Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 9

Description

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.

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 * 10^4
• -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.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> twoSum(vector<int>& numbers, int target) {
    int begin = 0;
    int end = numbers.size() - 1;
    while(begin < end && numbers[begin] + numbers[end] != target) {
      if(numbers[begin] + numbers[end] < target) {
        begin += 1;
      } else {
        end -= 1;
      }
    }
    return {begin + 1, end + 1};
  }
};
// Accepted
// 19/19 cases passed (4 ms)
// Your runtime beats 88.71 % of cpp submissions
// Your memory usage beats 75.51 % of cpp submissions (9.6 MB)