bkwebhero - 1 year ago 135
Swift Question

# Is there a way to shuffle an array so that no two consecutive values are the same?

I have an array of colors that will populate a pie chart to act as a game spinner. I don't want the same colors to appear next to each other, making one huge chunk in the circle.

My array looks something like this:

``````var colors = ["blue", "red", "green", "red", "blue", "blue", "blue", "green"]
``````

The problem is of course that there are three blues together. Is there anything built into Swift that will allow me to equally (or as close to equally as possible) separate my colors?

I can test for a match with the following code, but rearranging them proves to be a bit more difficult.

``````var lastColor = "white"

for color in colors {
if color == lastColor {
print("match")
}
lastColor = color
}
``````

UPDATE:

To make my
`colors`
array, I start out with the number of spaces for each color. It looks something like this:

``````let numberOfReds = 2
let numberOfGreens = 2
let numberOfBlues = 4

let spaces = numberOfReds + numberOfGreens + numberOfBlues

for _ in 0..< spaces {
if numberOfReds > 0 {
numberOfReds -= 1
colors.append("red")
}
if numberOfGreens > 0 {
numberOfGreens -= 1
colors.append("green")
}
if numberOfBlues > 0 {
numberOfBlues -= 1
colors.append("blue")
}
}
``````

Which ends up spiting out:

``````colors = ["red", "green", "blue", "red", "green", "blue", "blue", "blue" ]
``````

Disclaimer: In order to generate a "random" solution I am going to use backtracking. This approach is NOT fast and is NOT cheap by a space point of view.

Infact both Time And Space Complexity are O(n!)... and this is HUGE!

However it gives you a valid solution as random as possible.

## Backtracking

So you want a random combination of a list of values with the condition that the solution is valid if there are not be 2 consecutive equals elements.

In order to have a real random solution I suggest the following approach.

I generate every possible valid combination. For this I'm using a backtracking approach

``````func combinations<Element:Equatable>(unusedElms: [Element], sequence:[Element] = []) -> [[Element]] {
// continue if the current sequence doesn't contain adjacent equal elms
guard !Array(zip(sequence.dropFirst(), sequence)).contains(==) else { return [] }

// continue if there are more elms to add
guard !unusedElms.isEmpty else { return [sequence] }

// try every possible way of completing this sequence
var results = [[Element]]()
for i in 0..<unusedElms.count {
var unusedElms = unusedElms
let newElm = unusedElms.removeAtIndex(i)
let newSequence = sequence + [newElm]
results += combinations(unusedElms, sequence: newSequence)
}
return results
}
``````

Now given a list of colors

``````let colors = ["blue", "red", "green", "red", "blue", "blue", "blue", "green"]
``````

I can generate every valid possible combination

``````let combs = combinations(colors)

[["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "red", "blue", "green", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "green", "blue", "green", "blue", "red", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "blue", "green", "blue", "green"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "red", "blue", "green", "blue"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "green", "blue", "red", "blue", "green"], ["blue", "red", "blue", "green", "blue", "red", "green", "blue"], ["blue", "red", "blue", "green", "blue", "green", "red", "blue"], ["blue", "red", "blue", "green", "blue", "green", "blue", "red"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], ["blue", "red", "blue", "red", "green", "blue", "green", "blue"], …, ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "green", "blue", "red", "blue", "red", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "red", "blue", "green", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"], ["green", "blue", "red", "blue", "green", "blue", "red", "blue"]]
``````

Finally I just need to pick one of these combinations

``````let comb = combs[Int(arc4random_uniform(UInt32(combs.count)))]
// ["red", "blue", "green", "blue", "green", "blue", "red", "blue"]
``````

## Improvements

If you don't need a true random solution, but simply a permutation that doesn't have 2 consecutive equal elements we can change the previous function in order to return the first valid solution.

``````func combination<Element:Equatable>(unusedElms: [Element], sequence:[Element] = []) -> [Element]? {
guard !Array(zip(sequence.dropFirst(), sequence)).contains(==) else { return nil }
guard !unusedElms.isEmpty else { return sequence }

for i in 0..<unusedElms.count {
var unusedElms = unusedElms
let newElm = unusedElms.removeAtIndex(i)
let newSequence = sequence + [newElm]
if let solution = combination(unusedElms, sequence: newSequence) {
return solution
}
}
return nil
}
``````

Now you can simply write

``````combination(["blue", "red", "green", "red", "blue", "blue", "blue", "green"])
``````

to get a valid solution (if it does exists)

``````["blue", "red", "green", "blue", "red", "blue", "green", "blue"]
``````

This approach can be much faster (when the solution does exist) however the worst case scenario is still O(n!) for both space and time complexity.

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