Forem: Anurag Pandey

Python: import csv

Anurag Pandey — Sat, 20 Nov 2021 08:57:23 +0000

Q: What is CSV?

A: Comma separated values or CSV is file format defined for exchanging information where comma(,) acts as delimiter and each row/record is separated by line break.

For example, consider this capital.csv file.

country,capital
India,New Delhi
Ireland,Dublin
Germany,Berlin
Austria,Vienna

Adding a header is optional, but it very common to find one in a csv file as it provides context on the data.

Q: What are the benefits that I receive from using this format?

A: Firstly, it is quite easy to understand for a human. If you have used spreadsheets sometime in your life, you have interacted with some form of csv. Secondly, it is useful for storing structured data, where in each column is some data point, similar to a relational database.

Q: Okay, now tell me how can I read a csv file in Python?

A: You can read a csv file by using the standard csv module, as follows:

import csv

with open('capital.csv') as f:
    reader = csv.reader(f)

    header = next(reader)

    for row in reader:
        print(row)

Output:

['India', 'New Delhi']
['Ireland', 'Dublin']
['Germany', 'Berlin']
['Austria', 'Vienna']

import csv


with open('capital.csv') as f:
    reader = csv.DictReader(f)

    for row in reader:
        print('Country:', row['country'])

Output:

Country: India
Country: Ireland
Country: Germany
Country: Austria

By using the DictReader we get the output in dictionary format, using which we can access the members as row['country'], or row['capital']. Field names (header) can be retrieved as reader.fieldnames.

Q: Is it mandatory for the delimiter to be a comma or can we use a delimiter of our own choice?

A: Good question! There is no specific guideline for the same. Python allows to customize the delimiter and also provides several other options. A collection of such options is called a dialect.

You can look up the different dialects:

import csv

print(csv.list_dialects())

Output:

['excel', 'excel-tab', 'unix']

The different options are:

delimiter
doublequote
escapechar
lineterminator
quotechar
quoting
skipinitialspace
strict

Most of them are self explanatory. You can read more about them here.

You can either create a separate dialect and pass it to the reader function, or if you require only few options, and creating a separate dialect sounds overkill for you then you can pass these options directly to the reader function.

For example, consider the sample capital.csv file with some minor changes.

country,  capital
India,    New Delhi
Ireland,  Dublin
Germany,  Berlin
Austria,  Vienna

The file is maintained with some spaces to be able to read it properly, but but but, we don't require these spaces while parsing it. Guess which option can we use to solve this? Yes, you have guessed it right! skipinitialspace it is.

import csv

with open('captial.csv') as f:
    reader = csv.reader(f, skipinitialspace=True)
    header = next(reader)

    for row in reader:
        print(row)

Output:

['India', 'New Delhi']
['Ireland', 'Dublin']
['Germany', 'Berlin']
['Austria', 'Vienna']

Try playing with the rest of the options.

And if you want to create a separate dialect, you can use csv.Dialect for it, and pass it to the reader function as csv.reader(f, dialect=dialect).

...

my_dialect = csv.excel()
my_dialect.skipinitialspace = True


with open('capital.csv') as f:
    reader = csv.reader(f, dialect=my_dialect)

...

Here I am using the default csv.excel dialect, with a tweak in setting skipinitialspace option as true.

There is yet another option,

...

my_dialect = csv.excel()
my_dialect.skipinitialspace = True

csv.register_dialect('my_dialect', my_dialect)


with open('capital.csv') as f:
    reader = csv.reader(f, dialect='my_dialect')

...

This way you can register a dialect and then use it throughout the program.

Q: That was quite insightful! Before I leave, can you elaborate on how can I create a csv file if I have some data? I am guessing we have a writer function complementing the reader?

A: You are absolutely right! Similar to reader and DictReader, we have writer and DictWriter.

import csv

header = ["country", "capital"]

data = [("India", "New Delhi"), ("Ireland", "Dublin"), ("Germany", "Berlin"), ("Austria", "Vienna")]

with open('new_captial.csv', 'w') as f:
    writer = csv.writer(f)
    writer.writerow(header)

    for row in data:
        writer.writerow(row)

Rather than writing our own "for loop" for writing row, we can also use the writerows method of the writer.

...

with open('new_captial.csv', 'w') as f:
    writer = csv.writer(f)
    writer.writerow(header)

    writer.writerows(data)

csv.DictWriter is useful when working with json data in a particular format.

import csv

header = ["country", "capital"]

data = [
    {"country": "India", "capital": "New Delhi"},
    {"country": "Ireland", "capital": "Dublin"},
    {"country": "Germany", "capital": "Berlin"},
    {"country": "Austria", "capital": "Vienna"},
]

with open('new_captial.csv', 'w') as f:
    writer = csv.DictWriter(f, fieldnames=header)
    writer.writeheader()

    writer.writerows(data)

You can also pass the dialect to writer function to change the output format.

To Loop or Not to Loop

Anurag Pandey — Sun, 19 Sep 2021 10:04:32 +0000

Keep this thought in the back of your mind, as you go through this article.

Consider the following example:
We have a list of players, with their name and country, and we want to extract all Indian players.

players = [{'name': 'P V Sindhu', 'country': 'India'},
           {'name': 'Michael Phelps', 'country': 'USA'},
           {'name': 'Usain Bolt', 'country': 'Jamaica'},
           {'name': 'Manika Batra', 'country': 'India'}]

indian_players = []
for player in players:
  if player['country'] == 'India':
    indian_players.append(player)

Now, in the above code, each line is easily understood, but you have to cache four lines in the brain to understand the whole context.

indian_players = [player for player in players if player['country'] == 'India']

Using list comprehension we can write more expressible code.
The above code also doesn't violate the ITM [Initialize Then Modify] antipattern, where the variable is initialized first and then modified immediately after, as we did in the first example.

LOOP, IF LOOP, PROPER LOOP

Python provides some good abstractions over the looping construct, making code more expressible.

Accessing the elements in a list:

Bad ❌

names = ['P V Sindhu', 'Usain Bolt', 'Michael Phelps', 'Manika Batra']

for i in range(len(names)):
  print(names[i])

Good ✔️

for name in names:
  print(name)

But what if indices are required?🤔

for i in range(len(names)):
   print(i+1, names[i])

No! ❌

Better way ✔️

for idx, name in enumerate(names, start=1):
  print(idx, name)

Now, why is the second code better?
Firstly, using indices creates a level of indirection. If the value is required why bother about the index?

Secondly, what if names was not a list but a dictionary, or some other data structure where indexing is not possible?
In that case, the first code would fail as we are indexing in the collection, but the second one would still work correctly.

Some more examples

Iterating over keys and values in a dictionary:

turing_award_winners = {1966: 'Alan Perlis', 
                        1967: 'Maurice Wilkes',
                        1968: 'Richard Hamming',
                        1969: 'Marvin Minsky'}

for year in turing_award_winners:
  print(year, turing_award_winners[year])

Getting year, and then looking for name in dictionary again causes a level of indirection.

Instead:

for year, name in turing_award_winners.items():
  print(year, name)

It also provides clear meaning.

NOT LOOP

Many a time, looping should not be considered at all.
Consider a simple example of finding the sum of all numbers in a list.

Conventional:

nums = [1, 5, 11, 17, 23, 29]

ans = 0
for num in nums:
  ans += num

Instead:

ans = sum(nums)

The code conveys the idea, that ans contains the sum of all numbers in the list nums, and is also very concise.

Conclusion

Write more expressible code, by utilizing the different constructs that the language has to offer. Avoid raw loops whenever you can.

I highly recommend to watch the videos below.

Further Watch:

Loop like a native: while, for, iterators, generators by Ned Batchelder.
Beautiful Python Refactoring by Conor Hoekstra.
C++ Seasoning by Sean Parent.

Day 34: Slowly Building Up

Anurag Pandey — Mon, 16 Aug 2021 17:15:14 +0000

Disclaimer:

Confession

I know, I am being super inconsistent. Today is Day 34, it has been 25 days till last coding day.
Nevertheless, let us get back on the horse now.

Problem 1: Coin Rows

Main points:

The player can move only right or down
We need to minimize the score, i.e., the coins Bob collects.

First solution that comes in mind is to maximize the coins Alice collects, but that is not the goal.
More score for Alice doesn't imply overall minimum score for Bob.

My solution:

#define T int tt; cin >> t; while(tt--)

main() {
  T {
    int n; cin >> n;
    vvl v(2, vl(n));
    for(auto& row: v) for(auto& e: row) cin >> e;
    ll sum1 = accumulate(all(v[0]), 0L);
    ll sum2 = 0;
    ll bob_score = INT_MAX;
    forn(i, n) {
        sum1 -= v[0][i];
        bob_score = min(bob_score, max(sum1, sum2));
        sum2 += v[1][i];
    }
    cout << bob_score << endl;
  }
}

Official solution:

#define T int tt; cin >> t; while(tt--)
#define all(x) begin(x), end(x)

main() {
  T {
    int n; cin >> n;
    vvl v(2, vl(n));
    for(auto& row: v) for(auto& e: row) cin >> e;
    ll sum1 = accumulate(all(v[0]), 0L);
    ll sum2 = 0;
    ll bob_score = INT_MAX;
    forn(i, n) {
        sum1 -= v[0][i];
        bob_score = min(bob_score, max(sum1, sum2));
        sum2 += v[1][i];
    }
    cout << bob_score << endl;
  }
}

Things to learn:

On Algorithmic thinking:

Try to make a guess on what could be possible.
Check for restrictions if any.

On Technical point:

Write little code as possible.
Use small variable name.
Practice to type fast.

In this particular problem we just needed to iterate through all possible solution for Bob and find the
minimum from it.

Problem 2: Mikasa

The gist of the problem is:

n, m are given and we need to find smallest non-negative number not present in the sequence
(n^0, n^1...n^m)

My first submission: (TLE) [O(n) solution/test case]

main() {
  T {
    ll n, m;
    cin >> n >> m;
    forn(i, max(n, m)+1) {
        ll val = i^n;
        if(val > m) {
            cout << i << endl;
            break;
        }
    }
  }
}

My solution after reading the editorial (not seeing the code)

main() {
  T {
    ll n, m;
    cin >> n >> m;
    if(m < n) cout << 0 << endl;
    else {
        ll p = m+1;
        ll ans = 0;
        for(int i = 31; i >= 0; i--) {
            ll t = (1L<<i);
            if((ll(t&p) > 0) && (ll(t&n) == 0)) {
                ans = ans^t;
            } else {
                if(ll(ans^n) >= p) break;
                if(ll(p&t) > 0) {
                    if(ll(n&t) > 0) continue;
                    else {
                        ans = ans^t;
                    }
                }
            }
        }
        cout << ans << endl;
    }
  }
}

This is the official editorial solution. Still I need to wrap my head around it.

main() {
 int t; cin >> t;
  while (t--) {
    int n, m; cin >> n >> m;
    ++m;
    int ans = 0;
    for (int k = 30; k >= 0 and n < m; k--) {
      if ((n >> k & 1) == (m >> k & 1)) continue;
      if (m >> k & 1) ans |= 1 << k, n |= 1 << k;
    }
    cout << ans << '\n';
  }
}

Things to learn:

On Technical point:

Know C++ well. Some time went in debugging as I was comparing t&n > 0, it didn't work out.
Learn ways to debug better.

Algorithm

We didn't learn any new algorithm of sort, but we did studied how to use C++ STL algorithm, such as

accumulate
partial_sum
reduce

As Sean Parent said, we should know our algorithms and idioms.

Type Conversion in C++

Anurag Pandey — Sat, 24 Jul 2021 15:31:46 +0000

Disclaimer:

char to int

main() {
  char c = '5';
  int num = c - '0'; 
  cout << num << endl;

  // OR

  int number = atoi(&c);
  cout << number << endl;
}

int to char [only 0-9]

main() {
  int num = 5;
  char c = num + '0';
  cout << c << endl;  // output: 5
}

string to int

main() {
  string s = "256alphabet";
  int num = stoi(s);
  cout << num << endl;  // output: 256
}

stoi will parse any string and return integer. Additonally takes the base as parameter.

main() {
  string s = "001010";
  int num = stoi(s, nullptr, 2);
  cout << num << endl; // output: 10
}

int to string

main() {
  int num = 256;
  string s = to_string(num);
  cout << s << endl;  // output: 256
}

string to vector<char>

#define all(x) begin(x), end(x)

main() {
  string s = "5555";
  vector<char> v(all(s));
  for(auto& c: v) cout << c << ' ';  // output: 5 5 5 5
}

vector<char> to string

#define all(x) begin(x), end(x)

main() {
  vector<char> v = {'1', '2', '3', '4', '5'};
  string s(all(v));
  cout << s << endl;  // output: 12345
}

vector<int> to string

main() {
  vector<int> v = {12345, 54321};
  string s;
  for(auto& e: v) s.append(to_string(e));
  cout << s << endl;
}

string to vector<int>

main() {
  string s = "12345";
  vector<int> v;
  for(auto& c: s) v.push_back(c-'0');

  // OR

  vector<int> v_new;
  transform(all(s), back_inserter(v_new), [](auto& c) { return c-'0'; });
}

Okay! Now we understand basic transformation from one form to another.

Now let us see some base conversions.

base 10 to base 2 [decimal to binary]

main() {
  int num = 5; // binary : 0101
  string binary = bitset<8>(num).to_string();
  cout << binary << endl;  // output: 00000101
}

decimal to hex|oct [ int -> string ]

main() {
  int num = 16; // hex: 10
  stringstream ss;
  ss << std::hex << num;
  cout << ss.str() << endl;  // output: 10

  // for oct
  ss << std::dec << num;
  cout << ss.str() << endl;  // output: 20
}

some base to another base(dec/hex/oct) [ string -> string ]

main() {
  string num = "16";
  stringstream ss;
  // consider s to be decimal
  // decimal to hex
  int initial_base = 10;
  ss << std::hex << stoi(num, nullptr, initial_base);
  cout << ss.str() << endl; // output: 10
}

Sources:

Decimal to Binary

Further Read:

Base 10 to Base N conversion

Day 9: Uping the problem rating

Anurag Pandey — Thu, 22 Jul 2021 16:01:49 +0000

Disclaimer:

Problem 1: Taxi

using vi = vector<int>;

#define all(x) begin(x), end(x)
#define forn(i, n) for(int i = 0; i < n; i++)

main() {
  int n;
  cin >> n;
  vi v(5, 0);
  forn(_, n) {
    int g;
    cin >> g;
    v[g]++;
  }
  // every four will require one car
  int four = v[4]; 
  // every three will require one car
  int three = v[3];
  // it can take groups of one too
  v[1] -= min(v[1], v[3]);
  // a car can take two groups of two and change
  int two = v[2]/2 + v[2]%2;
  int rem = v[2]%2;
  v[1] -= min(v[1], rem*2);
  int one = v[1]/4 + (v[1]%4 != 0);

  int ans = one + two + three + four;
  cout << ans;
}

The above solution was the first accepted solution I could come up with.

This one is more cleaner [Greedy Approach]

using vi = vector<int>;

#define forn(i, n) for(int i = 0; i < n; i++)
#define all(v) begin(v), end(v)
#define endl '\n'

main() {
  int n;
  cin >> n;
  vi v(n);
  for(auto& e: v) cin >> e;
  sort(all(v));
  int i = 0, j = n-1;
  int ans = 0;
  while(i <= j) {
    int curr = v[j];
    while(i < j && curr+v[i] <= 4) {
      curr += v[i++];
    }
    j--;
    ans++;
  }
  cout << ans;
}

Problem 2: Interesting Drink

I was initially afraid to solve the problem seeing dp in the tags. But then I hid the tags, and went for the
problem. lol.

Just check the number of places whose booze price is less than or equal to the price the man can pay.
Binary Search.

using vi = vector<int>;

#define forn(i, n) for(int i = 0; i < n; i++)
#define all(v) begin(v), end(v)
#define endl '\n'

main() {
  int n;
  cin >> n;
  vi v(n);
  for(auto& e: v) cin >> e;
  sort(all(v));
  int q;
  cin >> q;
  forn(_, q) {
    int p;
    cin >> p;
    auto it = upper_bound(all(v), p);
    cout << distance(begin(v), it) << endl;
  }
}

Algorithm

We didn't learn any new algorithm today.

Day 8: Stacking up

Anurag Pandey — Wed, 21 Jul 2021 16:11:42 +0000

Disclaimer:

Problem 1: cAPS lOCK

Based on some property, give answer.

#define all(x) begin(x), end(x)

main() {
  string s;
  cin >> s;

  int n = s.length();
  bool first = islower(s[0]);
  bool c = count_if(all(s), [](char c){ return isupper(c); });
  // check condition
  if((c == n) || (first && (c == n-1))) {
    for(auto& e: s) {
      if(isupper(e)) cout << tolower(e);
      else cout << toupper(e);
    }  
  } else {
    cout << s;
  }
}

To toggle case we can also use bitwise operators. It grossly simplyfies the code.

for(auto& c: s) {
  cout << (char) (c^32); // toggle case
}

New cpp standard algorithm used:

count_if

Problem 2: Xenia and Ringroad

Simple math problem.

#define forn(i, n) for(int i = 0; i < n; i++)

main() {
  ll n, m;
  cin >> n >> m;
  ll ans = 0, curr = 1; 
  forn(_, m) {
    ll next;
    cin >> next;
    if(next >= curr) ans += (next-curr);
    else {
      ans += (n-curr)+next;
    }
    curr = next;
  }
  cout << ans;
}

Algorithm

We didn't learn any new algorithm today.
I should practice more problems, so that when I get stuck somewhere, I could try to learn a new algorithm.
Also knowing the basics should help, rather than learning all complex algorithms at once.
Sticking to learn new algorithm if and when required and doubling the questions to be practiced.

Sources

Toggle Case: geeksforgeeks

CP Template

Anurag Pandey — Wed, 21 Jul 2021 11:09:22 +0000

Disclaimer:

My CP Template

It is always fun to write and maintain your own template file for competitive programming, just like maintaining .virmc file.
I primarily program in C++. So here is my cpp template file:

Now how to keep improving this template,

Experiment with different things, and make it customize to your needs.
Look at other people template and learn.

The same thing could be said for editing your .vimrc file, that is if you use vim :-)

Competitive Programming Goals

Anurag Pandey — Sun, 18 Jul 2021 16:38:24 +0000

Let me write down where I want to be, with regards to my competitive programming skills.
I will divide it into two parts:

Maths
Art

Maths: End goals. Quantitative. Win/Loss.

2021: Base building.

Learning the basics algorithms and data structures.
Codeforces: Specialist by the end of year.

2022: Introducing yourself.

Master on codeforces.
Google CodeJam Top 2000.

2023: Attacking

Legendary GrandMaster
1st on the world ranking Codeforces, Atcoder.
Google CodeJam Winner.
Facebook HackerCup Winner.

Art: The journey. Road less taken.

Document the journey of learning and being the best in the world.
Consistently practicing and enjoying new algorithms.
Writing some real life code example.
Learning Parallel algorithms.
Inspire atleast one person to take up competitive programming.

The goals are looking far bigger right now, as it should. Further we practice, better will be the chances of achieving them.
We will always keep adding or subtracting to this goal list. And at last, if we don't achieve the goal we will label it as an experiment :-) But let us not think ahead now. Keeping my head down and starting practicing.

P.S: The end goal is to beat tourist. xD

Day 5: Inconsistent days

Anurag Pandey — Sun, 18 Jul 2021 16:15:37 +0000

Disclaimer:

Two problems a day, keeps doctor away!

Problem 1: Brain's Photos

Simple problem of taking input and comparing it with different values.

main() {
  int n, m; cin >> n >> m;
  bool bw = true;
  for(int i = 0; i < n; i++) {
    for(int j = 0; j < m; j++) {
      char c;
      cin >> c;
      if(c == 'C' || c == 'M' || c == 'Y') bw = false;
    }
  }
  if(bw) cout << "#Black&White";
  else cout << "#Color";
}

This is better style of writing for the given problem. No need to write two for loops. Cleaner code.

main() {
  int n, m; cin >> n >> m;
  bool bw = true;
  char c;
  while(cin >> c) {
    if(c == 'C' || c == 'M' || c == 'Y') bw = false;
  }
  if(bw) cout << "#Black&White";
  else cout << "#Color";
}

Problem 2: Sereja and Dima

Implementing the greedy take and find score of each player.

main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for(auto& e: v) cin >> e;
  int i = 0, j = n-1;
  int s = 0, d = 0;
  bool ser = true;
  while(i <= j) {
    if(v[i] > v[j]) {
      if(ser) s += v[i];
      else d += v[i];
      i++;
    } else {
      if(ser) s += v[j];
      else d += v[j];
      j--;
    }
    ser = !ser;
  }
  cout << s << ' ' << d;
}

More cleaner code and easy to understand.

main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for(auto& e: v) cin >> e;
  int i = 0, j = n-1;
  int s = 0, d = 0;
  bool ser = true;
  while(i <= j) {
    int val = max(v[i], v[j]);
    if(ser) s += val;
    else d += val;
    ser = !ser;
    if(val == v[i]) i++;
    else j--;
  }
  cout << s << ' ' << d;
}

Algorithm

We didn't learn any new algorithm today :-(

Dated: 17th July 2021

Day 1: Rebooting the competitive programming drive

Anurag Pandey — Sat, 17 Jul 2021 13:36:01 +0000

Disclaimer:

It has been a long time now that I practiced my competitive programming skills. Not super good at it. Last I practiced thoroughly was during interview preparation in July 2020.

Phase 1: Base Building

I have planned to solve at least two implementation based problems everyday for a month starting today and additionally learning one new algorithm daily. I personally like string algorithms so I will start from there.

Problem 1: Do Not Be Distracted!

main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  s.erase(unique(begin(s), end(s)), end(s));
  sort(begin(s), end(s));
  auto it = unique(begin(s), end(s));
  if(it != end(s)) cout << "NO";
  else cout << "YES";
}

Time Complexity: O(nlog(n))
Space Complexity: O(1) # not considering the space to hold the data

Another solution:

main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  unordered_set<char> se;
  bool yes = true;
  for(int i = 0; i < n; i++) {
    if(i && s[i] == s[i-1]) continue;
    if(se.count(s[i])) {
      yes = false;
      break;
    }
    se.insert(s[i]);
  }
  if(yes) cout << "YES";
  else cout << "NO";
}

Time Complexity: O(n)
Space Complexity: O(n) # the unordered set used to contain
elements for checking.

Problem 2: Black Square

main() {
  vector<int> v(4);
  for(auto& e: v) cin >> e;
  string s;
  cin >> s;
  long long total = 0;
  for(auto& c: s) {
    int p = c - '0' - 1;  // additional -1 for zero based indexing
    total += v[p];
  }
  cout << total;
}

This would be more fun to write in Python.

v = list(map(int, input().split()))
s = input()
s = [int(c)-1 for c in s]  #  -1 for zero based indexing
total = sum(v[i] for i in s)
print(total)

I am sure we could use find STL algorithm for the C++ code.
That is all for today 2 implementation problem solving.

Let us hop on to learning one new/old string algorithm.

String Algorithm

The basic question when it comes to string is about, whether a given pattern of length m occurs in a text of length n.
The problem can be solved using naive method in O(n*m) time.

main() {
  string text = "abcabc";
  string pattern = "abc";
  int n = text.length(), m = pattern.length();
  bool found = false;
  // looping through all starting position of the pattern
  for(int i = 0; i < n-m+1; i++) {
    // checking if the substring matches
    if(text.substr(i, m) == pattern) {
      found = true;
      break;
    }
  }
  if(found) cout << "YES";
  else cout << "NO";
}

Now there are algorithms that can solve the above pattern matching in O(n+m) time. The first one that comes in mind is the famous
Knuth Morris Pratt or better known as KMP algorithm.


vector<int> prefix_func(string s) {
  int n = s.length();
  vector<int> pi(n, 0);
  for(int i = 1; i < n; i++) {
    int j = pi[i-1];
    while(j > 0 && s[i] != s[j])
        j = pi[j-1];
    if(s[i] == s[j])
        j++;
    pi[i] = j;
  }
  return pi;
}

To use KMP algorithm, pre-compute the table using above prefix_func.
So for given pattern and text, the string that should be passed to prefix_func is pattern + x + text, where 'x' is some character which doesn't belong to neither pattern nor text.

The different questions that can be answered are:

Does the pattern occur in the text?
How many times does the pattern occur in the text?
If pattern occurs, find the starting position in text.
Find maximum non-overlapping occurrence of the pattern in the text. And many more.

Most of the time, pattern matching is used as a part for solving a problem, and not a whole problem.

Sources:

prefix_function :- cp-algorithms

Resources for additonal learning and practicing:

usaco.guide

Dated: 13th July 2021

Codeforces Tracker

Anurag Pandey — Tue, 11 May 2021 03:43:21 +0000

What is this site?

cf-tracker is a website for monitoring progress on codeforces.com in terms on which questions were solved in contests.
There are two types of questions to keep track on:

The ones that were solved during the contest.
The ones that were not solved during or after the contest[problems to upsolve later].

It also provides feature to filter contests based on division[div1, div2, div3, Educational Rounds], and is pretty damn quick about it.

How can one use it?

To upsolve problems.
To find contests that you haven't participated in before and practice on them.
Filter contests based on divisions for practice.

How is this site built?

The site is built entirely using Vue.js.
It uses codeforces api for fetching information.
The contests information is statically stored as a json file, for improving the speed.
To update the contests information there is python script written.
The site is hosted using Github Pages.

Can I look at the source code?

Sure, the site is completely open-source. You can find it here.

What features are coming next on it?

To keep the search and filter bar of the top steady if when you scroll down.
Refresh button to get latest data on problems solved by the user.
Proper error message for user, in case the codeforces website is down.
Improve the UI.
Pagination of problems.

Suggestions to improve the website are always welcome. If any issues found please raise it on the github repository.

Thank you! Enjoy the site!

Copy venv from one folder to another and still be able to use it?

Anurag Pandey — Sun, 02 May 2021 13:10:12 +0000

If you have ever copied the python virtual environment folder from one location to another and faced problem using it, this article will help you understand how you can solve this problem.

Let us first create a folder, named example and then create a virtual environment and install flask using pip

$ mkdir example
$ cd example
$ python -m venv venv
$ source venv/bin/activate
$ pip install Flask

Now you have a virtual env with flask installed.

After doing this python sets an environment variable VIRTUAL_ENV, let us check its value.

$ echo $VIRTUAL_ENV
/home/username/example/venv

We will now move this venv folder to some other folder.

$ mkdir example2
$ mv venv/ example2/
$ cd example2
$ source venv/bin/activate

Now let us check the value of VIRTUAL_ENV variable

$ echo $VIRTUAL_ENV
/home/username/example/venv

Wow, this is strange. I have moved the folder but still the value of VIRTUAL_ENV variable is the same?
This is because when you create the virtual environment, the location is hardcoded based on current directory.
You can check this inside activate file we have previously used to activate the virtual environment.
You will find something like VIRTUAL_ENV='/home/username/example/venv'.

The solution for using the virtual environment at the new location is simple. Just change this path in all program inside venv/bin directory to the new path :-)
Please don't do it manually!!!
Use tools like sed on unix or similar tools for the same.

Example using sed:

$ old_path='/home/username/example/venv'
$ new_path='/home/username/example/example2/venv'
$ cd venv/bin/
$ sed -i "s|$old_path|$new_path|g" *

The last command is replace in-place the value of old_path with new_path. * signifies all files in current directory(not exactly, but for this example consider it). Make sure you run this command while present in venv/bin folder.

After doing this reactivate the virtual env.

$ source venv/bin/activate

And voila!

Reference:

For sed command used: https://stackoverflow.com/questions/10309968/sed-search-and-replace-strings-containing