yitonghe00’s gists

yitonghe00 / Eloquent JavaScript 0501 Flattening.js

Created April 17, 2020 05:10

https://eloquentjavascript.net/05_higher_order.html#i_aIOczlLyX1

	let arrays = [[1, 2, 3], [4, 5], [6]];
	// Your code here.
	const flat = (arrays) => {
	const ret = [];
	arrays.forEach((array) => {
	array.forEach((num) => ret.push(num));
	});
	return ret;
	};

yitonghe00 / Eloquent JavaScript 0404 Deep comparison.js

Last active April 12, 2020 01:53

Excises of Eloquent JavaScript 4th edition. Chapter 04 Problem 04. https://eloquentjavascript.net/04_data.html#i_IJBU+aXOIC

	// Your code here.
	const deepEqual = (a, b) => {
	if (typeof a != 'object' \|\| typeof b != 'object') return a === b;
	if (a == null \|\| b == null) return a === b;

	const aKeys = Object.keys(a);
	const bKeys = Object.keys(b);
	if (aKeys.length != bKeys.length) return false;

	for (let i = 0 ; i < aKeys.length; i++) {

yitonghe00 / Eloquent JavaScript 0403 A list.js

Created April 12, 2020 01:16

Excises of Eloquent JavaScript 4th edition. Chapter 04 Problem 03. https://eloquentjavascript.net/04_data.html#i_6xTmjj4Rf5

	// Your code here.
	const arrayToList = (arr) => {
	return arrayToListR(arr, 0);
	}

	const arrayToListR = (arr, start) => {
	if (start >= arr.length) return null;

	const ans = {};
	ans.value = arr[start];

yitonghe00 / Eloquent JavaScript 0402 Reversing an array.js

Created April 12, 2020 01:15

Excises of Eloquent JavaScript 4th edition. Chapter 04 Problem 02. https://eloquentjavascript.net/04_data.html#i_6xTmjj4Rf5

	// Your code here.
	const reverseArray = (arr) => {
	const ans = [];
	for (let i = arr.length - 1; i >= 0; i--) {
	ans.push(arr[i]);
	}
	return ans;
	}

	const reverseArrayInPlace = (arr) => {

yitonghe00 / Eloquent JavaScript 0401 The sum of a range.js

Created April 12, 2020 01:14

Excises of Eloquent JavaScript 4th edition. Chapter 04 Problem 01. https://eloquentjavascript.net/04_data.html#i_8ZspxiCEC/

	// Your code here.
	const range = (start, end, step = 1) => {
	const ans = [];
	if (step > 0 && start < end) {
	for (let i = start; i <= end; i += step) {
	ans.push(i);
	}
	} else {
	if (step > 0) step = -1;
	for (let i = start; i >= end; i += step) {

yitonghe00 / 7. Reverse Integer (#1 Math).java

Created February 11, 2020 09:26

7. Reverse Integer (https://leetcode.com/problems/reverse-integer/): Given a 32-bit signed integer, reverse digits of an integer.

	// Math solution: Pull the digits and take care of the overflow
	// Time: O(logx), 1ms
	// Space: O(1), 36.6mb
	class Solution {
	public int reverse(int x) {
	int ans = 0;

	while(x != 0) {
	int tail = x % 10;
	int newAns = ans * 10 + tail;

yitonghe00 / 724. Find Pivot Index (#1 Array + DP).java

Created February 11, 2020 08:54

724. Find Pivot Index (https://leetcode.com/problems/find-pivot-index/): Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, …

	// Array + DP solution: store the prefix sum in array and scan from left
	// Time: O(n), 1ms
	// Space: O(n), 42.4mb
	class Solution {
	public int pivotIndex(int[] nums) {
	// Count the prefix sum of nums: prefix[i] = nums[0] + nums[1] + .. + nums[i - 1]
	int len = nums.length, sum = 0;
	int[] prefix = new int[len];
	for(int i = 0; i < len; i++) {
	prefix[i] = sum;

yitonghe00 / 724. Find Pivot Index (#1 Array + Hash table).java

Last active September 8, 2020 12:08

724. Find Pivot Index (https://leetcode.com/problems/find-pivot-index/): Given an array of integers nums, write a method that returns the "pivot" index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, …

	// Array + Hash table solution: store the prefix sum in array and scan from left
	// Time: O(n), 1ms
	// Space: O(n), 42.4mb
	class Solution {
	public int pivotIndex(int[] nums) {
	// Count the prefix sum of nums: prefix[i] = nums[0] + nums[1] + .. + nums[i - 1]
	int len = nums.length, sum = 0;
	int[] prefix = new int[len];
	for(int i = 0; i < len; i++) {
	prefix[i] = sum;

yitonghe00 / 974. Subarray Sums Divisible by K (#1 Array + Hash table).java

Last active February 11, 2020 08:55

974. Subarray Sums Divisible by K (https://leetcode.com/problems/subarray-sums-divisible-by-k/): Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

	// Array + Hash table solution: store prefix sum mod in table as we counting; combination of 560 & 523
	// Time: O(n), 19ms
	// Space: O(k), 45.2mb
	class Solution {
	public int subarraysDivByK(int[] A, int K) {
	int ans = 0;

	// Init map of <sum % K, count>
	Map<Integer, Integer> map = new HashMap<>();
	map.put(0, 1);

yitonghe00 / 523. Continuous Subarray Sum (#1 Array + Brutal force).java

Created February 11, 2020 02:21

523. Continuous Subarray Sum (https://leetcode.com/problems/continuous-subarray-sum/): Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to a multiple of k, that is, sums up to n*k where n is also an integer.

	// Array + Brutal force solution
	// Time: O(n ^ 2), 16ms
	// Space: O(1), 42.2mb
	class Solution {
	public boolean checkSubarraySum(int[] nums, int k) {
	for(int i = 0; i < nums.length; i++) {
	int sum = nums[i];

	for(int j = 1; i + j < nums.length; j++) {
	sum += nums[i + j];

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