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[yoonthecoder] Week 5 #870
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,27 @@ | ||
var threeSum = function (nums) { | ||
nums.sort((a, b) => a - b); | ||
const results = []; | ||
for (let i = 0; i < nums.length - 2; i++) { | ||
if (i > 0 && nums[i] === nums[i - 1]) continue; | ||
let left = i + 1; | ||
let right = nums.length - 1; | ||
while (left < right) { | ||
const currSum = nums[i] + nums[left] + nums[right]; | ||
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if (currSum == 0) { | ||
results.push([nums[i], nums[left], nums[right]]); | ||
left++; | ||
right--; | ||
// to avoid duplicates | ||
while (left < right && nums[left] === nums[left - 1]) left++; | ||
while (left < right && nums[right] === nums[right + 1]) right--; | ||
} else if (currSum < 0) { | ||
left++; | ||
} else right--; | ||
} | ||
} | ||
return results; | ||
}; | ||
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// Time complexity: O(n^2); | ||
// Space complexity: O(n) |
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,13 @@ | ||
var maxProfit = function (prices) { | ||
// set the initial value to Infinity so that it can always return minimum value | ||
let minPrice = Infinity; | ||
let maxPrice = 0; | ||
for (i = 0; i < prices.length; i++) { | ||
minPrice = Math.min(prices[i], minPrice); | ||
maxPrice = Math.max(maxPrice, prices[i + 1] - minPrice); | ||
} | ||
return maxPrice; | ||
}; | ||
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// Time complexity: O(n) - Iterating through the array | ||
// Space complexity: O(1) |
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var groupAnagrams = function (strs) { | ||
const map = new Map(); | ||
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for (const str of strs) { | ||
// split the string into each character (returns an array) => sort it => convert it to string | ||
const sortedStr = str.split('').sort().join(''); | ||
// use the sorted Str as unique keys in the map | ||
if (map.has(sortedStr)) { | ||
map.get(sortedStr).push(str); | ||
} else map.set(sortedStr, [str]); | ||
} | ||
// convert values into an array | ||
return [...map.values()]; | ||
}; |
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minPrice를 Infinity로 지정하신건
for문의 시작값을 0으로 하여 첫 번째 가격도 최소 값이 될수 있도록 하신걸까요?
저는 인덱스를 1부터 시작했는데 이런 방법도 있었군요~!👍