leetcode/work/Medium/Array/167.two-sum-ii-input-array-is-sorted.js

/*
 * 167. Two Sum II - Input Array Is Sorted
 * Difficulty: Medium
 * https://leetcode.com/problems/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, each
 * incremented 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.
*/

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    let i = 0, j = numbers.length - 1;

    while (i < j) {
        const curr = numbers[i] + numbers[j];
        if (curr === target){
            return [i + 1, j + 1];
        }else{
            if (curr < target){
                i++;
            }else{
                j--;
            }
        }
    }
    return [];
};