Two Sum II with a sorted array is another straightforward two-pointers style problem. The key is acknowledging that the sorted array provides us the opportunity of an O(2n) or O(n) solution as the would-be O(n log n) sort has already been done for us.
The main intuition for solving this problem is understanding where to start the pointers and how to perform the traversal. In my solution, we initialize the left pointer on the first element and the right pointer on the last element. Then, we loop continuously, calculating the step's sum, returning if the solution is found, otherwise adjusting our pointers for the subsequent loop.
Two Sum II - Input Array Is Sorted
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Medium
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Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use the same element twice.
Your solution must use only constant extra space.
Example 1:
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. We return [1, 2].
Example 2:
Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
Example 3:
Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
The backtracking-based function uses backtracking to progressively explore the solution set. The key insight is the creation and maintenance of the current subset, which is added to, and then removed from, as the traversal takes place across the solution space.
func twoSum(numbers []int, target int) []int {
// setup our two pointers
left := 0
right := len(numbers) - 1
// continuously multiply these values and move left/right accordingly
for left < right {
sum := numbers[left] + numbers[right]
if sum == target {
return []int{left + 1, right + 1}
} else if sum > target {
right--
} else {
left++
}
}
return nil // not allowed
}
Breakdown: