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C16 - Afina Walton #1

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49 changes: 42 additions & 7 deletions lib/exercises.js
Original file line number Diff line number Diff line change
@@ -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)

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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

// 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);
}

for (let value of Object.values(counter)) {
grouped.push(value)
}

return grouped
}

// 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 = {};

for (let num of arr) {
if (counter[num]) counter[num]++;
else counter[num] = 1;
}

return counter
}

function sortObject(counter) {
let counterArray = Object.entries(counter).map(element => +element[0]);

counterArray.sort((a, b) => b[1] - a[1]);

return counterArray
}

function topKFrequentElements(list, k) {

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✨ Nice! Since you are sorting the array, it would be O(n log n) time complexity

const counter = createCounter(list);
const sorted = sortObject(counter);

return sorted.slice(0, k);

}

// This method will return true if the table is still
// a valid sudoku table.
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