167. Two Sum II – Input Array Is Sorted using JavaScript

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.

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.

Solution

First attempt : Brute force (Can be optimized further)

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    let output = [];
    for(let i = 0; i <numbers.length ; i++){
        for(let k = i+1; k < numbers.length; k++){
            if((numbers[i] + numbers[k]) == target){
                output.push(i+1,k+1);
                break;
            }
        }
    }
    return output;
};

Second attempt : Two pointer method

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    let output = [];
    let leftPointer = 0;
    let rightPointer = numbers.length - 1;
    while(rightPointer > leftPointer){
        let sum = numbers[rightPointer] + numbers[leftPointer];
        if(sum == target){
            return [leftPointer+1, rightPointer+1]
        }
        else if( sum > target){
            rightPointer--;
        }
        else{
            leftPointer++;
        }
    }
    return output;
};