Two Sum II - Input Array Is Sorted
Problem Description
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].
Constraints:
- 2 <= numbers.length <= 3 * 104
- -1000 <= numbers[i] <= 1000
numbersis sorted in non-decreasing order.- -1000 <= target <= 1000
- The tests are generated such that there is exactly one solution.
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The simplest way to solve this problem is by checking every possible pair in the array to see if it adds up to the target. Since the array is sorted, once the sum exceeds the target, we can stop checking further elements with the current element.
Code:
- Time Complexity: O(n^2) - We are checking every possible pair.
- Space Complexity: O(1) - No additional space is used.
2. Two Pointers
Intuition:
Since the array is sorted, we can use a two-pointer technique. Start with one pointer at the beginning and the other at the end of the array. Check the sum of the values at these pointers:
- If the sum equals the target, return the 1-indexed positions.
- If the sum is less than the target, move the left pointer to the right to increase the sum.
- If the sum is more than the target, move the right pointer to the left to decrease the sum.
Code:
- Time Complexity: O(n) - Each element is visited at most once by either of the pointers.
- Space Complexity: O(1) - No additional space is used.