Contents.swift

/**
 Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

 Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

  

 Example 1:
 Input: nums = [1,2,2,3,1]
 Output: 2
 Explanation:
 The input array has a degree of 2 because both elements 1 and 2 appear twice.
 Of the subarrays that have the same degree:
 [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
 The shortest length is 2. So return 2.
 
 Example 2:
 Input: nums = [1,2,2,3,1,4,2]
 Output: 6
 Explanation:
 The degree is 3 because the element 2 is repeated 3 times.
 So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
  

 Constraints:
 - nums.length will be between 1 and 50,000.
 - nums[i] will be an integer between 0 and 49,999.
 */
class Solution {
    func findShortestSubArray(_ nums: [Int]) -> Int {
        var degreeMap = [Int:[Int]]()
        for (i, c) in nums.enumerated() {
            if var timesArr = degreeMap[c] {
                timesArr.append(i)
                degreeMap[c] = timesArr
            } else {
                degreeMap[c] = [i]
            }
        }
        
        var sortedArr = degreeMap.sorted { $0.value.count > $1.value.count }
        var maxTimes = sortedArr.first!.value.count
        var minimum = Int.max
        for c in sortedArr {
            if c.value.count == maxTimes {
                maxTimes = c.value.count
                let delta = c.value.last! - c.value.first!
                if delta < minimum {
                    minimum = delta
                }
            } else {
                break
            }
        }
        return minimum + 1
    }
}

let s = Solution()
let r = s.findShortestSubArray([1,2,2,3,1,4,2])
print(r)