leetcode-347-top-k-frequent-elements.swift

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https://leetcode.com/problems/top-k-frequent-elements/


Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:

Input: nums = [1], k = 1
Output: [1]
Note:

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
You can return the answer in any order.

*/

/**
Big O Annotation
Time complexity O(n log n) because of the sorting overhead.
Space complexity O(n) where n is the amount of elements in nums.
*/
func topKFrequent(_ nums: [Int], _ k: Int) -> [Int] {
    
    /**
    Hash map of nums elements and 
    their frequency
    */
    var hash: [Int: Int] = [:]
    
    // Iterate through nums
    for element in nums {
        
        /**
        Check if we have already mapped 
        the given element..
        */
        if let mapped = hash[element] {

            // ... if so, add 1
            hash[element] = mapped + 1

        } else {

            // If not, set 1
            hash[element] = 1
        }
    }
    
    // Sort the map by value descending
    let sorted = hash.sorted { element1, element2 in 
        return element1.value > element2.value
    }
    
    // Map to only have array of nums elements
    let mapped = sorted.map { element in
        return element.key
    }
    
    /**
    Return the desired elements 
    as a new array from 0 up to but not including k.

    Safe operation as stated in the notes:
    "You may assume k is always valid, 1 ≤ k ≤ number of unique elements."
    */
    return Array(mapped[0..<k])
}