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[heesun-task] WEEK 01 solutions #2388
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,26 @@ | ||
| /** | ||
| * @param {number[]} nums | ||
| * @return {boolean} | ||
| */ | ||
| /* | ||
| Goal: return true if duplicated #, false if no duplicated number | ||
|
|
||
| Plan: | ||
| - create a Set | ||
| - loop num in nums | ||
| - num exists in the Set -> return true | ||
| - not exsits -> add the num in the Set, continue to the next loop | ||
| - return false | ||
|
|
||
| space complexity: O(n) | ||
| time complexity: O(n) | ||
| */ | ||
| var containsDuplicate = function(nums) { | ||
| const seen = new Set(); | ||
|
|
||
| for(const num of nums) { | ||
| if (seen.has(num)) return true; | ||
| seen.add(num); | ||
| } | ||
| return false; | ||
| }; | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,33 @@ | ||
| /* | ||
| Goal: return indices of the two numbers such that they add up to target | ||
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|
||
| Plan: | ||
| - create a hash map (dict) to store number -> index | ||
| - loop through nums with index i | ||
| - get currentNum = nums[i] | ||
| - compute pairNum = target - currentNum | ||
| - if pairNum exists in dict | ||
| -> return [dict[pairNum], i] | ||
| - otherwise store currentNum in dict with its index | ||
| - if no pair is found, return [-1, -1] | ||
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| space complexity: O(n) | ||
| time complexity: O(n) | ||
| */ | ||
|
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| var twoSum = function(nums, target) { | ||
| const dict = {}; | ||
|
|
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| for (let i = 0; i < nums.length; i++) { | ||
| const currentNum = nums[i]; | ||
| const pairNum = target - currentNum; | ||
|
|
||
| if (typeof dict[pairNum] === 'number') { | ||
| return [dict[pairNum], i]; | ||
| } else { | ||
| dict[currentNum] = i; | ||
| } | ||
| } | ||
|
|
||
| return [-1, -1]; | ||
| }; |
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저는
new Set(nums).size !== nums.length으로 풀었는데,이 방식은 중복을 발견하는 순간 바로 return true로 종료되니 평균적으로 더 빠를 것 같네요 👍🏻
고생하셨습니다!