Count Pairs of Strings with Disjoint Character Sets

Question

The question asks the candidate to determine the number of pairs of strings from a given list that have disjoint character sets, meaning they share no common characters.

Codemia · Accepted Answer

To solve the problem of counting pairs of strings with disjoint character sets, we can represent each string's character set using a bitmask. Each bit in the bitmask corresponds to a character in the alphabet. If a character is present in a string, the corresponding bit is set to 1; otherwise, it is 0. This allows us to efficiently check if two strings have disjoint character sets using a bitwise AND operation.

1. Initialize a variable to count the valid pairs.
2. Create a list to store the bitmask for each string.
3. For each string, compute its bitmask by iterating through its characters and setting the corresponding bits.
4. Use nested loops to compare each pair of bitmasks. If the bitwise AND of two masks is 0, increment the count.
5. Return the count of valid pairs.

The time complexity of this approach is O(n^2), where n is the number of strings. This is because we compare each pair of strings. The space complexity is O(n) for storing the bitmasks. This solution is efficient for a moderate number of strings but may need optimization for larger datasets.

```python
def count_disjoint_pairs(strings):
    n = len(strings)
    bitmasks = [0] * n
    for i, s in enumerate(strings):
        for char in s:
            bitmasks[i] |= 1 << (ord(char) - ord('a'))
    count = 0
    for i in range(n):
        for j in range(i + 1, n):
            if bitmasks[i] & bitmasks[j] == 0:
                count += 1
    return count
```
This code calculates the number of pairs of strings with disjoint character sets.

Consider edge cases such as an empty list of strings, strings that are empty, or strings with all characters being the same. It is important to ensure that the algorithm handles these scenarios gracefully and returns the correct count.

Count Pairs of Strings with Disjoint Character Sets

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