 addValues(arr2)
 
 
-// create function that determines n power of n
+// create function that determines n to the power of n
 func powerOfN(number: Int, power: Int) -> Int {
     var sum = 1
     
-    if power == 0 {
+    if power == 0 { // O(1)
         return 1
     }
     

    
    return sum
}
powerOfN(2, power: 4)
powerOfN(0, power: 4) // 0
powerOfN(1, power: 4) // 1
powerOfN(2, power: 1) // 2
powerOfN(2, power: 0) // 1














// check array for missing number in consecutive numbers, return that number (to reuse numbers)

// assign that number to a customer


// take in string, return letter that appears the most frequently
//func mostCommonLetter(word: String) -> String? {
//    
//    var tracker: Dictionary<String,Int> = [:]
//    
//    var count: Int = 1
//    let wordArray = word.characters.map {String($0)} // convert to array of strings
//    
//    var letter: String? = nil
//    var largestNumber = 0
//    
//    for letter in wordArray {
//        
//        if !tracker.keys.contains(letter) {
//            tracker[letter] = count
//        } else {
//            
//        }
//  
//    }
//    
//    print(tracker)
//
//    return letter
//}
//    
//
//mostCommonLetter("heellloe")



//adding / multiplying / traversing multi dimensional arrays
//caesar cypher
//tic tac toe
//Check to see if string is balanced (())
//Given three numbers return the median
//Fibonacci - given an idx return the fib value at idx
//given a jumble: String and knownWords: [String] return list of potential words that the string may be.

    
//    numbers = numbers.sort() // 1) if we can use a sort function: O(n log n)
//    numbers.minElement() // 2) or just ask for min element: O(n)
//    let minNum = numbers.reduce(Int.max, combine: { min($0, $1) }) // 3) or use reduce
    
    for i in 0...numbers.count-1 {
        for j in 0...numbers.count-1 { // O(n^2)
