## Tech Blog

This blog is intended to talk about technology such as my technical experience, those things that I work that I would like to remember related to stuff I learn, or things I fixed, or tools I'm using...

This blog is intended to talk about technology such as my technical experience, those things that I work that I would like to remember related to stuff I learn, or things I fixed, or tools I'm using...

he Serverless Framework makes it easy to build applications using AWS Lambda. It is multi-provider framework, which means you can use it to build Serverless applications using other providers as well. For AWS, Serverless relies on CloudFormation to do the provisioning. It also scaffolds the project structure and takes care of deploying functions.

https://serverless.com/blog/node-rest-api-with-serverless-lambda-and-dynamodb/

`netsh wlan show profile`

`netsh wlan show profile “NETWORK NAME” key=clear`

Sometimes I forget this theory stuff, so here it is...

Something to remember as is easy to forget...

https://www.vojtechruzicka.com/bit-manipulation-java-bitwise-bit-shift-operations/

You are probably familiar with compound assignment operators for arithmetic operations such as `+=`

, `-=`

or `*=`

. But in addition to these, Java also offers variants for bitwise operators:

Operator | Example | Is equivalent to |
---|---|---|

|= | x |= 5 | x = x | 5 |

^= | x ^= 5 | x = x ^ 5 |

&= | x &= 5 | x = x & 5 |

<<= | x <<= 5 | x = x << 5 |

>>= | x >>= 5 | x = x >> 5 |

>>>= | x >>>= 5 | x = x >>> 5 |

I'm studying quick sort algorithm, which looks to be the most efficient one, I see it from this link

https://www.geeksforgeeks.org/quick-sort/

arr[] = {10, 80, 30, 90, 40, 50, 70} Indexes: 0 1 2 3 4 5 6 low = 0, high = 6, pivot = arr[h] = 70 Initialize index of smaller element,i = -1Traverse elements from j = low to high-1j = 0: Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])i = 0arr[] = {10, 80, 30, 90, 40, 50, 70} // No change as i and j // are samej = 1: Since arr[j] > pivot, do nothing // No change in i and arr[]j = 2: Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])i = 1arr[] = {10, 30, 80, 90, 40, 50, 70} // We swap 80 and 30j = 3: Since arr[j] > pivot, do nothing // No change in i and arr[]j = 4: Since arr[j] <= pivot, do i++ and swap(arr[i], arr[j])i = 2arr[] = {10, 30, 40, 90, 80, 50, 70} // 80 and 40 Swappedj = 5: Since arr[j] <= pivot, do i++ and swap arr[i] with arr[j]i = 3arr[] = {10, 30, 40, 50, 80, 90, 70} // 90 and 50 Swapped We come out of loop because j is now equal to high-1.Finally we place pivot at correct position by swapping arr[i+1] and arr[high] (or pivot)arr[] = {10, 30, 40, 50, 70, 90, 80} // 80 and 70 Swapped Now 70 is at its correct place. All elements smaller than 70 are before it and all elements greater than 70 are after it.