Two Sum

Description

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

 

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

Solution(javascript)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    const hashmap = {}
    
    // Iterate through the numbers in the array,
    for (let i=0; i<nums.length; i++) {
        // Check the difference between the target numbers, and nums[i]
        const difference = target - nums[i]
        
        // Since we know the difference between target and nums[i],
        // we can check to see if we've seen it already, by checking if hashmap[difference] exists.
        //
        // If we have seen the difference before, we return the current index, 
        // and the index at which we saw the difference.
        if (hashmap.hasOwnProperty(difference)) {
            return [i, hashmap[difference]]
        }
        
        // Otherwise, store the current number in the hashmap, 
        // along with its index
        hashmap[nums[i]] = i
    }
};