Subsequence With the Minimum Score

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StringsTwo PointersDynamic Programming

You are given two strings s and t.

You are allowed to remove any number of characters from the string t.

The score of the string is 0 if no characters are removed from the string t, otherwise:

Let left be the minimum index among all removed characters.
Let right be the maximum index among all removed characters.

Then the score of the string is right - left + 1.

Return the minimum possible score to make t a subsequence of s.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:

Input: s = "abacaba", t = "bzaa"
Output: 1
Explanation: In this example, we remove the character "z" at index 1 (0-indexed).
The string t becomes "baa" which is a subsequence of the string "abacaba" and the score is 1 - 1 + 1 = 1.
It can be proven that 1 is the minimum score that we can achieve.

Example 2:

Input: s = "cde", t = "xyz"
Output: 3
Explanation: In this example, we remove characters "x", "y" and "z" at indices 0, 1, and 2 (0-indexed).
The string t becomes "" which is a subsequence of the string "cde" and the score is 2 - 0 + 1 = 3.
It can be proven that 3 is the minimum score that we can achieve.

Constraints:

1 <= s.length, t.length <= 10⁵
s and t consist of only lowercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible lengths of strings `s` and `t`, and what is the character set they contain?
Can string `t` be an empty string or longer than string `s`? If `t` is empty, what should the score be?
If `t` cannot be formed as a subsequence of `s`, what should the function return?
If there are multiple possible subsequences of `s` that equal `t`, should I return the minimum score among all of them, or is there a tie-breaking condition?
Do the indices in the score calculation start from 0 or 1?

Brute Force Solution

Approach

The brute force method systematically checks every possible subsequence. It calculates the score for each subsequence and keeps track of the minimum score found. It's like trying every single combination to see which one gives you the best result.

Here's how the algorithm would work step-by-step:

First, we'll look at every possible subsequence. This means taking some elements from the original sequence and keeping their original order, but we don't have to take all of them.
For each subsequence we find, we will calculate its 'score'.
We need to keep track of the lowest score we have found so far. Initially, this can be a very large number.
As we calculate the score for each new subsequence, we compare it to the lowest score we've found so far.
If the new score is lower than the lowest score we've found so far, we update our 'lowest score' value.
After we have gone through every possible subsequence and calculated all their scores, the 'lowest score' will be the answer.

Code Implementation

def find_minimum_score_subsequence(sequence):
    number_of_elements = len(sequence)
    minimum_score = float('inf')

    # Iterate through all possible subsequences
    for i in range(1 << number_of_elements):
        subsequence = []
        for j in range(number_of_elements):
            # Check if the j-th element is included in the current subsequence
            if (i >> j) & 1:
                subsequence.append(sequence[j])

        # Calculate the score of the subsequence
        if subsequence:
            subsequence_score = sum(subsequence)

            # Update the minimum score if necessary
            if subsequence_score < minimum_score:

                minimum_score = subsequence_score

    # Handle the case where the input sequence is empty
    if minimum_score == float('inf'):
        return 0

    return minimum_score

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves examining every possible subsequence of the input sequence. A sequence of n elements has 2^n possible subsequences (each element can either be included or excluded). For each subsequence, calculating the score takes O(n) time in the worst case where the entire original sequence is chosen as a subsequence. However, since the number of subsequences dominates, the overall time complexity is driven by the enumeration of subsequences, leading to O(2^n) time complexity as the score calculation per subsequence becomes insignificant when compared to the exponential number of subsequences.

Space Complexity

O(1) – The brute force approach, as described, explores every subsequence and calculates its score. It maintains a variable to store the minimum score found so far. No auxiliary data structures that scale with the input size (N) are used. The space used is constant regardless of the size of the input sequence, thus the space complexity is O(1).

Optimal Solution

Approach

The goal is to find a specific part of a list that gives you the smallest possible score. We can achieve this efficiently by focusing on two special parts of the list instead of checking every single possibility.

Here's how the algorithm would work step-by-step:

Think of the score as being mainly influenced by the elements at the very beginning and the very end of your chosen part of the list.
Consider every possible element that could be the very first element in your subsequence.
For each possible first element, smartly figure out what the best possible last element would be to minimize your score.
Keep track of the smallest score you find while exploring these possibilities.
The smallest score you find is the answer to the problem.

Code Implementation

def find_minimum_score_subsequence(sequence):
    minimum_score = float('inf')
    
    for start_index in range(len(sequence)): 
        # Iterate through each possible start of the subsequence
        
        for end_index in range(start_index, len(sequence)): 
            # Consider every possible end index.
            
            current_score = sequence[start_index] + sequence[end_index]

            minimum_score = min(minimum_score, current_score)

    return minimum_score

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each element in the input list of size n, considering it as the potential starting element of the subsequence. For each starting element, the algorithm finds the best possible ending element, which might involve iterating through the remaining elements. Therefore, in the worst-case scenario, for each of the n starting elements, we potentially check up to n elements to find the optimal ending element. This results in roughly n * n comparisons or operations. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The plain English explanation outlines an approach that iterates through possible first elements and then finds the best last element for each. It describes keeping track of the smallest score found. This implies the use of a constant number of variables, such as one for the current first element, one for the best last element so far for that first element, and one to store the minimum score. Therefore, the space used is constant and independent of the input size N, where N is the length of the input list.

Edge Cases

Case	How to Handle
s is empty, t is not empty	Return -1, indicating no valid subsequence can be formed since t cannot be a subsequence of an empty s.
t is empty	Return 0, as an empty subsequence of s matches t, and there are no characters from s not in the subsequence.
s and t are both empty	Return 0 as there are no characters in s and thus no score.
s and t are identical strings	Return 0, since all characters in s are used in the subsequence (which is s itself).
t is not a subsequence of s	Return -1, indicating no valid subsequence can be formed.
Long strings for s and t approaching maximum allowed string length	Ensure the algorithm uses efficient data structures (e.g., dynamic programming) to avoid time limit exceeded errors and potential memory overflow.
t contains duplicate characters, and s contains these characters multiple times	The algorithm needs to select the indices in s that minimize the score while maintaining the subsequence property.
s and t contain extended ASCII or Unicode characters.	The solution should work correctly with different character encodings.

s is empty, t is not empty

How to Handle:

Return -1, indicating no valid subsequence can be formed since t cannot be a subsequence of an empty s.

t is empty

How to Handle:

Return 0, as an empty subsequence of s matches t, and there are no characters from s not in the subsequence.

s and t are both empty

How to Handle:

Return 0 as there are no characters in s and thus no score.

s and t are identical strings

How to Handle:

Return 0, since all characters in s are used in the subsequence (which is s itself).

t is not a subsequence of s

How to Handle:

Return -1, indicating no valid subsequence can be formed.

Long strings for s and t approaching maximum allowed string length

How to Handle:

Ensure the algorithm uses efficient data structures (e.g., dynamic programming) to avoid time limit exceeded errors and potential memory overflow.

t contains duplicate characters, and s contains these characters multiple times

How to Handle:

The algorithm needs to select the indices in s that minimize the score while maintaining the subsequence property.

s and t contain extended ASCII or Unicode characters.

How to Handle:

The solution should work correctly with different character encodings.