leetcode-1431-kids-with-the-greatest-number-of-candies.swift

/**

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https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/


Given the array candies and the integer extraCandies, where candies[i] represents the number of candies that the ith kid has.

For each kid check if there is a way to distribute extraCandies among the kids such that he or she can have the greatest number of candies among them. Notice that multiple kids can have the greatest number of candies.

 

Example 1:

Input: candies = [2,3,5,1,3], extraCandies = 3
Output: [true,true,true,false,true] 
Explanation: 
Kid 1 has 2 candies and if he or she receives all extra candies (3) will have 5 candies --- the greatest number of candies among the kids. 
Kid 2 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. 
Kid 3 has 5 candies and this is already the greatest number of candies among the kids. 
Kid 4 has 1 candy and even if he or she receives all extra candies will only have 4 candies. 
Kid 5 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. 
Example 2:

Input: candies = [4,2,1,1,2], extraCandies = 1
Output: [true,false,false,false,false] 
Explanation: There is only 1 extra candy, therefore only kid 1 will have the greatest number of candies among the kids regardless of who takes the extra candy.
Example 3:

Input: candies = [12,1,12], extraCandies = 10
Output: [true,false,true]

*/

/**
Big O Annotation
Time complexity O(n log n) because of the sorting overhead.
Space complexity O(n) as we're creating the sorted array.

Space could be optimized by not sorting first,
but iterating through the candies, getting the largest amount
and then iterating again to calculate.
*/
func kidsWithCandies(_ candies: [Int], _ extraCandies: Int) -> [Bool] {
        
    /**
    If we don't have any candies then
    there's nothing to supply
    */ 
    guard candies.count > 0 else {
        return []
    }
    
    /**
    If we only have one candy,
    then that is automatically the largest.
    */
    guard candies.count > 1 else {
        return [true]
    }
    
    // Sort so we have the largest at the end
    let sorted = candies.sorted()

    /**
    We can savely force unwrap the last item
    as we already checked if we have at least 2 items.
    */
    let largestAmountOfCandy = sorted.last!

    // Results array
    var result: [Bool] = []
    
    // Iterate through the candies
    for candy in candies {
        
        /**
        Calculate if the current candy + all the extra candies
        is at least equal to the largest amount of candy.
        */
        let willHaveLargestAmount = (candy + extraCandies) >= largestAmountOfCandy
        result.append(willHaveLargestAmount)
    }
    
    return result
}