Problem

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
  var hash = {};
  var len = nums.length;
  for (var i = 0; i < len; i++) {
    if (nums[i] in hash) return [hash[nums[i]], i];
    hash[target - nums[i]] = i
  }
  return [-1, -1];
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).