Steps

There are only three steps and step 2 is optional.

1. Creating a Slack App via https://api.slack.com/messaging/webhooks. What you want is the Webhook URLs under Incoming Webhooks which starts with https://hooks.slack.com/services/, so that you can send the notification message to it.
2. Aware of the Rate Limit of GitHub API and create a personal access token or creating GitHub Apps depending on your needs.
3. With help of ChatGPT and a few fixes, we derived at this script

import requests
import argparse
import time
import re

# Could be none or personal access token https://docs.github.com/en/rest/using-the-rest-api/rate-limits-for-the-rest-api
GITHUB_TOKEN="<your_github_token>"
# On how to get webhooks token https://api.slack.com/messaging/webhooks
SLACK_WEBHOOK="https://hooks.slack.com/services/XXXXXXXXXXX/YYYYYYYYYY/ZZZZZZZZZZZZZZ"
WAIT_DURATION_SEC=600

def parse_args():
    parser = argparse.ArgumentParser(description='Check if a GitHub pull request was merged and send a Slack message.')
    parser.add_argument('pr_url', help='GitHub pull request URL.')
    return parser.parse_args()

def parse_pr_url(pr_url):
    """
    Parses the PR URL and returns the owner, repository, and PR number.

    Example URL: https://github.com/owner/repo/pull/123
    """
    pattern = r'https?://github\.com/(?P<owner>[^/]+)/(?P<repo>[^/]+)/pull/(?P<pr_number>\d+)'
    match = re.match(pattern, pr_url)
    if match:
        return match.group('owner'), match.group('repo'), int(match.group('pr_number'))
    else:
        raise ValueError('Invalid GitHub pull request URL.')

def check_if_pr_merged(owner, repo, pr_number, github_token=None):
    url = f'https://api.github.com/repos/{owner}/{repo}/pulls/{pr_number}'
    headers = {'Accept': 'application/vnd.github.v3+json'}
    if github_token:
        headers['Authorization'] = f'token {github_token}'
    response = requests.get(url, headers=headers)
    if response.status_code == 200:
        pr_data = response.json()
        return pr_data.get('merged', False)
    else:
        print(f'Error fetching PR data: {url} {response.status_code} {response.reason}')
        return False

def send_slack_message(slack_webhook_url, message):
    payload = {'text': message}
    response = requests.post(slack_webhook_url, json=payload)
    if response.status_code != 200:
        print(f'Error sending Slack message: {slack_webhook_url} {response.status_code} {response.reason}')

def main():
    args = parse_args()
    try:
        owner, repo, pr_number = parse_pr_url(args.pr_url)
    except ValueError as e:
        print(e)
        return

    print(f"Monitoring PR #{pr_number} in repository '{owner}/{repo}' for merge status.")

    while True:
        is_merged = check_if_pr_merged(owner, repo, pr_number, GITHUB_TOKEN)

        if is_merged:
            message = f'Pull request {args.pr_url} has been merged.'
            send_slack_message(SLACK_WEBHOOK, message)
            break
        else:
            time.sleep(WAIT_DURATION_SEC)

if __name__ == '__main__':
    main()

After aliasing, you could do ntfslk https://github.com/owner/repo/pull/123 & to put it in the background. When the script scans and finds out the PR is merged, a notification will pop up in the channel.

Weekly Contest 156

1207. Unique Number of Occurrences

https://leetcode.com/contest/weekly-contest-156/problems/unique-number-of-occurrences/

Brute force. Record total occurrence of each number and iterate over it to see if there is any duplication.

class Solution {
public:
    bool uniqueOccurrences(vector<int>& arr) {
        unordered_map<int, int> occ;
        unordered_map<int, bool> flag;
        
        for (int i = 0; i < arr.size(); ++i) {
            occ[arr[i]]++;
        }
        
        for (const auto x : occ) {
            if (flag.find(x.second) != flag.end()) {
                if (flag[x.second]) return false;
            } else {
                flag[x.second] = true;
            }
        }
        
        return true;
    }
};

1208. Get Equal Substrings Within Budget

https://leetcode.com/contest/weekly-contest-156/problems/get-equal-substrings-within-budget/

Weekly Contest 155

https://leetcode.com/contest/weekly-contest-155

1200. Minimum Absolute Difference

https://leetcode.com/contest/weekly-contest-155/problems/minimum-absolute-difference/

Brute force. Find the minimum absolute difference and then iterate the list again to output the pairs with minimum absolute difference.

class Solution {
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
        vector<vector<int>> ans;
        if (arr.size() == 1) return ans;
        sort(arr.begin(), arr.end());
        
        int mini = 0x3f3f3f3f;
        for (int i = 1; i < arr.size(); ++i) {
            mini = min(mini, abs(arr[i] - arr[i-1]));
        }
        
        for (int i = 1; i < arr.size(); ++i) {
            if (abs(arr[i]- arr[i-1]) == mini) {
                if (arr[i] < arr[i-1]) {
                    ans.push_back({arr[i], arr[i-1]});
                } else {
                    ans.push_back({arr[i-1], arr[i]});
                }
            }
        }
        return ans;
    }
};

1201. Ugly Number III

https://leetcode.com/contest/weekly-contest-155/problems/ugly-number-iii/

This is my first project using Perl. I feel it is tedious to type formatted definitions to Quizlet(A website which can make flashcards for you) and I’m too lazy to open browser and online dictionary pages. Why not combining these two?

It uses following APIs:

• Merriam-Webster Learner

• Merriam-Webster Collegiate

For each entry, it contains:

• Pronunciation: IPA(International Phonetic Alphabet)

• Part of Speech

  ]]>https://pwzxxm.com/static-linked-list/Tue, 29 May 2018 00:00:00 +0000patrick.weizhixu@gmail.com (PwzXxm)https://pwzxxm.com/static-linked-list/Static Linked List is a data structure that stores linked list in static arrays. It is usually used to represent graphs. It is very interesting that its Chinese name literally translated as “Linked Forward Star”. You have two choices of paths to understand this.

  • Start from Forward Star.

  • Start from Adjacency List.

  However, I would recommend to explore both ideas to have a better understanding. If you know some of it or you just don’t care, you can jump to here straight away.

  ]]>https://pwzxxm.com/poj-3279/Mon, 23 Apr 2018 00:00:00 +0000patrick.weizhixu@gmail.com (PwzXxm)https://pwzxxm.com/poj-3279/There are \(M \times N\) \((1 \le M, N \le 15)\)square tiles. Each tile can be flipped and the color of tile can change between black(1) and white(0).

  When you flip a tile, 4 adjacent tiles will also be flipped. Note that the four adjacent flipped tiles will NOT cause their adjacent tiles to flip.

  Given a configuration, find the minimum number of flips so that all square tiles become white. If having the minimum number is the same, choose the least lexicographical one¹.

  ]]>https://pwzxxm.com/proxy-server-cn-netease-music/Fri, 23 Mar 2018 15:39:00 +1100patrick.weizhixu@gmail.com (PwzXxm)https://pwzxxm.com/proxy-server-cn-netease-music/Sometimes we need to access Chinese content, like Youku Video, Netease Music (Cloud Music) and QQ music. It is very annoying to get “Can Only Be Streamed in Mainland China” or similar messages. Being tired of that, fortunately I have got a VPS from Aliyun and I build a proxy server using Shadowsocks on it and everything works smoothly now.

  VPS

  You need to have a VPS with Chinese IP address, there are tons of choices, like Aliyun(Alibaba Cloud) and Tencent Cloud etc.

  ]]>https://pwzxxm.com/usaco-lamps/Wed, 31 Aug 2016 00:00:00 +0000patrick.weizhixu@gmail.com (PwzXxm)https://pwzxxm.com/usaco-lamps/A set of \(N\) \((10 \le N \le 100)\) lamps, numbered from 1 to N.

  Four types of button:

  Button type	Usage
  1	flip all lamps (On to Off, Off to On)
  2	flip odd numbered lamps (e.g. 1, 3, 5)
  3	flip even numbered lamps (e.g. 2, 4, 6)
  4	flip \(3k+1 ; with ; k \ge 0\) numbered lamps (e.g. 1, 4, 7)

  
  

  Given \(C\) (number of button presses, \(0 \le C \le 10000\)) and final state of some of the lamps, find all possible distinct configuration.

  ]]>https://pwzxxm.com/poj-3268/Sat, 14 May 2016 00:00:00 +0000patrick.weizhixu@gmail.com (PwzXxm)https://pwzxxm.com/poj-3268/<p>There are one cow from each \(N\) (\(1 \le N \le 1000\)) farms want to go to the number X farm to have a party, hurrah! One cow from each farm need to go to the party and go back. There are \(M\) (\(1 \le M \le 100,000\)) weighted (represents time) one-direction roads connects pairs of roads. Cows are smart though, they want to go via the shortest path. The question is, what is the longest time the cow will take.</p>