Wrote grouped_anagrams method. #1
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| // This method will return an array of arrays. | ||
| // Each subarray will have strings which are anagrams of each other | ||
| // Time Complexity: O(n) where n is the length of the array passed in as the argument "strings" because each item in the array is visited once | ||
| // Space Complexity: O(1) (***Chris, is this right? It takes 2 times the space of 'strings' which reduces to 1. Or is it O(n) because the space taken is is 2 x n where n is the length of the 'strings' array?***) | ||
| function groupedAnagrams(strings) { | ||
| let anagrams = {}; | ||
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| strings.forEach(str => { | ||
| let tempStr = str.split('').sort().join(''); // alphabetize str | ||
| anagrams[tempStr] ? anagrams[tempStr].push(str) : anagrams[tempStr] = [str]; | ||
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| }); | ||
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| return Object.values(anagrams); | ||
| } | ||
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| // This method will return the k most common elements. | ||
| // In the case of a tie it will select the first occuring element. | ||
| // Time Complexity: ? | ||
| // Space Complexity: ? | ||
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| // Clarifying questions: | ||
| // 1) What happens if there are < k elements in the list array? | ||
| // 2) How big are list and k likely to be? (Optimize for list or for k)? | ||
| // 3) Are the elements in list likely to be anything that can't be used as a key? (Will they always be numbers/strings/something that can be coorced into a string? Will they ever be null or undefined?) | ||
| function topKFrequentElements(list, k) { | ||
| if (list.length < k){ | ||
| return list; | ||
| } // this code might not be necessary once the function is complete | ||
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| let elementCounts = {}; | ||
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| // iterate through each item in list (array) | ||
| // if the item is a key in elementCounts, add 1 to is value | ||
| // if it's not a key already, add it as a key with value 1 | ||
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| // find highest count in arrary | ||
| const maxCount = elementCounts.values.max; | ||
| let solutions = []; | ||
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| obj = { name: 'Bobo' } | ||
| obj.somethingElse = obj.name | ||
| delete obj.name | ||
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| // OPTION 1 | ||
| // k.times { // *** O(k) --> O(1) ? | ||
| // loop through the list array // *** O(n*k) | ||
| // if the item's matching value === maxCount, add the key to solutions array | ||
| // update the maxCount to the next item in the array | ||
| // } | ||
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| // OPTION 2 | ||
| // const eleCountsInverted = a new object switching keys and values // *** O(n) ? | ||
| // now counts are keys, values are the items from the list array | ||
| let highestCountsArray = Object.keys(eleCountsInverted); // O(1) ? | ||
| // k.times { | ||
| // loop through highestCountsArray // *** O(k) --> O(1) ? | ||
| // add the values associated with these keys to the solutions array | ||
| // } | ||
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| // return an array with the first k keys with highest values in any order | ||
| return solutions; | ||
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| } | ||
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| // This method will return true if the table is a valid sudoku table. | ||
| // Each element can either be a ".", or a digit 1-9 | ||
| // The same digit cannot appear twice or more in the same | ||
| // row, column or 3x3 subgrid | ||
| // Time Complexity: ? | ||
| // Space Complexity: ? | ||
| function validSudoku(table) { | ||
| throw new Error("Method hasn't been implemented yet!"); | ||
| } | ||
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| module.exports = { | ||
| groupedAnagrams, | ||
| topKFrequentElements, | ||
| validSudoku | ||
| }; | ||
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I would say O(n) because you are building a JavaScript object which scales based on the size of the array.