Day 6 of #100DaysOfCode | 28-03-2024
Ashish Prajapati
Posted on March 29, 2024
What I learned?
I learned the following topics:
- call in javascript
- classes in javascript
- Inheritance insights in javascript
- How to use map data structure to solve problem
What I developed-solved?
- solved 1 leetcode problem 485. Max Consecutive Ones
- Solved 1 leetcode problem 136. Single number
Difficulties I am facing.
- I am facing difficulties in figuring out the time complexity of map based data structure
- I am facing difficulties in understading context(current execting context and global executing context) in javascript
Code snippet-screenshots-notes
- Leetcode 485. Max consecutive ones
- problem statement: Given a binary array
nums
, return the maximum number of consecutive1
's in the array.
class Solution {
public:
int findMaxConsecutiveOnes(vector<int>& nums) {
int size = nums.size();
int cnt = 0, maxi = 0;
for (int i = 0; i < size; i++) {
/*we maintain a count variable to count the number of consecutive ones
and when we encountered a 0, the count variable becomes 0 again
*/
if (nums[i] == 1) {
cnt++;
/*maintaining maxi variable to store maximum number of
consecutive ones
*/
if (cnt > maxi) {
maxi = cnt;
}
}
else{
cnt = 0;
}
}
return maxi;
}
};
//Time complexity: O(n), n = number of elements in an array
//Space Complexity: O(1), because we are not using any extra space
- Leetcode 136. Single number
- Problem statement: Given a non-empty array of integers
nums
, every element appears twice except for one. Find that single one. - example.
Input: nums = [2,2,1]
Output: 1
Input: nums = [4,1,2,1,2]
Output: 4
Input: nums = [1]
Output: 1
- Brute force approach
class Solution {
public:
int singleNumber(vector<int>& nums) {
int size = nums.size();
for(int i = 0; i<nums.size(); i++){
/*1. grab a single element
2. find that element in the entire array
3. if that element appears for the second time increase cnt count
4. if that element doesn't appear more than ones, the
cnt remains 0 and that's our final answer
*/
int number = nums[i];
int cnt = 0;
for(int j = 0; j<nums.size(); j++){
if(nums[j] == number && i != j){
cnt++;
}
}
if(cnt==0){
return number;
}
}
return -1;
}
};
//Time Complexity: O(n)*O(n)=O(n^2)
//Space Comlexity: O(1),
- better approach using (map data structure)
class Solution {
public:
int singleNumber(vector<int>& nums) {
map<long long, int> m;
for(int i = 0; i<nums.size(); i++){
m[nums[i]]++;
}
/* 1.map all the elements in key-value pairs
2.Iterate through the map and search for the value of key to be 1
3.return value's key which is our final answer
*/
for(auto it: m){
if(it.second == 1){
return it.first;
}
}
return -1;
}
};
/* Time complexity: O(N*logM) + O(M), where M = size of the map(i.e. m = (n/2)+1 )
N = size of an array
-> we are inserting N element in map, and insertion takes logM time
Space complexity: O(M) where M = size of the map ( m = (n/2)+1 )
we are using map data structure to store elements in key-value pairs
*/
- Optimal Solution
class Solution {
public:
int singleNumber(vector<int>& nums) {
int ans = 0;
for(int i = 0; i<nums.size(); i++){
ans = (ans ^ nums[i]);
}
return ans;
}
};
/* Example: nums = [4, 1, 2, 1, 2]
= 1 ^ 1 ^ 2 ^ 2 ^ 4
= ( 0 ) ^ ( 0 ) ^ 4
=> 0^0 = 0
=> 0^4 = 4
-> because 0 ^ number = (number itself),
number ^ same number = 0
*/
/* Time complexity: O(n), where n = number of elements in an array
Space complexity: O(1), because we are not using any extra space
*/
JavaScript call
it calls a method of an object, substituing another object for a current object
function SetUsername(username) {
this.username = username
}
function createuser(username, email, password) {
SetUsername.call(this, username)
this.email = email
this.password = password
}
const user1 = new createuser("user1", "user1@gmail.com", "something")
console.log(user1.username) //output: user1
Classes in javaScript
class User {
constructor(username, email, password) {
this.username = username
this.email = email
this.password = password
}
encryptPassword() {
return `${this.password}ads`
}
changeUsername() {
return `${this.username.toUpperCase()}`
}
}
const userDetails = new User("user", "user@gmail.com", "12345")
console.log(userDetails.encryptPassword()) //12345ads
console.log(userDetails.changeUsername()) //USER
// Behind the scenes
function User2(username, email, password) {
this.username = username
this.email = email
this.password = password
}
User2.prototype.encryptPassword = function () {
return `${this.password}ads`
}
User2.prototype.changeUsername = function () {
return `${this.username.toUpperCase()}`
}
const user2 = new User2("user2", "user2@gmail.com", "userPwd")
console.log(user2.encryptPassword()) //userPwdads
console.log(user2.changeUsername()) //user2 to USER2
Inheritance
**class User {
constructor(username) {
this.username = username
}
login() {
return `${this.username} logged in successfully!`
}
}
//'Teacher' inherits (acquires) properties of base class which is 'User')
class Teacher extends User {
constructor(username, email, password) {
super(username)
this.email = email
this.password = password
}
addCourse() {
console.log(`A new course was added by ${this.username}`)
}
}
const teacher = new Teacher("teacher1", "teacher1@gmail.com", "teacherPWD")
console.log(teacher.login()) //teacher1 logged in successfully!
const user = new User("user1")
console.log(user.username) //user1
console.log(teacher instanceof Teacher) //true
console.log(user instanceof User) //true
console.log(teacher instanceof User) //true**
Static properties
class user {
constructor(username) {
this.username = username
}
logme() {
console.log(`Username is: ${this.username}`)
}
}
const Goblin = new user("something new")
Goblin.logme() //Username is: something new
💖 💪 🙅 🚩
Ashish Prajapati
Posted on March 29, 2024
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