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
}
};