
# Two Sum

## Descrption

[original description](https://leetcode.com/problems/two-sum/)

## Solutions

### Brute Force

```javascript
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
  for (let i = 0; i < nums.length - 1; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target) {       // check every possible case
        return [i, j];
      }
    }
  }
  return [-1, -1] // will never touch
};
```

- Time Complexity: O(n^2)
- Space Complexity: O(1)

### Cache Diff by Map

```javascript
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
  const cache = {}
  for (let i = 0; i < nums.length; i++) {
    const cachedIndex = cache[target - nums[i]];
    if (cachedIndex !== undefined) {              // found cached diff, return directly
      return [cachedIndex, i];        
    }

    cache[nums[i]] = i;                           // otherwise cache it for later use
  }

  return [-1, -1];
};
```

- Time Complexity: O(n)
- Space Complexity: O(n)

