This is my public algorithms and data structures repository at GitHub. Feel free to inspect my code and propose, or submit, any changes you want.

A rope (or cord) is a data structure composed of smaller strings that is used for efficiently storing and manipulating a very big piece of text. Text editors may make use of this data structure so that insertions and deletions performed by the user can be done efficiently.

See more about it on its own page.

Finds the greatest common divisor (GCD) between two integers.

```
N gcd(N a, N b) {
while (a %= b) {
swap(a, b);
}
return b;
}
```

Finds how many set bits there are in an integer.

```
int count_set_bits(unsigned u) {
int count = 0;
while (u) { // Will loop until u has no more set bits.
u &= u - 1;
count++;
}
return count;
}
```

Subtracting one from an unsigned integer toggles all bits until the rightmost
set one (`10111000`

– `00000001`

= `10110111`

).

Therefore, the result of a bitwise and operation between `n`

and `n – 1`

, is
`n`

with the rightmost set bit unset. For instance, ```
10111000 & 10110111 =
1011000
```

.

If `n`

has `x`

set bits, it will take `x`

operations to make `n = 0`

.

If the argument is not too big, this can easily be solved by using logarithms.

See that the number of digits of `n!`

is `ceil(log10(n!))`

for all positive `n`

.

By definition,

```
n! = n * (n - 1) * ... * 1
```

Therefore,

```
log(n!) = log(n) + log(n - 1) + ... + log(1)
```

Therefore, a simple and fast solution in Python is:

```
def factorial_digit_count(n):
return ceil(sum(log10(x) for x in range(1, n + 1)))
```

Note: I have tested this solution with inputs as big as 1M (1,000,000,000) and it gives a correct result. If you determine its limits, please let me know. Also, feel free to share any performance improvements you think I am missing.

This algorithm detects whether a singly linked list has a cycle in linear time. It uses a moving rabbit and a stationary, then teleporting (at powers of two), turtle. Both turtle and rabbit start at the first node of the list. Each iteration moves rabbit one node forward. If it is then at the same position as the stationary turtle, there is obviously a loop. If it reaches the end of the list, there is no loop.

Lambda denotes the length of the loop. It keeps track of how many steps the rabbit took since we last teleported the turtle.

Mu denotes the index of the element where the loop starts, to find it, we reset the turtle to the first element of the list and the rabbit to lambda elements after the mu. When the rabbit and the turtle meet, the turtle is at mu.

```
def brent(first_node):
"""
Detects a loop in a linked list in O(n) time and O(1) space complexity.
Returns a tuple of the form (mu, la), where
mu denotes the index of the list where the loop starts and
la is the length of the loop.
If this list does not have a loop, this function returns (-1, -1).
"""
power = 1 # The power of two we are working with.
la = 1
turtle = first_node
rabbit = first_node.next
while rabbit is not None and rabbit != turtle:
if la == power: # Get the next power of two if lambda is big enough.
la = 0
power *= 2
turtle = rabbit
rabbit = rabbit.next
la += 1
if rabbit == turtle: # Determine where the loop starts.
mu = 0
turtle = rabbit = first_node
for _ in range(la): # Let the rabbit advance λ nodes (a full cycle).
rabbit = rabbit.next
while rabbit != turtle: # Increment both until they meet.
turtle = turtle.next
rabbit = rabbit.next
mu += 1
return mu, la
else:
return -1, -1
```

The longest common subsequence problem is the problem of finding the longest subsequence common to all sequences of a set of sequences. A subsequence is any ordered subset of a sequence. Therefore, AC is a subsequence of ABC.

This is a classic problem of computer science and is solved to generate useful difference lists between files by version control systems.

Here I will present the intuition behind the solution of the algorithm for the case where the sequence set has only two sequences.

For an arbitrary number of sequences the problem is NP-hard and for two sequences the dynamic programming solution has both time and space complexity of O(n × m).

The dynamic programming solution can be understood by a recursive definition:

```
LCS(X[:i], Y[:j]) =
0 if i == 0 or j == 0
LCS(X[:i - 1], Y[:j - 1]) + 1 if X[i] == Y[j]
max(LCS(X[:i], Y[:j - 1]), LCS(X[:i - 1], Y[:j])) otherwise
```

If X_{i} = Y_{j}, the LCS is the LCS of the prefixes (up to, but
not including, the last elements) plus one (to account for the new common
element).

Otherwise, do not account for the last element into the LCS and take the longest of the two possible candidates.

The simplest way to implement the dynamic programming solution based on the abovementioned recursive solution uses a matrix where each element corresponds to the longest common subsequence up to that point.

The longest common substring problem is quite similar to the longest common subsequence problem. It differs because the definition of a substring is different to that of a subsequence as it does not include non-contiguous subsequences. Therefore, AC is a subsequence of ABC, but not a substring. AB, for instance, is both a substring and a subsequence of ABC.

The recursive solution is similar to that of the LCS problem. The only change is that instead of taking the maximum of the two possible candidates when there is a mismatch, one should use zero. This is because there can be no mismatches in the middle of a substring.

Too big to be here. Got its own page.

This is a must-known shuffling algorithm. It has its own page.