How do you solve Append Characters to String to Make Subsequence on LeetCode?

Append Characters to String to Make Subsequence (LeetCode #2486) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding subsequences is crucial; a subsequence can skip characters.

What is the time complexity of Append Characters to String to Make Subsequence?

The optimal solution for Append Characters to String to Make Subsequence runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse each string at most once. The space complexity is O(1) since we only use a few variables for tracking indices.

What is the brute force solution for Append Characters to String to Make Subsequence?

The brute force approach for Append Characters to String to Make Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible subsequences of the string `s` to see if `t` can be formed. This method is straightforward but inefficient, as it requires examining many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Append Characters to String to Make Subsequence?

Append Characters to String to Make Subsequence uses the following concepts: Two Pointers, String, Greedy. The recommended patterns to study are: Two Pointers, Greedy.

What are the interview tips for Append Characters to String to Make Subsequence?

Clarify the definition of a subsequence before diving into coding. Think about edge cases, such as when `t` is empty or when `s` is much shorter than `t`. Practice the two-pointer technique on similar problems to build familiarity.

What are common mistakes when solving Append Characters to String to Make Subsequence?

Failing to realize that not all characters in `s` need to be used. Not properly handling cases where `t` is already a subsequence of `s`.

#2486

Append Characters to String to Make Subsequence

Medium

Two PointersStringGreedyTwo PointersGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to efficiently find the longest prefix of `t` that is a subsequence of `s`. This approach minimizes unnecessary checks and directly calculates how many characters need to be appended.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one for `s` and one for `t`.
2Step 2: Iterate through `s` and `t` simultaneously. If characters match, move both pointers forward.
3Step 3: If characters do not match, only move the pointer for `s`.
4Step 4: The number of characters to append is the length of `t` minus the number of matched characters.

solution.py10 lines

1# Full working Python code
2
3def min_chars_to_append(s, t):
4    i, j = 0, 0
5    while i < len(s) and j < len(t):
6        if s[i] == t[j]:
7            j += 1
8        i += 1
9    return len(t) - j
10

ℹ

Complexity note: The time complexity is O(n) because we traverse each string at most once. The space complexity is O(1) since we only use a few variables for tracking indices.

1Understanding subsequences is crucial; a subsequence can skip characters.
2Using two pointers can significantly reduce the complexity of the problem.

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