LeetCode/167.two-sum-ii-input-array-is-sorted.cpp at master · samba9274/LeetCode · GitHub

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/*
 * @lc app=leetcode id=167 lang=cpp
 *
 * [167] Two Sum II - Input Array Is Sorted
 *
 * https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
 *
 * algorithms
 * Medium (59.76%)
 * Likes:    7391
 * Dislikes: 1052
 * Total Accepted:    1.2M
 * Total Submissions: 2M
 * Testcase Example:  '[2,7,11,15]\n9'
 *
 * 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.
 *
 *
 */

// @lc code=start
class Solution
{
public:
    vector<int> twoSum(vector<int> &numbers, int target)
    {
        int i = 0, j = numbers.size() - 1;
        vector<int> result;
        while (i < j)
        {
            if ((numbers[i] + numbers[j]) == target)
            {
                result.push_back(i + 1);
                result.push_back(j + 1);
                return result;
            }
            if ((numbers[i] + numbers[j]) < target)
                i++;
            else
                j--;
        }
        return result;
    }
};
// @lc code=end