0167. Two Sum II - Input Array Is Sorted¶
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.
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] <= 1000numbersis sorted in non-decreasing order.-1000 <= target <= 1000- The tests are generated such that there is exactly one solution.
Analysis¶
Since the input array is sorted, to find the the target sum we can use two pointers to reach the optimal complexity. The comparsion condition is by comparing the current sum and target, if current sum is greater than target then we should minimize the sum, and the only way by doing so is moving the bigger (right pointer) to a smaller value (move to left). Note that left pointer can only move to right and right pointer can only move to left, or the algorithm will fail (infinite loop due to duplicating check).
- Time: O(n)
- Space: O(1)
Code¶
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for (int l = 0, r = numbers.size() - 1; l < r;) {
int curr = numbers[l] + numbers[r];
if (target == curr) return {l + 1, r + 1};
else if (target < curr) r --;
else l ++;
}
return {};
}
};