining entries with 0 to get a $(2N-1)times(2N-1)$ matrix. With the example input, the result so far is

   
   
   

    3 5 0
    4 16 8
    0 12 11
   

   
   


4. 
   

   For each remaining input matrix, stagger it so that its upper-left coincides with the lower-right of the accumulated result matrix so far. In the example, including the third input matrix gives

   
   
   

    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   



5. The ouput is the $((N−1)P+1)times((N−1)P+1)$ matrix obtained after including the last input matrix.

Additional rules and clarifications

• 
  $N$ and $P$ are positive integers.


• You can optionally take $N$ and $P$ as additional inputs.


• Input and output can be taken by any reasonable means. Their format is flexible as usual.


• Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.


• Shortest code in bytes wins.

Test cases:

In each case, input matrices are shown first, then the output.

1. 
   

   $N=2$, $P=3$:

   
   
   

    3 5
    4 10
   
    6 8
   12 11
   
    2 0
    9 1
   
    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   


2. 
   

   $N=2$, $P=1$:

   
   
   

    3 5
    4 10
   
    3 5
    4 10
   

   
   


3. 
   

   $N=1$, $P=4$:

   
   
   

    4
   
    7
   
   23
   
    5
   
   39
   

   
   


4. 
   

   $N=3$, $P=2$:

   
   
   

   11 11 8
    6 8 12
   11 0 4
   
    4 1 13
    9 19 11
   13 4 2
   
   11 11 8 0 0
    6 8 12 0 0
   11 0 8 1 13
    0 0 9 19 11
    0 0 13 4 2
   

   
   


5. 
   

   $N=2$, $P=4$:

   
   
   

   14 13
   10 0
   
   13 20
   21 3
   
    9 22
    0 8
   
   17 3
   19 16
   
   14 13 0 0 0
   10 13 20 0 0
    0 21 12 22 0
    0 0 0 25 3
    0 0 0 19 16
   

   
   

code-golf number array-manipulation integer

share|improve this question

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

Inspired by this Stack Overflow question.

The challenge

Input

An array of square matrices containing non-negative integers.

Output

A square matrix built from the input matrices as follows.

Let $N times N$ be the size of each input matrix, and $P$ the number of input matrices.

For clarity, consider the following example input matrices ($N=2$, $P=3$):

 3 5
 4 10

 6 8
12 11

 2 0
 9 1

1. Start with the first input matrix.


2. Shift the second input matrix N−1 steps down and N−1 steps right, so that its upper-left entry coincides with the lower-right entry of the previous one.


3. 
   

   Imagine the second, shifted matrix as if it were stacked on top of the first. Sum the two values at the coincident entry. Write the other values, and fill the remaining entries with 0 to get a $(2N-1)times(2N-1)$ matrix. With the example input, the result so far is

   
   
   

    3 5 0
    4 16 8
    0 12 11
   

   
   


4. 
   

   For each remaining input matrix, stagger it so that its upper-left coincides with the lower-right of the accumulated result matrix so far. In the example, including the third input matrix gives

   
   
   

    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   



5. The ouput is the $((N−1)P+1)times((N−1)P+1)$ matrix obtained after including the last input matrix.

Additional rules and clarifications

• 
  $N$ and $P$ are positive integers.


• You can optionally take $N$ and $P$ as additional inputs.


• Input and output can be taken by any reasonable means. Their format is flexible as usual.


• Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.


• Shortest code in bytes wins.

Test cases:

In each case, input matrices are shown first, then the output.

1. 
   

   $N=2$, $P=3$:

   
   
   

    3 5
    4 10
   
    6 8
   12 11
   
    2 0
    9 1
   
    3 5 0 0
    4 16 8 0
    0 12 13 0
    0 0 9 1
   

   
   


2. 
   

   $N=2$, $P=1$:

   
   
   

    3 5
    4 10
   
    3 5
    4 10
   

   
   


3. 
   

   $N=1$, $P=4$:

   
   
   

    4
   
    7
   
   23
   
    5
   
   39
   

   
   


4. 
   

   $N=3$, $P=2$:

   
   
   

   11 11 8
    6 8 12
   11 0 4
   
    4 1 13
    9 19 11
   13 4 2
   
   11 11 8 0 0
    6 8 12 0 0
   11 0 8 1 13
    0 0 9 19 11
    0 0 13 4 2
   

   
   


5. 
   

   $N=2$, $P=4$:

   
   
   

   14 13
   10 0
   
   13 20
   21 3
   
    9 22
    0 8
   
   17 3
   19 16
   
   14 13 0 0 0
   10 13 20 0 0
    0 21 12 22 0
    0 0 0 25 3
    0 0 0 19 16
   

   
   

code-golf number array-manipulation integer

code-golf number array-manipulation integer

share|improve this question

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

share|improve this question

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

share|improve this question

share|improve this question

edited 30 mins ago

edited 30 mins ago

edited 30 mins ago

asked 1 hour ago

Luis Mendo

72.9k885284

asked 1 hour ago

Luis Mendo

72.9k885284

asked 1 hour ago

Luis Mendo

72.9k885284

72.9k885284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How long is your MATL solution for this?
  – Giuseppe
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Thanks! Corrected
  – Luis Mendo
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
  – Luis Mendo
  28 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How long is your MATL solution for this?
  – Giuseppe
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Thanks! Corrected
  – Luis Mendo
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
  – Luis Mendo
  28 mins ago
  
  

  
  
  
  

How long is your MATL solution for this?
– Giuseppe
1 hour ago

How long is your MATL solution for this?
– Giuseppe
1 hour ago

@Arnauld Thanks! Corrected
– Luis Mendo
30 mins ago

@Arnauld Thanks! Corrected
– Luis Mendo
30 mins ago

@Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
– Luis Mendo
28 mins ago

@Giuseppe I haven’t tried it in MATL. For the test cases I used the MATLAB code from my answer in the linked question
– Luis Mendo
28 mins ago

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote

R, 88 81 bytes

function(A,N,P,o=0*diag(P*(N-1)+1))for(i in 1:P)o[x,x]=o[x<-1:N+i-1,x]+A[[i]];o

Try it online!

Takes a list of matrices, A, N, and P.

Builds the requisite matrix of zeros o and adds elementwise the contents of A to the appropriate submatrices in o.

share|improve this answer

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

add a comment | 

up vote
1
down vote

JavaScript (ES6), 104 bytes

Takes input as (n,p,a).

(n,p,a)=>[...Array(n--+n*~-p)].map((_,y,r)=>r.map((_,x)=>a.map((a,i)=>s+=(a[y-i*n]||0)[x-i*n]|0,s=0)|s))

Try it online!

How?

Redimensioning matrices filled with a default constant ($0$ in that case) is neither very easy nor very short in JS, so we just build a square matrix with the correct width $w$ right away:

$$w=(n-1)times(p-1)+n$$

For each cell at $(x,y)$, we compute:

$$s_x,y=sum_i=0^p-1a_i(x-itimes (n-1),y-itimes (n-1))$$

where undefined cells are replaced with zeros.

share|improve this answer

edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

add a comment | 

up vote
0
down vote

Python 2, 124 bytes

def f(m,N,P):w=~-N*P+1;a=[w*[0]for _ in' '*w];g=0;exec"x=g/N/N*~-N;a[x+g/N%N][x+g%N]+=m[x][g/N%N][g%N];g+=1;"*P*N*N;return a

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

add a comment | 

up vote
0
down vote

Jelly, 20 bytes

µ;€0Zð⁺0ṁ+ṚU$}ṚU+⁸µ/

Try it online!

Bah, Jelly has an attitude today...

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

R, 88 81 bytes

function(A,N,P,o=0*diag(P*(N-1)+1))for(i in 1:P)o[x,x]=o[x<-1:N+i-1,x]+A[[i]];o

Try it online!

Takes a list of matrices, A, N, and P.

Builds the requisite matrix of zeros o and adds elementwise the contents of A to the appropriate submatrices in o.

share|improve this answer

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

add a comment | 

up vote
2
down vote

R, 88 81 bytes

function(A,N,P,o=0*diag(P*(N-1)+1))for(i in 1:P)o[x,x]=o[x<-1:N+i-1,x]+A[[i]];o

Try it online!

Takes a list of matrices, A, N, and P.

Builds the requisite matrix of zeros o and adds elementwise the contents of A to the appropriate submatrices in o.

share|improve this answer

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

add a comment | 

up vote
2
down vote

up vote
2
down vote

R, 88 81 bytes

function(A,N,P,o=0*diag(P*(N-1)+1))for(i in 1:P)o[x,x]=o[x<-1:N+i-1,x]+A[[i]];o

Try it online!

Takes a list of matrices, A, N, and P.

Builds the requisite matrix of zeros o and adds elementwise the contents of A to the appropriate submatrices in o.

share|improve this answer

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

R, 88 81 bytes

function(A,N,P,o=0*diag(P*(N-1)+1))for(i in 1:P)o[x,x]=o[x<-1:N+i-1,x]+A[[i]];o

Try it online!

Takes a list of matrices, A, N, and P.

Builds the requisite matrix of zeros o and adds elementwise the contents of A to the appropriate submatrices in o.

share|improve this answer

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

share|improve this answer

share|improve this answer

edited 1 min ago

edited 1 min ago

edited 1 min ago

answered 33 mins ago

Giuseppe

15.1k31051

answered 33 mins ago

Giuseppe

15.1k31051

answered 33 mins ago

Giuseppe

15.1k31051

15.1k31051

add a comment | 

add a comment | 

up vote
1
down vote

JavaScript (ES6), 104 bytes

Takes input as (n,p,a).

(n,p,a)=>[...Array(n--+n*~-p)].map((_,y,r)=>r.map((_,x)=>a.map((a,i)=>s+=(a[y-i*n]||0)[x-i*n]|0,s=0)|s))

Try it online!

How?

Redimensioning matrices filled with a default constant ($0$ in that case) is neither very easy nor very short in JS, so we just build a square matrix with the correct width $w$ right away:

$$w=(n-1)times(p-1)+n$$

For each cell at $(x,y)$, we compute:

$$s_x,y=sum_i=0^p-1a_i(x-itimes (n-1),y-itimes (n-1))$$

where undefined cells are replaced with zeros.

share|improve this answer

edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

add a comment | 

up vote
1
down vote

JavaScript (ES6), 104 bytes

Takes input as (n,p,a).

(n,p,a)=>[...Array(n--+n*~-p)].map((_,y,r)=>r.map((_,x)=>a.map((a,i)=>s+=(a[y-i*n]||0)[x-i*n]|0,s=0)|s))

Try it online!

How?

Redimensioning matrices filled with a default constant ($0$ in that case) is neither very easy nor very short in JS, so we just build a square matrix with the correct width $w$ right away:

$$w=(n-1)times(p-1)+n$$

For each cell at $(x,y)$, we compute:

$$s_x,y=sum_i=0^p-1a_i(x-itimes (n-1),y-itimes (n-1))$$

where undefined cells are replaced with zeros.

share|improve this answer

edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (ES6), 104 bytes

Takes input as (n,p,a).

(n,p,a)=>[...Array(n--+n*~-p)].map((_,y,r)=>r.map((_,x)=>a.map((a,i)=>s+=(a[y-i*n]||0)[x-i*n]|0,s=0)|s))

Try it online!

How?

Redimensioning matrices filled with a default constant ($0$ in that case) is neither very easy nor very short in JS, so we just build a square matrix with the correct width $w$ right away:

$$w=(n-1)times(p-1)+n$$

For each cell at $(x,y)$, we compute:

$$s_x,y=sum_i=0^p-1a_i(x-itimes (n-1),y-itimes (n-1))$$

where undefined cells are replaced with zeros.

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edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

JavaScript (ES6), 104 bytes

Takes input as (n,p,a).

(n,p,a)=>[...Array(n--+n*~-p)].map((_,y,r)=>r.map((_,x)=>a.map((a,i)=>s+=(a[y-i*n]||0)[x-i*n]|0,s=0)|s))

Try it online!

How?

Redimensioning matrices filled with a default constant ($0$ in that case) is neither very easy nor very short in JS, so we just build a square matrix with the correct width $w$ right away:

$$w=(n-1)times(p-1)+n$$

For each cell at $(x,y)$, we compute:

$$s_x,y=sum_i=0^p-1a_i(x-itimes (n-1),y-itimes (n-1))$$

where undefined cells are replaced with zeros.

share|improve this answer

edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

share|improve this answer

share|improve this answer

edited 16 mins ago

edited 16 mins ago

edited 16 mins ago

answered 42 mins ago

Arnauld

65.9k583278

answered 42 mins ago

Arnauld

65.9k583278

answered 42 mins ago

Arnauld

65.9k583278

65.9k583278

add a comment | 

add a comment | 

up vote
0
down vote

Python 2, 124 bytes

def f(m,N,P):w=~-N*P+1;a=[w*[0]for _ in' '*w];g=0;exec"x=g/N/N*~-N;a[x+g/N%N][x+g%N]+=m[x][g/N%N][g%N];g+=1;"*P*N*N;return a

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

add a comment | 

up vote
0
down vote

Python 2, 124 bytes

def f(m,N,P):w=~-N*P+1;a=[w*[0]for _ in' '*w];g=0;exec"x=g/N/N*~-N;a[x+g/N%N][x+g%N]+=m[x][g/N%N][g%N];g+=1;"*P*N*N;return a

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

add a comment | 

up vote
0
down vote

up vote
0
down vote

Python 2, 124 bytes

def f(m,N,P):w=~-N*P+1;a=[w*[0]for _ in' '*w];g=0;exec"x=g/N/N*~-N;a[x+g/N%N][x+g%N]+=m[x][g/N%N][g%N];g+=1;"*P*N*N;return a

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

Python 2, 124 bytes

def f(m,N,P):w=~-N*P+1;a=[w*[0]for _ in' '*w];g=0;exec"x=g/N/N*~-N;a[x+g/N%N][x+g%N]+=m[x][g/N%N][g%N];g+=1;"*P*N*N;return a

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

share|improve this answer

share|improve this answer

answered 12 mins ago

ovs

17.5k21058

answered 12 mins ago

ovs

17.5k21058

answered 12 mins ago

ovs

17.5k21058

17.5k21058

add a comment | 

add a comment | 

up vote
0
down vote

Jelly, 20 bytes

µ;€0Zð⁺0ṁ+ṚU$}ṚU+⁸µ/

Try it online!

Bah, Jelly has an attitude today...

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

add a comment | 

up vote
0
down vote

Jelly, 20 bytes

µ;€0Zð⁺0ṁ+ṚU$}ṚU+⁸µ/

Try it online!

Bah, Jelly has an attitude today...

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

add a comment | 

up vote
0
down vote

up vote
0
down vote

Jelly, 20 bytes

µ;€0Zð⁺0ṁ+ṚU$}ṚU+⁸µ/

Try it online!

Bah, Jelly has an attitude today...

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

Jelly, 20 bytes

µ;€0Zð⁺0ṁ+ṚU$}ṚU+⁸µ/

Try it online!

Bah, Jelly has an attitude today...

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

share

share

answered 1 min ago

Erik the Outgolfer

29.6k42899

answered 1 min ago

Erik the Outgolfer

29.6k42899

answered 1 min ago

Erik the Outgolfer

29.6k42899

29.6k42899

add a comment | 

add a comment | 

 

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