swift subset of array match the duplicate elements-CodePudding

Scene:

let superTrue = [1,2,3,3]
let superFalse = [1,2,3]

let sub = [2,3,3]

Requirements:

sub is subset of superTrue but not superFalse.

The wrong code:

print(Set(sub).isSubset(of: superTrue)) // true
print(Set(sub).isSubset(of: superFalse)) // true

because Set(sub) is [2,3], removes the duplicated 3

Don't remove the duplicate, and compute subset, any ideas?

CodePudding user response：

Set will always remove duplicates by its nature.

One of the options would be to count occurrences of all elements in all three arrays:

let subCounts = sub.reduce(into: [:]) { $0[$1, default: 0]  = 1 }
let superTrueCounts = superTrue.reduce(into: [:]) { $0[$1, default: 0]  = 1 }
let superFalseCounts = superFalse.reduce(into: [:]) { $0[$1, default: 0]  = 1 }

And then to check if values for all keys of the wannabe subarray equals or are less than the values of the same keys of the "super" array:

subCounts
  .keys
  .allSatisfy { subCounts[$0, default: 0] <= superTrueCounts[$0, default: 0] }
subCounts
  .keys
  .allSatisfy { subCounts[$0, default: 0] <= superFalseCounts[$0, default: 0] }

This will gives the output you expect (as far as I understood the questions).

Note 1. Since you originally tried to use Set, I assumed the order doesn't matter and you only care about specific elements and the number of their occurrences.

Note 2. The elements should be Equatable.