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LESSON 121 OF 678 · ARRAYS - CHALLENGES

First N Items of an Array in Swift

Practice Read the full explanation

You get an array of integers nums and a number n. Return an array with the first n numbers of nums.

Keep them in the same order. When n is 0, return an empty array.

 

Example 1:

Input: nums = [3, 8, 1, 9, 4], n = 3
Output: [3, 8, 1]
Explanation: The first 3 numbers are 3, 8 and 1.
The 9 and the 4 are left out.

Example 2:

Input: nums = [5, 6, 7], n = 0
Output: []
Explanation: No numbers are taken, so the result is empty.

Example 3:

Input: nums = [2, 4], n = 2
Output: [2, 4]
Explanation: n is the length of nums, so every number is kept.

 

Constraints:

  • nums has 1 to 10 numbers.
  • Every number is an integer from -1000 to 1000.
  • n is from 0 to the length of nums.
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    SWIFT · ARRAYS - CHALLENGES

    First N Items in Swift, explained

    Problem overview

    The method firstN receives an array of integers, nums, and a number n. It returns a new array holding the first n numbers, in the same order. When n is 0 the result is an empty array, and when n equals the length of nums every number is kept.

    With nums = [3, 8, 1, 9, 4] and n = 3, the answer is [3, 8, 1]: the 9 and the 4 are left out. The constraints promise that n is between 0 and the length of nums, so you never have to handle a number that is too big.

    Approach

    The first n items sit at indexes 0 up to n - 1. The half-open range 0..<n stops just before n, so it covers exactly those n indexes. Putting that range in the array's subscript picks the items. No loop is needed.

    Step by step

    1. Take the items of nums in the range 0..<n. The result of that subscript is an ArraySlice, a view of part of nums, not an array. The method must return [Int], so wrap the slice in Array(...), which copies the items into a new array. Store it in result.
    2. Return result.

    The solution in Swift

    class Solution {
    func firstN(_ nums: [Int], _ n: Int) -> [Int] {
        let result = Array(nums[0..<n])
        return result
    }
}

    Running it on the three examples:

    [3, 8, 1]
[]
[2, 4]

    Those lines come from nums = [3, 8, 1, 9, 4] with n = 3, nums = [5, 6, 7] with n = 0, and nums = [2, 4] with n = 2.

    Language details

    • When n is 0, 0..<0 is an empty range. Slicing with it is allowed and gives an empty slice, so result is [].
    • When n equals nums.count, the range ends at the count, one past the last index. That is how a range reaches the end of an array.
    • The input array is not changed. result is a separate copy.

    Testing

    A few more inputs are worth trying, because they sit at the edges:

    • nums = [-5, 0, 7], n = 1 gives [-5]. Negative numbers are copied like any other.
    • nums = [10], n = 1 gives [10]: a single item, all kept.
    • nums = [10], n = 0 gives []: a single item, none kept.

    Slices matter here for one more reason. A slice keeps the indexes of the array it came from, so a slice that starts at 0, like this one, happens to look like an array, while a slice that starts later does not. Wrapping in Array(...) removes that surprise for every case.

    Edge cases and common mistakes

    • Forgetting Array(...). The method promises [Int], and an ArraySlice<Int> is a different type. The compiler stops with cannot convert return expression of type 'ArraySlice<Int>' to return type '[Int]'.
    • Using 0...n. The closed range includes index n, which gives n + 1 items, and it crashes when n is the full length. Use ..<.
    • Looking for a method that returns an optional. Nothing is optional here. The subscript gives items directly.

    Cost

    The work is one copy of at most n numbers, and the constraints keep the array to ten items. Nothing here depends on cleverness, only on picking the right range.

    Another way

    The library has prefix(_:), which also returns a slice. Array(nums.prefix(n)) gives the same result for these inputs, and it doesn't crash when n is larger than the array. The range version is the one this lesson practices.

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