-
Notifications
You must be signed in to change notification settings - Fork 0
/
TwoSum.ST.js
43 lines (33 loc) · 1.07 KB
/
TwoSum.ST.js
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
/*
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].
*/
// Naive: O(n^2) double for loop check every sum added to another
function naive(nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let j = i+1; j < nums.length; j++) {
if (nums[i]+nums[j] === target) return[nums[i],nums[j]]
}
}
return [];
}
// can we do this smarter?
// what if we use a cache? Do we need to loop twice?
// Having to check for 0 case tripped me up a little bit.
function twoSum(nums, target) {
let cache = {};
for (let i = 0; i < nums.length; i++) {
// cache that shit
if (typeof cache[nums[i]] === 'number') {
return [i, cache[nums[i]]];
} else {
cache[target-nums[i]] = i;
}
}
return [];
}
twoSum([2,7,11,15], 9)