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Kirsten #2
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,17 +1,24 @@ | ||
| // 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 log n), where n is the combined length of the input strings (since the strings are being sorted) | ||
| // Space Complexity: O(n), where n is the length of the input array. | ||
| function grouped_anagrams(strings) { | ||
| throw new Error("Method hasn't been implemented yet!"); | ||
| } | ||
| const indexHash = {}; | ||
| strings.forEach ((string) => { | ||
| const tempString = string.split('').sort().join(''); | ||
| if (!indexHash[tempString]) { | ||
| indexHash[tempString] = [string]; | ||
| } else { | ||
| indexHash[tempString].push(string); | ||
| } | ||
| }) | ||
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| const returnArray = [] | ||
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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 top_k_frequent_elements(list, k) { | ||
| throw new Error("Method hasn't been implemented yet!"); | ||
| for (let key in indexHash) { | ||
| returnArray.push(indexHash[key]) | ||
| } | ||
| return returnArray; | ||
| } | ||
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@@ -20,10 +27,56 @@ function top_k_frequent_elements(list, k) { | |
| // 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: ? | ||
| // Time Complexity: O(n), where n is the number of elements in the sudoku table | ||
|
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. I would say instead O(n2) where n is the length of a side of the sudoku table. |
||
| // Space Complexity: O(n), where n is the number of elements in the sudoku table | ||
| function valid_sudoku(table) { | ||
| throw new Error("Method hasn't been implemented yet!"); | ||
| // check rows | ||
|
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| for (let row = 0; row < table.length; row++) { | ||
| const rowHash = {}; | ||
| for (let i = 0; i < table[row].length; i++) { | ||
| let char = table[row][i] | ||
| if (char !== ".") { | ||
| if (rowHash[char]) { | ||
| return false; | ||
| } | ||
| rowHash[char] = 1; | ||
| } | ||
| } | ||
| } | ||
|
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| for (let i = 0; i < table[0].length; i++) { | ||
| const colHash = {}; | ||
| for (let col = 0; col < table.length; col++) { | ||
| let char = table[col][i]; | ||
| if (char !== '.') { | ||
| if (colHash[char]) { | ||
| return false; | ||
| } | ||
| colHash[char] = 1; | ||
| } | ||
| } | ||
| } | ||
|
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| const subboxes = [[0, 1, 2], [3, 4, 5], [6, 7, 8]]; | ||
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| for (let counter1 = 0; counter1 < 3; counter1++) { | ||
| for (let counter2 = 0; counter2 < 3; counter2++) { | ||
| const subboxHash = {}; | ||
| for (let counter3 = 0; counter3 < 3; counter3++) { | ||
| for (let counter4 = 0; counter4 < 3; counter4++) { | ||
| let char = table[subboxes[counter1][counter3]][subboxes[counter2][counter4]]; | ||
| if (char !== '.') { | ||
| if (subboxHash[char]) { | ||
| return false; | ||
| } | ||
| subboxHash[char] = 1; | ||
| } | ||
| } | ||
| } | ||
| } | ||
| } | ||
| return true; | ||
| } | ||
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| module.exports = { | ||
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But you're not sorting an array of all the characters in all the strings.
Instead I would say that the time complexity is O(n * m log m), where n is the number of strings and
mis the length of the largest string. If you could assume the all the strings were less than a certain number of characters you could then say it's O(n).