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Wow! The pointer solution is so clever! Simple and beautiful :)

I really think showing the linear solution with a set here would’ve been highly beneficial. Our last solution works only because the array is sorted and this problem in interviews is almost never sorted. The solution below is always linear. var seenSet: Set = [] for num in array { If seenSet.contains(sum - num) { return true } else { seenSet.insert(num) } } return false

Fast? Finally a video I don't need to speed up. :)

Thanks and I hope you realize how much value you are bringing to iOS developers. These algos are not easy to learn and you are doing a great job at chunking down the larger concept into code that is easy to absorb.

That was the best explanation I found on youtube until now.

@Sean If you look at the documentation the remove(at: Index) is of complexity O(n) it would add more computation in binarySearch flow /// - Complexity: O(*n*), where *n* is the length of the array. public mutating func remove(at index: Int) -> [Element].Element

In brute-force methode you can loop j starting from i+1 instead of from 0, because for j = 0...i you've done in the previous loops. The loop should say "for j in i+1..

First of all, thank you very much, your videos are very well explained and easy to understand. I think it's worth mentioning that the array need to be sorted for binary search and pointer move approach.

Thanks so much for your videos. I love these videos as I can grasp it easier if I am stepped through it. Thanks to your delegate tutorial, I was able to do my app which is on the app store now :) thanks so much for doing these videos please keep doing them.

Pointer solution is really good bro.. i also solved this problem with 2 for loops but whatever you suggested its simply short and sweet

Your last two videos explain starting with a sorted array. Do you plan on reviewing any sorting algorithms? Love your videos btw man. Easy to follow and you do a great job explaining concepts!

This was a long one.... Happy to help answer questions or clear up any confusion in the comments!

Thank you Sean!

great video series and explanations

perfect!!!! Can you please make video on how to calculate time complexity of a program.It would be really helpful Thanks In Advance.

in Brute Force method, we could even skip the last iteration because the last number(14) has already been compared to all other numbers, like: for i in 0..

Just today I had oral exam in data structures and algorithms and was talking about brute force, binary search, notations etc... It fantastic to see how to work with something like that in swift :D keep crushing it

nice one, Move pointer is definitely the easy one :D

Man! Thanks! Revisiting Algorithms on the Job Search! This explanation is dope. Any book Recommendations for Interview Questions?

Hey Sean Will Brute Force code work if we have multiple combinations for sum in array ?

Another great video.

The pointer solution doesn't work well in different scenarios like [3, 2, 4] and the target is 6

Nice Explanation....Are you planning to do a separate video on Time complexity and how to calculate it?

I like hidden basketball references like 23 lol

How would you do this if we are doing this with 2 arrays and need to find if elements in both equal a sum?

let's try twoSumPointers(array: [3,2,4], sum: 6)

Nice video Sean!

Why is the input ascending ordered? Else we would need to sort it first and which makes loop count higher

Hey Sean, I'm not sure if this is the right place to ask you questions about this sort of old post you have, but when I downloaded your source code and ran 2. Binary Search method, the code wouldn't run and stops at "let = hasCompliment = binarySearch(array: ~~~)" while saying "Use of unresolved identifier 'binarySearch'. By any means would you know why I'm getting this error? Thanks

Could you add a tutorial about Big O Notations?

Just saw this video and i wanted to confirm something. The Linear Solution you did for it ...... Its fine for the Array you took. But you are taking 3 presumptions to solve this. 1. Data in array is sorted, 2. There is ONLY 1 match to get the sum value 3. No Value in Array is repeated (this one goes for all 3) Can you confirm if I am thinking wrong here ? or am i right ?

Isn't "complement" written with "e" ? just to make sure I understand correctly. Anyway great video as always!

With brute force method, if array contains two or more same object and you add "where a != b" line, the algorithm will not work

Hey Sean. Any idea about solving 3 sum or 4 sum problem?

When you say array.count - 1 where is the - 1 coming from? i understand that you need to get the highIndex, but i dont understand how you're able to get the highIndex from - 1? This is for the last func twoSumPointers

if you change "let numbers3" to "var numbers3" and under it add "numbers3.sort()". Input: _______________________________________________________ var numbers3 = [1, 20, 31, 14, 15, 3] numbers3.sort() print(numbers3) Output: _______________________________________________________ [1, 3, 14, 15, 20, 31] Then no matter if the largest number is in the middle or 1st or anywhere it will sort them from low to high and you will get the right answer. Also thank you for this amazing video! really helped me a lot.

Nice tutorial Sean. I have added some code to your last method to filter and sort the array(I know it's sorted already. But if it wasn't here it is) as a bit of input from me to potentially reduce the iterations inside the while loop. func comparePointers(array: [Int], sum: Int) -> Bool { //Firstly apply a filter to eliminate all numbers greater than the sum and sort, in case numbers are not sorted. Thus reducing the amount of numbers the while loop has to iterate through. var filteredArray = array.filter({return $0 < sum}).sorted() print(filteredArray) var lowIndex = 0 var highIndex = filteredArray.count - 1 while lowIndex < highIndex { let sumOfItems = filteredArray[lowIndex] + filteredArray[highIndex] if sumOfItems == sum { print("Sum of \(filteredArray[lowIndex]) and \(filteredArray[highIndex]) = \(sum)") return true } else if sumOfItems < sum { lowIndex += 1 } else if sumOfItems > sum { highIndex -= 1 } } print("Pointers have crossed") return false }

Hey Sean, it's a request, i have struck at a coding problem that, i wanted to save a image which is picked from local gallery and then save it into sqlite database or any database.Thanks Keep posting...

Hi Sean, The pointer algorithm isn't sufficient for all condition, let say input array is [3,2,4] and the sum is 6.

The last solution isn't working if we looking for the index and not the value itself. For example I'm looking for `n` below: ``` // it's working if we looking for this n array[index] = n // but not working for array[n] = 10 ```

Shouldn't the tempArray be initiated outside of the loop in point 2? When you're removing the n-th element in each loop from tempArray, you're only removing this single element, because you have (var tempArray = array) in the line above. If you'd initiate tempArray outside of the loop you'd shorten tempArray by n elements for each binary search.

Hi Sean, I was testing your method Brute Force, and it has a bug, cause in your array you have 2 of number 7, and if you try bruteForceTwoSum(array:numbers, sum:14) it will return false, but it should return true, cause you have 2 unique number 7

In 15.25 sec output is 1 and 12 = 13. Why it's not showing 6 and 7 = 13?

For method 2, there is an assumption here that you can copy the array and then remove an element from the copy in constant time. Is that a good assumption? I would not expect so.

I guess the 3rd approach works only for a sorted array

For solutions 2 & 3, you are assuming that the array is sorted, this can' t be a given assumption. Sorting itself takes a best case O(NlogN) time, so you need to factor that as well when calculating the total running time.

If I run this code on leetcode It gives the wrong answer: Input: [3,2,4] Target: 6 The output I am getting: [1,1] Expected Output : [1,2]

at 0.75 speed you sound pretty hammered and it's hard to take a drunk programmer seriously... :-)

Wan't bruteforce solution give us false for 14?

Hi Sean , Great tutorial however solution #3 (index pointer) will not work if the input is negative number array ex- ([-1,-2,-3,-4,-5] , -8). In this scenario the else if logic will have to be reversed . else if sumOfTwoItems < sumValue { highIndex -= 1 }else if sumOfTwoItems > sumValue { lowIndex += 1 } Please comment . Btw great videos:)

This works if theres only one solution for the sum

this doesnt work for the two sum problem on leetcode

Hey sean! I just realized you can use ".contains" function in the tempArray in twoSumBinarySearch with compliment function. Hehe let numbers = [1, 3, 6, 7, 7, 12, 14] func twoSumBinarySearch(array: [Int], sum: Int) -> Bool { if array.count == 0 { return false } for i in 0..

I like your videos man. But one friendly advice; there are too many people here who are not native speakers. Please for them, speak a bit slower. Thanks for your videos.

your logic will fail with [3,2,4] with sum 6.

It will fix 6 and 7 = 13 //Find all two elements pair in the array that add up to given sum func findPairGiverSum(numbers: [Int], sum: Int) { let sortedNumber = numbers.sorted() // print(sortedNumber) var left = 0 var right = sortedNumber.count - 1 while(left < right){ // print("index (\(left), \(right))") if(left != 0 && sortedNumber[left] == sortedNumber[left - 1]){ left = left + 1 continue } let pSum = sortedNumber[left] + sortedNumber[right] if(pSum == sum){ print("(\(sortedNumber[left]), \(sortedNumber[right]))") right = right - 1 left = left + 1 }else if(pSum > sum){ right = right - 1 }else{ left = left + 1 } } } //findPairGiverSum(numbers: [2,3,4,3,4,1,6,6,6,7,5,9,1,8,9], sum: 10) //findPairGiverSum(numbers: [0,1,2,4,4,7,5,3,4,1,8], sum: 8) findPairGiverSum(numbers: [1,3,6,7,7,12,14], sum: 13)