Split array two array based on sum

RE: Split array two array based on sum

Dan Lluhi — Tue, 16 Apr 2024 19:03:35 GMT

You could probably do this using recursion assuming that your array of numbers doesn't get too long, and if you're worried about recursion then usually there's a way to replicate recursion using the fn!reduce function... but the logic would be something like this pseudocode:

inputs: A (set of solution arrays), B (current solution array that we are building), C (remaining numbers to process)

initiated by calling rule!recursion(A: {}, B: {}, C: { 10, 5, 6, 9, 8, 11, 7, 4, 2, 13, 16, 1, 5 })

if(
    isNullOrEmpty(C),
    {A,B},
    if( 
        sum(B, C[1]) > 20, 
        rule!recursion( A: {A,B}, B: {}, C: C[2:end]),
        rule!recursion( A: A, B: {B, C[1}, C: C[2:end]
    )
)

RE: Split array two array based on sum

Giuseppe Leo — Tue, 16 Apr 2024 13:57:26 GMT

Thanks,
but the lenght of subarray can be more than 2.
For this input:
{ 10, 5, 6, 9, 8, 11, 7, 4, 2, 13, 16, 1, 5 }
I want this output
{{10,5},{6,9},{8,11},{7, 4, 2},{13},{16,1},{5}}

RE: Split array two array based on sum

Stewart Burchell — Tue, 16 Apr 2024 13:48:43 GMT

Not sure if this is what you're looking for but the following code takes each item in the array and looks ahead across the array for a second number that when added to it is less than or equal to 20:

a!localVariables(
  local!nums: { 10, 5, 6, 9, 8, 11, 7, 4, 2, 13, 16, 1, 5 },
  local!indexes: fn!enumerate(fn!length(local!nums)) + 1,
  /* */
  local!reducedArrays: a!forEach(
    items: local!indexes,
    expression: if(
      fv!isLast,
      local!nums,
      fn!remove(local!nums, fn!enumerate(fv!item) + 1)
    )
  ),
  fn!reject(
    a!isNullOrEmpty,
    a!forEach(
      items: local!reducedArrays,
      expression: a!localVariables(
        local!currentArray: fv!item,
        a!forEach(
          items: local!currentArray,
          expression: if(
            fv!isFirst,
            null,
            if(
              local!currentArray[1] + fv!item <= 20,
              a!map(
                firstNumber: local!currentArray[1],
                secondNumber: fv!item
              ),
              null
            )
          )
        )
      )
    )
  )
)

If you're looking to see if you can take each number in turn and see if you can add one or more numbers from the rest of the array then that'll be a different solution.

RE: Split array two array based on sum

Giuseppe Leo — Tue, 16 Apr 2024 12:52:16 GMT

Hi,
each number can be used only one time.

RE: Split array two array based on sum

Stewart Burchell — Tue, 16 Apr 2024 12:49:19 GMT

What rules govern the split? Are you looking for all permutations (possible ways) of making a total of 20 or less? Using the numbers in the initial array once or as many times as you like? e.g. {10,5}, {10,6}, {10,9} etc.