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

The optimal solution has a time complexity of O(m+n).

What data structures are used to solve Subsequence With the Minimum Score?

This problem uses Two Pointers, String, Binary Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Subsequence With the Minimum Score?

Insight 1: The problem can be solved by finding the longest prefix and suffix of `t` that are subsequences of `s`. The remaining characters in `t` that are not part of the prefix or suffix need to be removed. The goal is to minimize the range of indices of these removed characters. Insight 2: Dynamic programming or two pointers can be used to efficiently find the longest prefix and suffix. This specific solution employs two-pointer approach which makes it efficient with O(m+n) time complexity.

Which companies ask LeetCode 2565. Subsequence With the Minimum Score in interviews?

This problem is frequently asked at DoorDash.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Subsequence With the Minimum Score - LeetCode 2565 Solution

Subsequence With the Minimum Score - Complete Solution Guide

Subsequence With the Minimum Score is LeetCode problem 2565, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 o

Detailed Explanation

The problem asks us to find the minimum possible score to make string `t` a subsequence of string `s` by removing characters from `t`. The score is calculated as `right - left + 1`, where `left` is the minimum index of removed characters and `right` is the maximum index of removed characters. If no characters are removed, the score is 0. A subsequence is a sequence of characters that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. The input consists of two strings, `s` and `t`, and the output is the minimum score.

Solution Approach

The solution uses a two-pointer approach to find the longest prefix and suffix of `t` that are subsequences of `s`. It calculates `prefix_match` which stores the index in `s` where the prefix of `t` ends. Similarly, `suffix_match` stores the index in `s` where the suffix of `t` starts. Then, it iterates through all possible prefixes of `t` and finds the corresponding suffix that can complete `t` as a subsequence of `s`. The minimum score is updated during this iteration.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `prefix_match` and `suffix_match`, to store the indices in `s` where the prefixes and suffixes of `t` end and start, respectively. `prefix_match[i]` stores the index in `s` where `t[:i]` is a subsequence, and `suffix_match[i]` stores the index in `s` where `t[i:]` is a subsequence.
Step 2: Calculate the `prefix_match` array. Iterate through `t` from left to right. Use a pointer `s_ptr` to iterate through `s`. If a character in `t` matches a character in `s`, update `prefix_match` with the index in `s` and increment both pointers. If the end of `s` is reached before the end of `t`, stop.
Step 3: Calculate the `suffix_match` array. Iterate through `t` from right to left. Use a pointer `s_ptr` to iterate through `s` from right to left. If a character in `t` matches a character in `s`, update `suffix_match` with the index in `s` and decrement both pointers. If the beginning of `s` is reached before the beginning of `t`, stop.
Step 4: Iterate through all possible prefixes of `t` (using index `i`). For each prefix, find the shortest suffix (starting from index `j` in `t`) such that the prefix and suffix together form a subsequence of `s`. In other words, find `j` such that `suffix_match[j]` is greater than or equal to `prefix_match[i]`.
Step 5: Calculate the score for removing the characters between the prefix and suffix (i.e., from index `i` to `j-1` in `t`). The score is `j - i`. Update the minimum score found so far.
Step 6: Return the minimum score.

Key Insights

Insight 1: The problem can be solved by finding the longest prefix and suffix of `t` that are subsequences of `s`. The remaining characters in `t` that are not part of the prefix or suffix need to be removed. The goal is to minimize the range of indices of these removed characters.
Insight 2: Dynamic programming or two pointers can be used to efficiently find the longest prefix and suffix. This specific solution employs two-pointer approach which makes it efficient with O(m+n) time complexity.
Insight 3: When calculating score, remember edge cases like when no characters are removed (score = 0) and when all characters are removed (score = length of t).

Complexity Analysis

Time Complexity: O(m+n)

Space Complexity: O(m)

Topics

This problem involves: Two Pointers, String, Binary Search.

Companies

Asked at: DoorDash.

Subsequence With the Minimum Score | LeetCode 2565

Hard

Two Pointers String Binary Search

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

Solution Code

Available in:

class Solution {
    public int minimumScore(String s, String t) {
        int n = s.length();
        int m = t.length();

        int[] prefix_match = new int[m + 1];
        for (int i = 0; i <= m; i++) {
            prefix_match[i] = n;
        }
        prefix_match[0] = -1;
        int s_ptr = 0;
        for (int i = 0; i < m; i++) {
            while (s_ptr < n && s.charAt(s_ptr) != t.charAt(i)) {
                s_ptr++;
            }
            if (s_ptr == n) {
                break;
            }
            prefix_match[i + 1] = s_ptr;
            s_ptr++;
        }

        int[] suffix_match = new int[m + 1];
        for (int i = 0; i <= m; i++) {
            suffix_match[i] = -1;
        }
        suffix_match[m] = n;
        s_ptr = n - 1;
        for (int i = m - 1; i >= 0; i--) {
            while (s_ptr >= 0 && s.charAt(s_ptr) != t.charAt(i)) {
                s_ptr--;
            }
            if (s_ptr < 0) {
                break;
            }
            suffix_match[i] = s_ptr;
            s_ptr--;
        }

        int min_score = m;
        int j = 0;
        for (int i = 0; i <= m; i++) {
            if (prefix_match[i] == n) {
                break;
            }

            if (j < i) {
                j = i;
            }

            while (j <= m && suffix_match[j] <= prefix_match[i]) {
                j++;
            }

            min_score = Math.min(min_score, j - i);
        }

        return min_score;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2565: Subsequence With the Minimum Score - Detailed explanation and solution approaches.

Topics

Two Pointers String Binary Search

Companies

DoorDash

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2565

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize two arrays, `prefix_match` and `suffix_match`, to store the indices in `s` where the prefixes and suffixes of `t` end and start, respectively. `prefix_match[i]` stores the index in `s` where `t[:i]` is a subsequence, and `suffix_match[i]` stores the index in `s` where `t[i:]` is a subsequence.
2
Step 2: Calculate the `prefix_match` array. Iterate through `t` from left to right. Use a pointer `s_ptr` to iterate through `s`. If a character in `t` matches a character in `s`, update `prefix_match` with the index in `s` and increment both pointers. If the end of `s` is reached before the end of `t`, stop.
3
Step 3: Calculate the `suffix_match` array. Iterate through `t` from right to left. Use a pointer `s_ptr` to iterate through `s` from right to left. If a character in `t` matches a character in `s`, update `suffix_match` with the index in `s` and decrement both pointers. If the beginning of `s` is reached before the beginning of `t`, stop.
4
Step 4: Iterate through all possible prefixes of `t` (using index `i`). For each prefix, find the shortest suffix (starting from index `j` in `t`) such that the prefix and suffix together form a subsequence of `s`. In other words, find `j` such that `suffix_match[j]` is greater than or equal to `prefix_match[i]`.
5
Step 5: Calculate the score for removing the characters between the prefix and suffix (i.e., from index `i` to `j-1` in `t`). The score is `j - i`. Update the minimum score found so far.
6
Step 6: Return the minimum score.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m+n)

💾

Space Complexity

O(m)

📝

Why?

The prefix and suffix matching processes each take O(m+n) time. The final loop determining the minimum score iterates at most m times, with an inner while loop that cumulatively (over all outer loop iterations) runs at most m times, due to j only ever increasing. The space complexity is O(m) because we are storing the prefix and suffix match indices, which depend on the length of the target string t.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A dynamic programming approach can be used to find the longest common subsequence of `s` and `t`. Then the characters in `t` that are not part of the LCS would be the characters to remove. This could potentially result in a more intuitive if possibly less efficient implementation.

Time: O(m*n)Space: O(m*n)

Trade-offs: The dynamic programming approach has higher time and space complexity than the two-pointer approach. However, it may be easier to understand and implement for some developers. The two-pointer approach is more efficient for large inputs.

✨

Key Insights

1Insight 1: The problem can be solved by finding the longest prefix and suffix of `t` that are subsequences of `s`. The remaining characters in `t` that are not part of the prefix or suffix need to be removed. The goal is to minimize the range of indices of these removed characters.
2Insight 2: Dynamic programming or two pointers can be used to efficiently find the longest prefix and suffix. This specific solution employs two-pointer approach which makes it efficient with O(m+n) time complexity.
3Insight 3: When calculating score, remember edge cases like when no characters are removed (score = 0) and when all characters are removed (score = length of t).

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: If `t` is already a subsequence of `s`, the score should be 0. The code handles this because when the entire `t` is a subsequence, the prefix match iterates all the way through without needing suffix removal, and min_score remains 0 initially.
Edge case 2: If no part of `t` is a subsequence of `s`, all characters must be removed, resulting in a score of `t.length`. This is handled correctly as the matching loops may not complete and the initialization of min_score to `t.length` will guarantee this becomes the solution.
Edge case 3: If s or t is empty. While not explicitly handled in the code, the loops will run either zero times or not at all so will not cause problems since only one loop advances index j, ensuring correct behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `t` is already a subsequence of `s` correctly, which results in a non-zero score.

✗Common mistake 2: Incorrectly calculating the score as `right - left` instead of `right - left + 1`.

✗Common mistake 3: Forgetting to initialize `min_score` to the length of `t` or a sufficiently high value. If no subsequence exists at all, then the score should be the entire length of `t`.

✗Common mistake 4: Incorrectly implementing the two pointer logic for prefix and suffix matching, leading to wrong indices being captured.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This algorithm can be used in text editing software to find the minimum number of changes needed to make one string a subsequence of another. This is useful for auto-correction and suggestion features.

🏦Real-world scenario 2: In bioinformatics, this algorithm can be used to compare DNA sequences and find the minimum number of mutations needed to transform one sequence into another, where a mutation is the deletion of a base pair.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Subsequence With the Minimum Score"?

For this hard-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?