notes/OJ notes/pages/Leetcode Two-Sum-II-Input-Array-Is-Sorted.md

Leetcode Two-Sum-II-Input-Array-Is-Sorted

2022-07-11 14:54

Algorithms:

#algorithm #two_pointers

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

---

Problem

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.

Examples

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
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

Thoughts

[!summary] This is a #two_pointers problem.

decrement the right pointer makes the sum smaller, increment the left pointer makes the sum bigger, and thud we get the answer.

Solution

class Solution {
public:
  vector<int> twoSum(vector<int> &numbers, int target) {
    // Sorted array, use two pointers
    // remember to subtract 1
    int r = numbers.size() - 1;
    int l = 0;
    int sum;

    while (l < r) {
      sum = numbers[l] + numbers[r];
      if (sum < target) {
        l++;
      } else if (sum > target) {
        r--;
      } else {
        return {l + 1, r + 1};
      }
    }

    // Shouldn't return this, since there is a solution
    return {0, 0};
  }
};