Happy to see you visting WebSlate.IO. I'm a JavaScript and web developer living in sunny Singapore creator of WebSlate.IO.
With a passion for teaching, I started making lot of training materials for free. during pandemic 2020, i started helping people who lose jobs through offline now its online called WebSlate.IO.
I'm just getting started.
Happy to see you visting WebSlate.IO. I'm a JavaScript and web developer living in sunny Singapore creator of WebSlate.IO.
With a passion for teaching, I started making lot of training materials for free. during pandemic 2020, i started helping people who lose jobs through offline now its online called WebSlate.IO.
I'm just getting started.
Big O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity. Big O is a member of a family of notations invented by Paul Bachmann.
Aug 8, 2021☕ 10 mins readBig O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity. Big O is a member of a family of notations invented by Paul Bachmann.
Aug 8, 2021☕ 10 mins readBig O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity.
n.| Constant | Logarithmic | Linear | N Log N | Quadratic | Cubic | Exponential | |
|---|---|---|---|---|---|---|---|
| N | O(1) | O(log n) | O(n) | O(n log n) | O(n2) | O(n3) | O(2n) |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 2 |
| 2 | 1 | 1 | 2 | 2 | 4 | 8 | 4 |
| 4 | 1 | 2 | 4 | 8 | 16 | 64 | 16 |
| 8 | 1 | 3 | 8 | 24 | 64 | 512 | 256 |
| 16 | 1 | 4 | 16 | 64 | 256 | 4,096 | 65536 |
| 32 | 1 | 5 | 32 | 160 | 1,024 | 32,768 | 4,294,967,296 |
| 64 | 1 | 6 | 64 | 384 | 4,069 | 262,144 | 1.84 X 1019 |
| 1024 | 1 | 10 | 1,024 | 10,240 | 1,048,576 | 1,073,741,824 | still curious? |
This takes constant / same time regardless of the number of inputs. always execute in same amount of time, doesn't matter about input size.
Example 1.1: function return whether number is odd or even.
1function isEvenOrOdd(n) {2 return n % 2 ? 'Odd' : 'Even';3}
Example 1.2: Look-up table - function return exist key's value.
1const greetTable = { en: 'Hi', fr: '', ta: 'வணக்கம்' }2function greet(locale = 'en') {3 return greetTable[locale];4}
Example 1.2: Find Square of number.
1function square(no) {2 return no * no3}
Example 2.1: function return logarithmic. value of n is halved on each iteration of the loop.
1function logarithmic(n) {2 let result = 03 while(n>1) {4 n = n / 2;5 result = result + 16 }7 return result8}
Simple multiply table if i pass multiply(2)
1function multiply(no) {2 for(let i=0;i<=no;i++) {3 console.log(i, ': ', i*no)4 }5}
In algorithmics, space and time are like two separate poles. Increasing speed will most often lead to increased memory consumption and vice-versa. for example: merge sort, which is extremely fast but requires a lot of memory and bubble sort, a slow algorithm but one that occupies minimal space.
Big O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity.
n.| Constant | Logarithmic | Linear | N Log N | Quadratic | Cubic | Exponential | |
|---|---|---|---|---|---|---|---|
| N | O(1) | O(log n) | O(n) | O(n log n) | O(n2) | O(n3) | O(2n) |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 2 |
| 2 | 1 | 1 | 2 | 2 | 4 | 8 | 4 |
| 4 | 1 | 2 | 4 | 8 | 16 | 64 | 16 |
| 8 | 1 | 3 | 8 | 24 | 64 | 512 | 256 |
| 16 | 1 | 4 | 16 | 64 | 256 | 4,096 | 65536 |
| 32 | 1 | 5 | 32 | 160 | 1,024 | 32,768 | 4,294,967,296 |
| 64 | 1 | 6 | 64 | 384 | 4,069 | 262,144 | 1.84 X 1019 |
| 1024 | 1 | 10 | 1,024 | 10,240 | 1,048,576 | 1,073,741,824 | still curious? |
This takes constant / same time regardless of the number of inputs. always execute in same amount of time, doesn't matter about input size.
Example 1.1: function return whether number is odd or even.
1function isEvenOrOdd(n) {2 return n % 2 ? 'Odd' : 'Even';3}
Example 1.2: Look-up table - function return exist key's value.
1const greetTable = { en: 'Hi', fr: '', ta: 'வணக்கம்' }2function greet(locale = 'en') {3 return greetTable[locale];4}
Example 1.2: Find Square of number.
1function square(no) {2 return no * no3}
Example 2.1: function return logarithmic. value of n is halved on each iteration of the loop.
1function logarithmic(n) {2 let result = 03 while(n>1) {4 n = n / 2;5 result = result + 16 }7 return result8}
Simple multiply table if i pass multiply(2)
1function multiply(no) {2 for(let i=0;i<=no;i++) {3 console.log(i, ': ', i*no)4 }5}
In algorithmics, space and time are like two separate poles. Increasing speed will most often lead to increased memory consumption and vice-versa. for example: merge sort, which is extremely fast but requires a lot of memory and bubble sort, a slow algorithm but one that occupies minimal space.
Stack is a linear data structure to store a one-dimensional list and store items in LIFO (Last-in, First-Out) or (FILO) First-In, Last-Out
Apr 16, 2020☕ 10 mins readTree is one of the useful data structures
Mar 16, 2020☕ 10 mins readStack is a linear data structure to store a one-dimensional list and store items in LIFO (Last-in, First-Out) or (FILO) First-In, Last-Out
Apr 16, 2020☕ 10 mins readTree is one of the useful data structures
Mar 16, 2020☕ 10 mins read1int[] findSignatureCounts(int[] arr)
Below is the complexity table for Stack.
| Method | Use | |
|---|---|---|
| 1 | isEmpty() | Return true when Stack is empty |
| 2 | push() | To add item to last |
| 3 | pop() | To remove item which is last |
| 4 | peek() | Return the last item |
| 4 | size() | Return the Stack size |
Below is the simple example in JavaScript.
1function Stack() {2 let count = 0;3 const storage = {};4 const push = (value) => {5 storage[count] = value;6 count++;7 }8 const pop = () => {9 if(count === 0) {10 return undefined;11 }12 count--;13 const lastValue = storage[count];14 delete storage[count]15 return lastValue;16 }17 const size = () => {18 return count;19 }20 const peek = () => {21 return storage[count-1]22 }23 return {24 push,25 pop,26 size,27 peek28 }29}
Below is the complexity table for Stack.
| Operation | Best | Average | Worst | |
|---|---|---|---|---|
| 1 | Access | O(n) | O(n) | |
| 2 | Search | O(n) | O(n) | |
| 3 | Insertion | O(n) | O(n) | |
| 4 | Deletion | O(n) | O(n) | |
| 5 | Space Complexity | O(n) |
We can use it for this problems.
We can use it for this problems.
Below is the problem and its solution.
1Code goes here...
We can use it for this problems.
Below is the Question and its solution.
1Code goes here...
1int[] findSignatureCounts(int[] arr)
Below is the complexity table for Stack.
| Method | Use | |
|---|---|---|
| 1 | isEmpty() | Return true when Stack is empty |
| 2 | push() | To add item to last |
| 3 | pop() | To remove item which is last |
| 4 | peek() | Return the last item |
| 4 | size() | Return the Stack size |
Below is the simple example in JavaScript.
1function Stack() {2 let count = 0;3 const storage = {};4 const push = (value) => {5 storage[count] = value;6 count++;7 }8 const pop = () => {9 if(count === 0) {10 return undefined;11 }12 count--;13 const lastValue = storage[count];14 delete storage[count]15 return lastValue;16 }17 const size = () => {18 return count;19 }20 const peek = () => {21 return storage[count-1]22 }23 return {24 push,25 pop,26 size,27 peek28 }29}
Below is the complexity table for Stack.
| Operation | Best | Average | Worst | |
|---|---|---|---|---|
| 1 | Access | O(n) | O(n) | |
| 2 | Search | O(n) | O(n) | |
| 3 | Insertion | O(n) | O(n) | |
| 4 | Deletion | O(n) | O(n) | |
| 5 | Space Complexity | O(n) |
We can use it for this problems.
We can use it for this problems.
Below is the problem and its solution.
1Code goes here...
We can use it for this problems.
Below is the Question and its solution.
1Code goes here...
1function findSignatureCounts(arr) {2 const ans = Array(arr.length).fill(1);3 // Iterate Array.4 for (let i = 0; i < arr.length; i++) {5 let k = i;6 while (arr[k] !== i + 1) {7 ans[i]++;8 k = arr[k] - 1;9 }10 }11 return ans;12}
On High level note, It worth to read in detail for a better understanding.
There are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
You're given a list of n integers arr[1..n], which is guaranteed to be a permutation of 1..n (in other words, it includes the integers from 1 to n exactly once each, in some order). The meaning of this list is described below.
Initially, each student is holding their own yearbook. The students will then repeat the following two steps each minute: Each student i will first sign the yearbook that they're currently holding (which may either belong to themselves or to another student), and then they'll pass it to student arr[i-1]. It's possible that arr[i-1] = i for any given i, in which case student i will pass their yearbook back to themselves. Once a student has received their own yearbook back, they will hold on to it and no longer participate in the passing process.
It's guaranteed that, for any possible valid input, each student will eventually receive their own yearbook back and will never end up holding more than one yearbook at a time.
You must compute a list of n integers output, whose element at i-1 is equal to the number of signatures that will be present in student i's yearbook once they receive it back.
1int[] findSignatureCounts(int[] arr)
Test Case 1 Explaination:
Test Case 2 Explaination:
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m) | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
1Code goes here...
1function findSignatureCounts(arr) {2 const ans = Array(arr.length).fill(1);3 // Iterate Array.4 for (let i = 0; i < arr.length; i++) {5 let k = i;6 while (arr[k] !== i + 1) {7 ans[i]++;8 k = arr[k] - 1;9 }10 }11 return ans;12}
On High level note, It worth to read in detail for a better understanding.
There are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
You're given a list of n integers arr[1..n], which is guaranteed to be a permutation of 1..n (in other words, it includes the integers from 1 to n exactly once each, in some order). The meaning of this list is described below.
Initially, each student is holding their own yearbook. The students will then repeat the following two steps each minute: Each student i will first sign the yearbook that they're currently holding (which may either belong to themselves or to another student), and then they'll pass it to student arr[i-1]. It's possible that arr[i-1] = i for any given i, in which case student i will pass their yearbook back to themselves. Once a student has received their own yearbook back, they will hold on to it and no longer participate in the passing process.
It's guaranteed that, for any possible valid input, each student will eventually receive their own yearbook back and will never end up holding more than one yearbook at a time.
You must compute a list of n integers output, whose element at i-1 is equal to the number of signatures that will be present in student i's yearbook once they receive it back.
1int[] findSignatureCounts(int[] arr)
Test Case 1 Explaination:
Test Case 2 Explaination:
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m) | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
1Code goes here...
learn more about Storage API (Local and Session Storage).
Aug 8, 2016☕ 2 mins readlearn more about Storage API (Local and Session Storage).
Aug 8, 2016☕ 2 mins readOne of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
In short, running continuation-passing style. Let me shoot few bullets with simple diagram on Promises below.
One of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
In short, running continuation-passing style. Let me shoot few bullets with simple diagram on Promises below.
I would like to continue with an examination of JavaScript Promise API. Let’s have a look at Promise evolution by various libraries. below three libraries which we are going to explore on this article.
Dec 5, 2016☕ 5 mins readOne of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
Oct 21, 2016☕ 2 mins readI would like to continue with an examination of JavaScript Promise API. Let’s have a look at Promise evolution by various libraries. below three libraries which we are going to explore on this article.
Dec 5, 2016☕ 5 mins readOne of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
Oct 21, 2016☕ 2 mins readI would like to continue with an examination of JavaScript Promise API. Let’s have a look at Promise evolution by various libraries. below three libraries which we are going to explore on this article.
The q.js is best one for promise implementation by Kris Kowal. It is more evident of promise evolution.
Lets have a look, What makes it special.
I would like to continue with an examination of JavaScript Promise API. Let’s have a look at Promise evolution by various libraries. below three libraries which we are going to explore on this article.
The q.js is best one for promise implementation by Kris Kowal. It is more evident of promise evolution.
Lets have a look, What makes it special.
One of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
In short, running continuation-passing style. Let me shoot few bullets with simple diagram on Promises below.
One of the good Features in ES6 is Promises Object and their useful methods and they are called software abstraction helps to works smoothly with asynchronous operations. Promise API followed Promises/A+ Specification prior to this, there was Promise/A.
In short, running continuation-passing style. Let me shoot few bullets with simple diagram on Promises below.
Understand deep links in 5 minutes
Jul 21, 2020☕ 2 mins readUnderstand deep links in 5 minutes
Jul 21, 2020☕ 2 mins readTricks to optimize Web Bundle File Size
Jul 21, 2020☕ 2 mins readTricks to optimize Web Bundle File Size
Jul 21, 2020☕ 2 mins readComing soon
Coming soon
$desc$
Feb 7, 2020☕ 10 mins readThere are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
Mar 16, 2020☕ 10 mins readGiven two arrays A and B of length N, determine if there is a way to make A equal to B by reversing any subarrays from array B any number of times.
Feb 7, 2020☕ 10 mins readFibonacci numbers, commonly denoted Fn sequence
Aug 7, 2021☕ 10 mins readFind the missing element in a given permutation.
Jul 21, 2020☕ 2 mins readPalindrome is in reverse also same
Jul 21, 2020☕ 2 mins readCount minimal number of jumps from position X to Y.
Jul 21, 2020☕ 2 mins readPalindrome is in reverse also same
Jul 21, 2020☕ 2 mins read$desc$
Feb 7, 2020☕ 10 mins readThere are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
Mar 16, 2020☕ 10 mins readGiven two arrays A and B of length N, determine if there is a way to make A equal to B by reversing any subarrays from array B any number of times.
Feb 7, 2020☕ 10 mins readFibonacci numbers, commonly denoted Fn sequence
Aug 7, 2021☕ 10 mins readFind the missing element in a given permutation.
Jul 21, 2020☕ 2 mins readPalindrome is in reverse also same
Jul 21, 2020☕ 2 mins readCount minimal number of jumps from position X to Y.
Jul 21, 2020☕ 2 mins readPalindrome is in reverse also same
Jul 21, 2020☕ 2 mins readAre you landed to this page and rushing for the immediate solution in javascript. here you go !!
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
On High level note, It worth to read in detail for a better understanding.
$desc$
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Are you landed to this page and rushing for the immediate solution in javascript. here you go !!
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
On High level note, It worth to read in detail for a better understanding.
$desc$
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Are you landed to this page and rushing for the immediate solution in javascript. here you go !!
1function findSignatureCounts(arr) {2 const ans = Array(arr.length).fill(1);3 // Iterate Array.4 for (let i = 0; i < arr.length; i++) {5 let k = i;6 while (arr[k] !== i + 1) {7 ans[i]++;8 k = arr[k] - 1;9 }10 }11 return ans;12}
On High level note, It worth to read in detail for a better understanding.
There are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
You're given a list of n integers arr[1..n], which is guaranteed to be a permutation of 1..n (in other words, it includes the integers from 1 to n exactly once each, in some order). The meaning of this list is described below.
Initially, each student is holding their own yearbook. The students will then repeat the following two steps each minute: Each student i will first sign the yearbook that they're currently holding (which may either belong to themselves or to another student), and then they'll pass it to student arr[i-1]. It's possible that arr[i-1] = i for any given i, in which case student i will pass their yearbook back to themselves. Once a student has received their own yearbook back, they will hold on to it and no longer participate in the passing process.
It's guaranteed that, for any possible valid input, each student will eventually receive their own yearbook back and will never end up holding more than one yearbook at a time.
You must compute a list of n integers output, whose element at i-1 is equal to the number of signatures that will be present in student i's yearbook once they receive it back.
1int[] findSignatureCounts(int[] arr)
Test Case 1 Explaination:
Test Case 2 Explaination:
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m) | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
1Code goes here...
Are you landed to this page and rushing for the immediate solution in javascript. here you go !!
1function findSignatureCounts(arr) {2 const ans = Array(arr.length).fill(1);3 // Iterate Array.4 for (let i = 0; i < arr.length; i++) {5 let k = i;6 while (arr[k] !== i + 1) {7 ans[i]++;8 k = arr[k] - 1;9 }10 }11 return ans;12}
On High level note, It worth to read in detail for a better understanding.
There are n students, numbered from 1 to n, each with their own yearbook. They would like to pass their yearbooks around and get them signed by other students.
You're given a list of n integers arr[1..n], which is guaranteed to be a permutation of 1..n (in other words, it includes the integers from 1 to n exactly once each, in some order). The meaning of this list is described below.
Initially, each student is holding their own yearbook. The students will then repeat the following two steps each minute: Each student i will first sign the yearbook that they're currently holding (which may either belong to themselves or to another student), and then they'll pass it to student arr[i-1]. It's possible that arr[i-1] = i for any given i, in which case student i will pass their yearbook back to themselves. Once a student has received their own yearbook back, they will hold on to it and no longer participate in the passing process.
It's guaranteed that, for any possible valid input, each student will eventually receive their own yearbook back and will never end up holding more than one yearbook at a time.
You must compute a list of n integers output, whose element at i-1 is equal to the number of signatures that will be present in student i's yearbook once they receive it back.
1int[] findSignatureCounts(int[] arr)
Test Case 1 Explaination:
Test Case 2 Explaination:
Lets see the list of approaches and their complexities.
| Approach | Time Complexity | Space Complexity | |
|---|---|---|---|
| 1 | Brute Force | O(n+m) | O(m+n) |
| 2 | $Approach 1$ | O(n) | O(n) |
| 3 | $Approach 2$ | O(n) | O(n) |
| 4 | Time Optimized | O(n) | O(n) |
| 5 | Memory Optimized | O(n) | O(n) |
With no further due, lets take a example of code solutions.
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
Description
1Code goes here...
1Code goes here...
Are you landed to this page and rushing for the immediate solution in javascript. here you go !!
1function areTheyEqual(array_a, array_b) {2const data = {};3for (let targetArrayIndex = 0; targetArrayIndex < array_a.length; ++targetArrayIndex) {4 data[array_a[targetArrayIndex]] = data[array_a[targetArrayIndex]] + 1 || 15}6// Iterating Array B.7for (let arrayIndex = 0; arrayIndex < array_b.length; ++arrayIndex) {8 const arrayBValueInDataHashMap = data[array_b[arrayIndex]];9 if (!arrayBValueInDataHashMap) return false;10 if (!!arrayBValueInDataHashMap) {11 if (arrayBValueInDataHashMap < 1) {12 return false;13 }14 data[array_b[arrayIndex]] = data[array_b[arrayIndex]] - 115 }16}17return true18};
On High level note, It worth to read in detail for a better understanding.
+Reverse to Make Equal \ No newline at end of file +As per, Leet CodeReverse to Make Equal
Feb 7, 2020☕ 10 mins readStraight to the Point !
Are you landed to this page and rushing for the immediate solution in javascript. here you go !!
1function areTheyEqual(array_a, array_b) {2const data = {};3for (let targetArrayIndex = 0; targetArrayIndex < array_a.length; ++targetArrayIndex) {4 data[array_a[targetArrayIndex]] = data[array_a[targetArrayIndex]] + 1 || 15}6// Iterating Array B.7for (let arrayIndex = 0; arrayIndex < array_b.length; ++arrayIndex) {8 const arrayBValueInDataHashMap = data[array_b[arrayIndex]];9 if (!arrayBValueInDataHashMap) return false;10 if (!!arrayBValueInDataHashMap) {11 if (arrayBValueInDataHashMap < 1) {12 return false;13 }14 data[array_b[arrayIndex]] = data[array_b[arrayIndex]] - 115 }16}17return true18};On High level note, It worth to read in detail for a better understanding.
This is one of the Facebook interview question and Leet code problem as well - 1460. Make Two Arrays Equal by Reversing Sub-arraysTable of Contents
Given two arrays A and B of length N, determine if there is a way to make A equal to B by reversing any subarrays from array B any number of times.
1bool areTheyEqual(int[] arr_a, int[] arr_b)Problem Statement
A = [1, 2, 3, 4] B = [1, 4, 3, 2] output = true
Constraints
All integers in array are in the range [0, 1,000,000,000].
-As per, Leet Code
- target.length == arr.length
- 1 <= target.length <= 1000
- 1 <= target[i] <= 1000
- 1 <= arr[i] <= 1000
Expected
Return true if B can be made equal to A, return false otherwise.
Foot Note
- Each element of target should have a corresponding element in arr, and if it doesn't have a corresponding element, return false.
- To solve it easiely you can sort the two arrays and check if they are equal.
Solution Intro
Lets see the list of approaches and their complexities.
Approach Time Complexity Space Complexity 1 Brute Force (Use HashMap with Two for...loop) O(n+m O(m+n) 2 Sort and Stringify both Array O(2n+2m) O(n+m) 3 Use HashMap with Two for...loop O(2n+m) O(n+m) 4 Time Optimized (2nd Solution) O(2n+2m) O(n+m) 5 Memory Optimized (3rd Solution) O(2n+m) O(n+m) Solutions
With no further due, lets take a example of code solutions.
Brute Force
Below solution sort both array then iterate the first array to check each index value is same, method returns false if there is no match with same index in both array.
1function(target, arr) {2 if (target.length === 0 || arr.length === 0) return false;3 if (target.length !== arr.length) return false;4 const sortedTarget = target.sort((a,b) => a -b);5 const sortedArray = arr.sort((a,b) => a -b);6 // Iterate Target Elements.7 for(let index = 0;index<sortedTarget.length;index++) {8 if(sortedTarget[index] !== sortedArray[index]) {9 return false;10 }11 }12 return true;13};Sort and Stringify both Array
Using JS predefined method sort and join, basically this Sort and Stringify both array elements.
1function(target, arr) {2 return target.sort().join("") === arr.sort().join("");3};Use HashMap with Two for...loop
Basically, this stores the elements value in hashmap as property name and value as number of occurence and then iterate the second array to check the hashmap is exist with greater than zero and then decrement the hashmap property value when occurence if exist in second array.
1function(target, arr) {2const data = {};3for (let targetArrayIndex = 0; targetArrayIndex < array_a.length; ++targetArrayIndex) {4 data[array_a[targetArrayIndex]] = data[array_a[targetArrayIndex]] + 1 || 15}6// Iterating Array B.7for (let arrayIndex = 0; arrayIndex < array_b.length; ++arrayIndex) {8 const arrayBValueInDataHashMap = data[array_b[arrayIndex]];9 if (!arrayBValueInDataHashMap) return false;10 if (!!arrayBValueInDataHashMap) {11 if (arrayBValueInDataHashMap < 1) {12 return false;13 }14 data[array_b[arrayIndex]] = data[array_b[arrayIndex]] - 115 }16}17return true18};Time Optimized
Time Optimized is Sort and Stringify both Array which is second solution.
Space Optimized
Space Optimized is Brute Force Code solution.
- target.length == arr.length
- 1 <= target.length <= 1000
- 1 <= target[i] <= 1000
- 1 <= arr[i] <= 1000
Expected
Return true if B can be made equal to A, return false otherwise.
Foot Note
- Each element of target should have a corresponding element in arr, and if it doesn't have a corresponding element, return false.
- To solve it easiely you can sort the two arrays and check if they are equal.
Solution Intro
Lets see the list of approaches and their complexities.
Approach Time Complexity Space Complexity 1 Brute Force (Use HashMap with Two for...loop) O(n+m O(m+n) 2 Sort and Stringify both Array O(2n+2m) O(n+m) 3 Use HashMap with Two for...loop O(2n+m) O(n+m) 4 Time Optimized (2nd Solution) O(2n+2m) O(n+m) 5 Memory Optimized (3rd Solution) O(2n+m) O(n+m) Solutions
With no further due, lets take a example of code solutions.
Brute Force
Below solution sort both array then iterate the first array to check each index value is same, method returns false if there is no match with same index in both array.
1function(target, arr) {2 if (target.length === 0 || arr.length === 0) return false;3 if (target.length !== arr.length) return false;4 const sortedTarget = target.sort((a,b) => a -b);5 const sortedArray = arr.sort((a,b) => a -b);6 // Iterate Target Elements.7 for(let index = 0;index<sortedTarget.length;index++) {8 if(sortedTarget[index] !== sortedArray[index]) {9 return false;10 }11 }12 return true;13};Sort and Stringify both Array
Using JS predefined method sort and join, basically this Sort and Stringify both array elements.
1function(target, arr) {2 return target.sort().join("") === arr.sort().join("");3};Use HashMap with Two for...loop
Basically, this stores the elements value in hashmap as property name and value as number of occurence and then iterate the second array to check the hashmap is exist with greater than zero and then decrement the hashmap property value when occurence if exist in second array.
1function(target, arr) {2const data = {};3for (let targetArrayIndex = 0; targetArrayIndex < array_a.length; ++targetArrayIndex) {4 data[array_a[targetArrayIndex]] = data[array_a[targetArrayIndex]] + 1 || 15}6// Iterating Array B.7for (let arrayIndex = 0; arrayIndex < array_b.length; ++arrayIndex) {8 const arrayBValueInDataHashMap = data[array_b[arrayIndex]];9 if (!arrayBValueInDataHashMap) return false;10 if (!!arrayBValueInDataHashMap) {11 if (arrayBValueInDataHashMap < 1) {12 return false;13 }14 data[array_b[arrayIndex]] = data[array_b[arrayIndex]] - 115 }16}17return true18};Time Optimized
Time Optimized is Sort and Stringify both Array which is second solution.
Space Optimized
Space Optimized is Brute Force Code solution.
The Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci sequence, and characterized by the fact that every number after the first two is the sum of the two preceding ones.
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
Problem is about a person who has a pair of newborn rabbits (different gender). here problem to determine the numbers of pairs after a year. at the end of each month, a newborn pair grows to maturity.
| Newborn | Matured | Total | |
|---|---|---|---|
| Jan 1 | 1 | 0 | 1 |
| Feb 1 | 0 | 1 | 1 |
| Mar 1 | 1 | 1 | 2 |
| Apr 1 | 1 | 2 | 3 |
| May 1 | 2 | 3 | 5 |
| Jun 1 | 3 | 5 | 8 |
| Jul 1 | 5 | 8 | 13 |
| Aug 1 | 8 | 13 | 21 |
| Sep 1 | 13 | 21 | 34 |
| Oct 1 | 21 | 34 | 55 |
| Nov 1 | 34 | 55 | 89 |
| Dec 1 | 55 | 89 | 144 |
| Jan 1 | 89 | 144 | 233 |
Finding the great common divisor of F5 = 5 and F6 = 8 is 1. This is due to the fact that only positive integer that divide F5 = 5 are 1 and 5 (denoted as gcd(F5, F6) = 1 likewise, F6 = 8 are 1,2,4 and 8 (denoted as gcd(F9, F10) = 1) there are common properties available. lets look at them.
Let's take a problem to find a N-th value of the Fibonacci sequence? +
The Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci sequence, and characterized by the fact that every number after the first two is the sum of the two preceding ones.
10, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
Problem is about a person who has a pair of newborn rabbits (different gender). here problem to determine the numbers of pairs after a year. at the end of each month, a newborn pair grows to maturity.
| Newborn | Matured | Total | |
|---|---|---|---|
| Jan 1 | 1 | 0 | 1 |
| Feb 1 | 0 | 1 | 1 |
| Mar 1 | 1 | 1 | 2 |
| Apr 1 | 1 | 2 | 3 |
| May 1 | 2 | 3 | 5 |
| Jun 1 | 3 | 5 | 8 |
| Jul 1 | 5 | 8 | 13 |
| Aug 1 | 8 | 13 | 21 |
| Sep 1 | 13 | 21 | 34 |
| Oct 1 | 21 | 34 | 55 |
| Nov 1 | 34 | 55 | 89 |
| Dec 1 | 55 | 89 | 144 |
| Jan 1 | 89 | 144 | 233 |
Finding the great common divisor of F5 = 5 and F6 = 8 is 1. This is due to the fact that only positive integer that divide F5 = 5 are 1 and 5 (denoted as gcd(F5, F6) = 1 likewise, F6 = 8 are 1,2,4 and 8 (denoted as gcd(F9, F10) = 1) there are common properties available. lets look at them.
Let's take a problem to find a N-th value of the Fibonacci sequence? There is various kinds of approaches in order to get this sequence and there is many solutions available. lets see each of them with solutions written in javascript. Before, lets look at the formulae.
Basically, this approach call itself creating more and more branches of the tree until it hits the base case. Below is the recursive solution in Javascript.
Time Complexity: O(2n) - Exponential1function fib_1(n) {2 if (n < 2) {3 return n4 }5 return fib_1(n - 1) + fib_1(n - 2)6}
Time Complexity: O(n).
Space Complexity: O(1).
1function fib_2(n) {2 let a = 0, b = 1, c, i;3 if (n == 0)4 return a;5 for (i = 2; i <= n; i++) {6 c = a + b;7 a = b;8 b = c;9 }10 return b;11}
Time Complexity: O(Logn).
Space Complexity: O(1) or O(Logn) on function call stack size consideration.
1function fib_3(n) {2 const F = [[1, 1], [1, 0]];3 if (n == 0)4 return 0;5 power(F, n - 1);6 return F[0][0];7}8// Helper function that multiplies 29// matrices F and M of size 2*2, and10// puts the multiplication result11// back to F[][]12function multiply(F, M) {13 const x = F[0][0] * M[0][0] + F[0][1] * M[1][0];14 const y = F[0][0] * M[0][1] + F[0][1] * M[1][1];15 const z = F[1][0] * M[0][0] + F[1][1] * M[1][0];16 const w = F[1][0] * M[0][1] + F[1][1] * M[1][1];17 F[0][0] = x;18 F[0][1] = y;19 F[1][0] = z;20 F[1][1] = w;21}22// Helper function that calculates F[][]23// raise to the power n and puts the24// result in F[][]25function power(F, n) {26 const M = [[1, 1], [1, 0]];27 // n - 1 times multiply the28 // matrix to {{1,0},{0,1}}29 for (let i = 2; i <= n; i++)30 multiply(F, M);31}32// Optimized version of power() in method 4 */33function power(F, n) {34 if (n == 0 || n == 1)35 return;36 const M = [[1, 1], [1, 0]];37 power(F, n / 2);38 multiply(F, F);39 if (n % 2 != 0)40 multiply(F, M);41}
Binet formula, sums, combinatorial representations and generating function of the generalized Fibonacci -numbers.
Time Complexity: O(logn), this is because calculating phi^n takes logn time.
-Space Complexity: O(1)
1function fib_4(n) {2 let phi = (1 + Math.sqrt(5)) / 2;3 return Math.round(Math.pow(phi, n) / Math.sqrt(5));4}
Let's look at the other facts where we used in real life.
Even our planet activities like Rotations period, Precession Period, Orbital Period are in fibonacci. Refer to Journal of Astronomy on Modeling Celestial Mechanics Using the Fibonacci Numbers
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities where F(n+1) / Fn is most of the time is 1.618
Golden ratio everywhere like example below.
Fibonacci extensions are a tool that traders can use to establish profit targets or estimate how far a price may travel after a retracement/pullback is finished. Extension levels are also possible areas where the price may reverse.
In finance, Fibonacci retracement is a method of technical analysis for determining support and resistance levels.[1] They are named after their use of the Fibonacci sequence.
1function fib_4(n) {2 let phi = (1 + Math.sqrt(5)) / 2;3 return Math.round(Math.pow(phi, n) / Math.sqrt(5));4}
Let's look at the other facts where we used in real life.
Even our planet activities like Rotations period, Precession Period, Orbital Period are in fibonacci. Refer to Journal of Astronomy on Modeling Celestial Mechanics Using the Fibonacci Numbers
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities where F(n+1) / Fn is most of the time is 1.618
Golden ratio everywhere like example below.
Fibonacci extensions are a tool that traders can use to establish profit targets or estimate how far a price may travel after a retracement/pullback is finished. Extension levels are also possible areas where the price may reverse.
In finance, Fibonacci retracement is a method of technical analysis for determining support and resistance levels.[1] They are named after their use of the Fibonacci sequence.
An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Write a function:
1function solution(A) {2}
that, given an array A, returns the value of the missing element.
For example, given array A such that:
the function should return 4, as it is the missing element.
Write an efficient algorithm for the following assumptions:
1.1
1function solution(A) {2 const b = A.sort()3 for(let i=0;i<=b.length;i++) {4 if(b[i+1] !== b[i] + 1) {5 return b[i] + 16 }7 }8 return ''9}
1.2
1function solution(A) {2 const n = A.length + 13 const sumOfN = Math.floor(n * (n + 1) / 2)4 const sumOfAll = A.reduce((a, b) => a + b, 0)5 return sumOfN - sumOfAll6}
An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Write a function:
1function solution(A) {2}
that, given an array A, returns the value of the missing element.
For example, given array A such that:
the function should return 4, as it is the missing element.
Write an efficient algorithm for the following assumptions:
1.1
1function solution(A) {2 const b = A.sort()3 for(let i=0;i<=b.length;i++) {4 if(b[i+1] !== b[i] + 1) {5 return b[i] + 16 }7 }8 return ''9}
1.2
1function solution(A) {2 const n = A.length + 13 const sumOfN = Math.floor(n * (n + 1) / 2)4 const sumOfAll = A.reduce((a, b) => a + b, 0)5 return sumOfN - sumOfAll6}
PALINDROMES means PALIN (Back again) and DROMES (Running) -Single Character is palindromic string.
Below is the basic coding solution written in javascript.
1function isPalindrome(S) {2 let iteration = Math.round(S.length / 2);3 let backward = S.length;4 for (let forward = 0; forward < iteration; forward++) {5 backward--;6 if (S[forward].toLowerCase() !== S[backward].toLowerCase()) {7 return "NO";8 }9 }10 return "YES";11}
xyxracecart
PALINDROMES means PALIN (Back again) and DROMES (Running) +Single Character is palindromic string.
Below is the basic coding solution written in javascript.
1function isPalindrome(S) {2 let iteration = Math.round(S.length / 2);3 let backward = S.length;4 for (let forward = 0; forward < iteration; forward++) {5 backward--;6 if (S[forward].toLowerCase() !== S[backward].toLowerCase()) {7 return "NO";8 }9 }10 return "YES";11}
xyxracecart
A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.
Count the minimal number of jumps that the small frog must perform to reach its target.
Write a function:
1function solution(X, Y, D) {2 // write your code in JavaScript (Node.js 8.9.4)3}
that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.
For example, given:
the function should return 3, because the frog will be positioned as follows:
Write an efficient algorithm for the following assumptions:
1function solution(x, y, d) {2 // write your code in JavaScript (Node.js 8.9.4)3 if ((y - x) % d === 0) {4 return (y - x) / d;5 }6 return Math.ceil((y - x) / d);7}
A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.
Count the minimal number of jumps that the small frog must perform to reach its target.
Write a function:
1function solution(X, Y, D) {2 // write your code in JavaScript (Node.js 8.9.4)3}
that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.
For example, given:
the function should return 3, because the frog will be positioned as follows:
Write an efficient algorithm for the following assumptions:
1function solution(x, y, d) {2 // write your code in JavaScript (Node.js 8.9.4)3 if ((y - x) % d === 0) {4 return (y - x) / d;5 }6 return Math.ceil((y - x) / d);7}
PALINDROMES means PALIN (Back again) and DROMES (Running)
Below is the basic coding solution written in javascript.
1function isPalindrome(S) {2 let iteration = Math.round(S.length / 2);3 let backward = S.length;4 for (let forward = 0; forward < iteration; forward++) {5 backward--;6 if (S[forward].toLowerCase() !== S[backward].toLowerCase()) {7 return "NO";8 }9 }10 return "YES";11}
PALINDROMES means PALIN (Back again) and DROMES (Running)
Below is the basic coding solution written in javascript.
1function isPalindrome(S) {2 let iteration = Math.round(S.length / 2);3 let backward = S.length;4 for (let forward = 0; forward < iteration; forward++) {5 backward--;6 if (S[forward].toLowerCase() !== S[backward].toLowerCase()) {7 return "NO";8 }9 }10 return "YES";11}
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Jul 21, 2020☕ 2 mins readSSRF stands for Server-Side Request Forgery. Its one type of exploit to server where hacker attempts to gather user information by various way in functionality which is not recommended.
There are mainly two types.
We have two subjects.
SSRF stands for Server-Side Request Forgery. Its one type of exploit to server where hacker attempts to gather user information by various way in functionality which is not recommended.
There are mainly two types.
We have two subjects.
