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‎JavaScript/chapter01/1.1 - Is Unique/solution.js

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@@ -4,9 +4,7 @@ Output: returns a deduped array of integers
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*/
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// Solution using Set
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const isUnique = (arr) => [...new Set(arr)];
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// Test Cases
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console.log(isUnique([1,1,1,2,2,2,2,3,3,3,3]) === [1,2,3]);

‎JavaScript/chapter01/1.2 - Check Perm/rroque98_sol.js

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/*Given two strings, write a method to decide
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if one is apermutation of the other permutation:*/
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const isPermutation = (str1, str2) =>{
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if (str1.length !== str2.length) {
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return false;
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}
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const obj1 = determineCharCount(str1);
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const obj2 = determineCharCount(str2);
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for (let char of str1) {
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if (obj1[char] !== obj2[char]) {
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return false;
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}
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}
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return true;
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//******** helper function ******
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function determineCharCount(string) {
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const obj = {};
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for (let char of string) {
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if (obj[char] !== undefined) {
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obj[char]++;
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} else {
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obj[char] = 1;
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}
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}
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return obj;
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}
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}
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// Tests:
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console.log(isPermutation('abc', 'abb') === false);
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console.log(isPermutation('abb', 'abc') === false);
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console.log(isPermutation('aaa', 'abc') === false);
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console.log(isPermutation('abc', 'abcd') === false);
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console.log(isPermutation('abc', 'bac') === true);
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console.log(isPermutation('', '') === true);
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console.log(isPermutation('12', '21') === true);

‎JavaScript/chapter01/1.2 - Check Perm/solution.js

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/* Given two strings, write a method to decide
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if one is permutation of the other */
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var checkPerm = function(stringOne, stringTwo) {
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//if different lengths, return false
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if (stringOne.length !== stringTwo.length ) {
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return false;
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}
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//else sort & compare
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else {
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var sortStringOne = stringOne.split('').sort('').join('');
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var sortStringTwo = stringTwo.split('').sort('').join('');
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return sortStringOne === sortStringTwo;
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}
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};

‎JavaScript/chapter01/1.3 - URLify/rroque98_sol.js

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// Write a method to replace all spaces within a string with '%20'
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const URLify = (string) => string.split(' ').join('%20');
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// Tests:
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console.log(URLify('Hello World') === 'Hello%20World');
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console.log(URLify('') === '');
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console.log(URLify('This is an example') === 'This%20is%20an%20example');

‎JavaScript/chapter01/1.3 - URLify/solution.js

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//testing
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let arr = ['M', 'r', ' ', 'J', 'o', 'h', 'n', ' ', 'S', 'm', 'i', 't', 'h', ' ', ' ', ' ', ' '];
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//before
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console.log(arr);
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let ans = URLify(arr, 13);
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//after
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console.log(ans);

‎JavaScript/chapter01/1.4 - PalinPerm/rroque98_sol.js

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// Given a string, write a function to check if it is a permutation of a palindrome
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const isPalindromePermutation = (str) => {
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const strNoSpaces = str.split(' ').join('');
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const obj = {};
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var oddCount = 0;
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for (let char of strNoSpaces) {
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if (obj[char] !== undefined) {
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obj[char]++;
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} else {
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obj[char] = 1;
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}
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}
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for (let key in obj) {
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if (obj[key] % 2) {
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oddCount++;
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}
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if (oddCount > 1) {
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return false;
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}
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}
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return true;
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}
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console.log(isPalindromePermutation('tact coa') === true);
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console.log(isPalindromePermutation('tact cooa') === true);
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console.log(isPalindromePermutation('tacr coa') === false);
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console.log(isPalindromePermutation('tactr coa') === false);

‎JavaScript/chapter01/1.5 - OneAway/rroque98_sol.js

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/* There are 3 types of edits that can be made on a string:
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insert character, remove character, replace a character
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Given 2 strings write a function that will check if it is
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1 or 0 edits away*/
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const isOneAway = (str1, str2) => {
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if (str1.length === str2.length) {
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let errorCount = 0;
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for (let i = 0; i < str1.length; i++) {
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if (str1[i] !== str2[i]) {
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errorCount++;
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}
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if (errorCount > 1) {
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return false;
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}
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}
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} else {
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let errorCount = 0;
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const longestStr = findLongestStr(str1, str2);
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let x = 0;
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for(let i = 0; i < longestStr; i++) {
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if (str1[i] !== str2[x]) {
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errorCount++;
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x++;
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}
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if (errorCount > 1) {
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return false;
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}
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x++;
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}
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// ***** Helper functions ********
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function findLongestStr(str1, str2) {
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str1 > str2 ? str1 : str2;
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}
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}
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return true;
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}
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// TESTS
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console.log(isOneAway('pale', 'ple') === true);
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console.log(isOneAway('pales', 'pale') === true);
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console.log(isOneAway('pale', 'bale') === true);
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console.log(isOneAway('pale', 'bake') === false);

‎JavaScript/chapter01/1.6 - String Compression/rroque98_sol.js

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/* Implement a method to perform basic string
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compression using the counts of repeated characters
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Ex: 'aabcccccaaa' would become a2b1c5a3
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*/
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const stringCompression = (str) => {
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if (!str.length) {
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return '';
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}
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var compStr = '';
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var count = 1;
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var currentChar = str[0];
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for (let i = 1; i < str.length; i++) {
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let char = str[i];
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if (char === currentChar) {
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count++;
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if (i === str.length - 1) {
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compStr += `${currentChar}${count}`;
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}
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} else {
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compStr += `${currentChar}${count}`;
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currentChar = char;
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count = 1;
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}
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}
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return compStr;
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}
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// TESTS
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console.log(stringCompression('aabcccccaaa') === 'a2b1c5a3');
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console.log(stringCompression('cccccccc') === 'c8');
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console.log(stringCompression('') === '');
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console.log(stringCompression('AabccCccaaa') === 'A1a1b1c2C1c2a3');

‎JavaScript/chapter01/1.9 - String Rotation/solution.js

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/* Assume you have a method isSubstring which checks if one word is a substring
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of another. Given two strings, sl and s2, write code to check if s2 is a rotation of sl using only one
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call to isSubstring (e.g., "waterbottle" is a rotation of"erbottlewat") */
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var StringRotate = function(string1, string2) {
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if (string1.length !== string2.length ){
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return false;
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}
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return ( string2 + string1 ).includes(string1); // one call of Substring
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};
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//Test
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console.log(StringRotate('waterbottle', 'erbottlewat'), true);

‎JavaScript/chapter02/2.2 - Return Kth to Last/solution.js

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/* Return Kth to Last: Implement an algorithm to find the kth to last element of a singly linked list. */
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var linkedList = function(value) {
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this.value = value;
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this.next = null;
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};
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var findKthToLast = function(k, head) {
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//do recursive
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if ( head == null || k < 1 ) {
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return;
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} else if ( k == 1 ) {
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console.log(head.value);
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findKthToLast(k, head.next)
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} else {
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findKthToLast( k-1, head.next);
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}
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};
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/* Tests */
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var a = new linkedList('1');
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var b = new linkedList('2');
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var c = new linkedList('3');
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var d = new linkedList('4');
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var e = new linkedList('5');
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var f = new linkedList('6');
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var g = new linkedList('7');
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a.next = b;
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b.next = c;
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c.next = d;
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d.next = e;
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e.next = f;
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f.next = g;
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findKthToLast(3, a);
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findKthToLast(10, a);
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findKthToLast(-1, a);
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findKthToLast(0, a);
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findKthToLast(1, a);

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