0001 - Two Sum

0001 - Two Sum

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.

Examples

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

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

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?

Java Solution

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if(map.containsKey(complement)) {
                return new int[] {map.get(complement), i};
            } 
            else map.put(nums[i], i);
        }
        return new int[] {};
    }
}

Javascript Solution

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let map = new Map();
    
    for(let i = 0; i < nums.length; i++) {
        if(map.get(target - nums[i]) !== undefined) return [map.get(target - nums[i]), i]
        else map.set(nums[i], i)
    }
};