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26 changes: 26 additions & 0 deletions contains-duplicate/heesun-task.js
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) {
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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!

const seen = new Set();

for(const num of nums) {
if (seen.has(num)) return true;
seen.add(num);
}
return false;
};
33 changes: 33 additions & 0 deletions two-sum/heesun-task.js
Original file line number Diff line number Diff line change
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/*
Goal: return indices of the two numbers such that they add up to target

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]

space complexity: O(n)
time complexity: O(n)
*/

var twoSum = function(nums, target) {
const dict = {};

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];
};