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leetcode-167-two-sum-II-input-array-is-sorted.swift
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/**
Copyright (c) 2020 David Seek, LLC
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM,
DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE
OR OTHER DEALINGS IN THE SOFTWARE.
https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
Note:
Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
*/
/**
Big O Annotation
Time complexity O(n) where n is the amount of elements in nums.
Space complexity O(n) where n is the amount of elements in nums.
*/
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
/**
Creating a map for every element in nums
with their respecitve indices.
*/
var map: [Int: Int] = [:]
// Iterate through nums
for (index, element) in nums.enumerated() {
/**
Check if we have a mapped element can be
created when substracting the current target
with the current element
*/
if let mapped = map[target - element] {
// And return their indices each +1
return [mapped + 1, index + 1]
}
// Add the current element to the map
map[element] = index
}
return []
}