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Refactor JS - Array & Hash#426
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Adding multiple solutions is being discussed in#189, It's also difficult to decide which refactors make the code easier to understand, even though I am partial to your use of array methods.
I don't know where@neetcode-gh wants to take this.
| } | ||
| }; | ||
| return [ -1, -1 ] |
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I don't thinkreturn [-1, -1] is needed since"You may assume that each input would haveexactly one solution" for this problem.
| /** | ||
| * @param {number[]} nums | ||
| * Time O(N) | Space O(N) | ||
| * @return {boolean} | ||
| */ | ||
| var containsDuplicate = function(nums) { | ||
| const numsSet = new Set() | ||
| for(const i of nums){ | ||
| if(numsSet.has(i)){ | ||
| return true | ||
| } | ||
| numsSet.add(i) | ||
| } | ||
| return false | ||
| return (new Set(nums)).size !== nums.length |
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This solution has already been proposed (#371)
| var isAnagram = function(s, t) { | ||
| let map = {}; | ||
| if (s.length !== t.length) return false; | ||
| const reOrder = (str) => str | ||
| .split('') | ||
| .sort((a, b) => a.localeCompare(b)) | ||
| .join(''); | ||
| return reOrder(s) === reOrder(t) | ||
| }; | ||
| /** | ||
| * @param {string} s | ||
| * @param {string} t | ||
| * Time O(N) | Space O(N) | ||
| * @return {boolean} | ||
| */ | ||
| var isAnagram = function(s, t, map = new Map()) { | ||
| if (s.length !== t.length) return false; | ||
| if (s.length !== t.length) { | ||
| return false; | ||
| } | ||
| addCharFrequency(s, map); | ||
| return subtractCharFrequency(t, map) | ||
| for (let i = 0; i < s.length; i++) { | ||
| if (map[s[i]]) { | ||
| map[s[i]]++; | ||
| } else { | ||
| map[s[i]] = 1; | ||
| } | ||
| }; |
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Adding multiple solutions is being discussed here (#189)
Closing for anotherPR |
No description provided.