How do you solve Subsequence With the Minimum Score on LeetCode?

Subsequence With the Minimum Score (LeetCode #2565) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(m) space complexity. Key insight: Understanding subsequences is crucial; they maintain the order of characters.

What is the time complexity of Subsequence With the Minimum Score?

The optimal solution for Subsequence With the Minimum Score runs in O(n + m) time and uses O(m) space. This complexity is efficient because we only traverse the strings a limited number of times to fill the leftmost and rightmost arrays.

What is the brute force solution for Subsequence With the Minimum Score?

The brute force approach for Subsequence With the Minimum Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of characters to remove from string t to see if the remaining characters form a subsequence of string s. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Subsequence With the Minimum Score?

Subsequence With the Minimum Score uses the following concepts: Two Pointers, String, Binary Search. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Subsequence With the Minimum Score?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases and how they might affect your solution. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Subsequence With the Minimum Score?

Failing to handle edge cases, such as when t is empty. Not considering the order of characters when checking for subsequences.

#2565

Subsequence With the Minimum Score

Hard

Two PointersStringBinary SearchTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(m)

💡

Intuition

Time O(n + m)Space O(m)

The optimal approach uses two pointers to find the leftmost and rightmost indices of characters in t that can be matched in s. This allows us to efficiently calculate the minimum score without generating all subsequences.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays, leftmost and rightmost, to store the minimum indices of characters in t that can be matched in s from the left and right respectively.
2Step 2: Use two pointers to fill the leftmost array by iterating through s and t from the left.
3Step 3: Use two pointers to fill the rightmost array by iterating through s and t from the right.
4Step 4: Calculate the minimum score using the leftmost and rightmost arrays.

solution.py26 lines

1# Full working Python code
2
3def min_score_subsequence(s, t):
4    n, m = len(s), len(t)
5    leftmost = [-1] * (m + 1)
6    rightmost = [-1] * (m + 1)
7
8    j = 0
9    for i in range(n):
10        if j < m and s[i] == t[j]:
11            leftmost[j] = i
12            j += 1
13
14    j = m - 1
15    for i in range(n - 1, -1, -1):
16        if j >= 0 and s[i] == t[j]:
17            rightmost[j] = i
18            j -= 1
19
20    min_score = float('inf')
21    for i in range(m + 1):
22        if leftmost[i] != -1 and rightmost[i] != -1:
23            min_score = min(min_score, rightmost[i] - leftmost[i] + 1)
24
25    return min_score
26

ℹ

Complexity note: This complexity is efficient because we only traverse the strings a limited number of times to fill the leftmost and rightmost arrays.

1Understanding subsequences is crucial; they maintain the order of characters.
2Using two pointers can significantly reduce the time complexity by avoiding unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.