1.two-sum.en.md

Problem

https://leetcode-cn.com/problems/two-sum

Problem Description

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

The easiest solution to come up with is Brute Force. We could write two for-loops to traverse every element, and find the target numbers that meet the requirement. However, the time complexity of this solution is O(N^2), while the space complexity is O(1). Apparently, we need to find a way to optimize this solution since the time complexity is too high. What we could do is to record the numbers we have traversed and the relevant index with a Map. Whenever we meet a new number during traversal, we go back to the Map and check whether the diff between this number and the target number appeared before. If it did, the problem has been solved and there's no need to continue.

Key Points

• Find the difference instead of the sum
• Connect every number with its index through the help of Map
• Less time by more space. Reduce the time complexity from O(N) to O(1)

Code

• Support Language: JS

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

Complexity Anlysis

• Time Complexity: O(N)
• Space Complexity：O(N)