Problem 1517. Put m balls into n boxes (again)

Created by Binbin Qi

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According to Cody Problem 1516, if I put 3 balls into 2 boxes then
   1     1
   1     2
   2     2
   2     2
The columns stand for a ball, the number stands for a box, and the row stands for a case.
Now we count the number of balls in each boxes.
so we get
   0
   1
   2
   3
the 1-th row stand for the 1-th case,3 balls are put into the first boxes,the others is empty the 2-th row stand for the 2-th case,2 balls are put into the first boxes,1 balls is put into the second boxes
You should write a function, input m and n, m stand for the number of balls ,n stand for the number of boxes,output all the cases.
Example
If m = 3,n = 4, you should output
   0     0     0
   1     0     0
   0     1     0
   0     0     1
   2     0     0
   1     1     0
   1     0     1
   0     2     0
   0     1     1
   0     0     2
   3     0     0
   2     1     0
   2     0     1
   1     2     0
   1     1     1
   1     0     2
   0     3     0
   0     2     1
   0     1     2
   0     0     3

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Last Solution submitted on Apr 20, 2025

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2 Comments

Piotr Lesniewski on 12 Jul 2019

I'm not sure what is the point of eliminating 'ifs'? At any rate they can be easily replaced by e.g. 'while' loops, so in my opinion you should either eliminate both of them or none.

Rafael S.T. Vieira on 17 Jul 2020

Again, the best way to generate such a sequence is using depth-first search or recursion. By prohibiting the usage of IFs, all solutions are inefficient (because we can't cut the tree, and are forced to follow all branches).

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Problem 1517. Put m balls into n boxes (again)

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Problem 1517. Put m balls into n boxes (again)

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