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C16 - Afina Walton #1
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Original file line number | Diff line number | Diff line change |
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@@ -1,19 +1,54 @@ | ||
// This method will return an array of arrays. | ||
// Each subarray will have strings which are anagrams of each other | ||
// Time Complexity: ? | ||
// Space Complexity: ? | ||
// Time Complexity: O(N^2) | ||
// Space Complexity: O(N^2) | ||
function groupedAnagrams(strings) { | ||
throw new Error("Method hasn't been implemented yet!"); | ||
let counter = {}; | ||
let grouped = [] | ||
for (let str of strings) { | ||
// separates each string into array - O(N^2) operation | ||
const sorted = [...str].sort().join(''); | ||
if (!counter[sorted]) counter[sorted] = [str]; | ||
else counter[sorted].push(str); | ||
} | ||
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for (let value of Object.values(counter)) { | ||
grouped.push(value) | ||
} | ||
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return grouped | ||
} | ||
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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: ? | ||
function topKFrequentElements(list, k) { | ||
throw new Error("Method hasn't been implemented yet!"); | ||
// Time Complexity: O(N) | ||
// Space Complexity: O(N) | ||
function createCounter(arr) { | ||
let counter = {}; | ||
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for (let num of arr) { | ||
if (counter[num]) counter[num]++; | ||
else counter[num] = 1; | ||
} | ||
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return counter | ||
} | ||
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function sortObject(counter) { | ||
let counterArray = Object.entries(counter).map(element => +element[0]); | ||
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counterArray.sort((a, b) => b[1] - a[1]); | ||
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return counterArray | ||
} | ||
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function topKFrequentElements(list, k) { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ Nice! Since you are sorting the array, it would be O(n log n) time complexity |
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const counter = createCounter(list); | ||
const sorted = sortObject(counter); | ||
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return sorted.slice(0, k); | ||
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} | ||
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// This method will return true if the table is still | ||
// a valid sudoku table. | ||
|
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✨If you want to include sorting the strings in your time complexity, we could say that the time complexity is O(n m log(m)) where m is the length of the string you are sorting and n is the length of
strings
Space complexity would also be O(n) where counter and grouped are each of size n