How to sort an integer array based on a duplicate values count. here less number of duplicates should come first.

input  [5, 2, 1, 2, 4, 4, 1, 1, 2, 3, 3, 6]
OutPut [5, 6, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1]

CodePudding user response：

This is actually really easy to do using the .sort/.sorted function like this:

let output = input.sorted { i1, i2 in
    i1 != i2
}

CodePudding user response：

let numbers = [5,2,1,2,4,4,1,1,2,3,3,6]
let sortedNumber = numbers.sorted()
print("Input: ",sortedNumber)

var dict = [Int: Int]()
for item in sortedNumber {
    let isExist = dict.contains(where: {$0.key == item})
    if !isExist {
        dict[item] = 1
    } else {
        if let value = dict[item] {
            dict[item] = value   1
        }
    }
}

var finalArray = [Int]()
let sortedArray = dict.sorted { (first, second) -> Bool in
    return first.value < second.value
}

for d in sortedArray {
    for _ in 1...d.value {
        finalArray.append(d.key)
    }
}
print("Output: ",finalArray)

Input:  [1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 6]
Output:  [5, 6, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1]