twoSumIIInputArrayIsSorted.js

////////////////////////////////////////////////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].
/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
 const twoSum2 = function(numbers, target) {
    const res = [];

    for (let i = 0; i < numbers.length; i++){
        for (let j = i + 1; j < numbers.length; j++){
            if(numbers[i] + numbers[j] === target){
                res.push(i + 1, j + 1);
            }
        }
    }

    return res;
};

// console.log(twoSum2([2,7,11,15], 9));
// console.log(twoSum2([2,3,4], 6));
// console.log(twoSum2([-1,0], -1));