What is the time complexity of Smallest K-Length Subsequence With Occurrences of a Letter?

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

What data structures are used to solve Smallest K-Length Subsequence With Occurrences of a Letter?

This problem uses String, Stack, Greedy, Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Smallest K-Length Subsequence With Occurrences of a Letter?

Insight 1: A monotonic stack can be used to efficiently build the lexicographically smallest subsequence. The stack helps in maintaining a sequence that always prioritizes smaller characters at the beginning. Insight 2: A greedy approach is suitable because at each step, we try to keep the smallest possible character while still being able to satisfy the length (`k`) and repetition requirements.

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Smallest K-Length Subsequence With Occurrences of a Letter - LeetCode 2030 Solution

Q: Which companies ask LeetCode 2030. Smallest K-Length Subsequence With Occurrences of a Letter in interviews?

This problem is frequently asked at Deutsche Bank.

Smallest K-Length Subsequence With Occurrences of a Letter - Complete Solution Guide

Smallest K-Length Subsequence With Occurrences of a Letter is LeetCode problem 2030, 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 a string s , an integer k , a letter letter , and an integer repetition . Return the lexicographically smallest subsequence of s of length k that has the letter letter appear at least repetition times . The test cases are generated so that the letter appears in s at least repetition times. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. A string a is lexicographically smaller

Detailed Explanation

The problem asks us to find the lexicographically smallest subsequence of a given string `s` that satisfies two conditions: the subsequence must have a length of `k`, and it must contain the character `letter` at least `repetition` times. Lexicographically smallest means we want the subsequence that would come first in alphabetical order. A subsequence is formed by deleting zero or more characters from the original string without changing the order of the remaining characters.

Solution Approach

The solution utilizes a greedy approach combined with a stack (or vector/StringBuilder in other languages) to construct the lexicographically smallest subsequence. The algorithm iterates through the input string `s`, maintaining a stack `res`. For each character `c` in `s`, the algorithm performs two main phases: a popping phase and a pushing phase. The popping phase aims to remove larger characters from the end of `res` to make space for smaller characters. The pushing phase adds `c` to `res` if doing so maintains the subsequence's length to be less than or equal to `k` and if it's still possible to meet the repetition criteria for the `letter` character. The algorithm also tracks the remaining counts of `letter` in `s` and the number of `letter` currently in the subsequence to enforce the constraints.

Step-by-Step Algorithm

Step 1: Initialize `letters_left` to the total count of `letter` in `s`. This variable keeps track of how many of the required letter are remaining in the unprocessed part of the original string.
Step 2: Initialize an empty stack `res` to store the subsequence being built and `letters_in_res` to 0, which counts the number of occurrences of `letter` currently in the stack.
Step 3: Iterate through the string `s` character by character.
Step 4: For each character `c`, enter the popping phase: while `res` is not empty, the last element of `res` is greater than `c`, and removing the last element of `res` still allows forming a subsequence of length `k`, pop the last element. When popping, carefully check to see if the element being popped is equal to `letter`. If it is, we need to check if we can still meet the repetition criteria. If popping would reduce the number of letters we've taken such that we can no longer meet the repetition criteria, we must stop popping.
Step 5: Enter the pushing phase: If `res` has length less than `k`, check the type of the charcter `c`. If `c` is equal to `letter`, push `c` to `res` and increment the `letters_in_res` count. If `c` is not equal to `letter`, check if adding `c` will leave enough space to add the minimum needed `letter` characters to reach the repetition constraint. If it does, add `c` to `res`.
Step 6: After processing each character, decrement `letters_left` if `c` was equal to `letter`.
Step 7: After iterating through all characters in `s`, convert `res` to a string and return.

Key Insights

Insight 1: A monotonic stack can be used to efficiently build the lexicographically smallest subsequence. The stack helps in maintaining a sequence that always prioritizes smaller characters at the beginning.
Insight 2: A greedy approach is suitable because at each step, we try to keep the smallest possible character while still being able to satisfy the length (`k`) and repetition requirements.
Insight 3: Maintaining a count of the occurrences of the `letter` remaining in the input string and in the resulting subsequence is critical to ensuring that the repetition constraint is met.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: String, Stack, Greedy, Monotonic Stack.

Companies

Asked at: Deutsche Bank.

Smallest K-Length Subsequence With Occurrences of a Letter | LeetCode 2030

Hard

String Stack Greedy Monotonic Stack

Time: O(n)Space: O(k)

Solution Code

Available in:

class Solution {
    public String smallestSubsequence(String s, int k, char letter, int repetition) {
        int n = s.length();
        int letters_left = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == letter) {
                letters_left++;
            }
        }

        StringBuilder res = new StringBuilder();
        int letters_in_res = 0;

        for (int i = 0; i < n; i++) {
            char c = s.charAt(i);

            // The popping phase: try to remove larger characters from the end of res
            while (res.length() > 0 && res.charAt(res.length() - 1) > c && (res.length() - 1 + n - i >= k)) {
                char popped = res.charAt(res.length() - 1);
                if (popped == letter) {
                    // If we pop a 'letter', check if we can still meet the repetition requirement
                    if (letters_in_res - 1 + letters_left < repetition) {
                        break;  // Cannot pop, we need this letter
                    }
                    letters_in_res--;
                }
                res.deleteCharAt(res.length() - 1);
            }

            // The pushing phase: add the current character if it fits the criteria
            if (res.length() < k) {
                if (c == letter) {
                    res.append(c);
                    letters_in_res++;
                } else {
                    // For a non-letter, check if there's enough space for the required letters
                    // Slots needed for other characters: k - len(res)
                    // Letters we must still take: repetition - letters_in_res
                    // If we take the current non-letter, slots left for letters: k - len(res) - 1
                    // This must be >= the number of letters we still need
                    if (k - res.length() - 1 >= repetition - letters_in_res) {
                        res.append(c);
                    }
                }
            }

            // Update the count of letters remaining in s
            if (c == letter) {
                letters_left--;
            }
        }

        return res.toString();
    }
}

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Problem Info

Description

Problem 2030: Smallest K-Length Subsequence With Occurrences of a Letter - Detailed explanation and solution approaches.

Topics

String Stack Greedy Monotonic Stack

Companies

Deutsche Bank

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2030

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

7 steps

1
Step 1: Initialize `letters_left` to the total count of `letter` in `s`. This variable keeps track of how many of the required letter are remaining in the unprocessed part of the original string.
2
Step 2: Initialize an empty stack `res` to store the subsequence being built and `letters_in_res` to 0, which counts the number of occurrences of `letter` currently in the stack.
3
Step 3: Iterate through the string `s` character by character.
4
Step 4: For each character `c`, enter the popping phase: while `res` is not empty, the last element of `res` is greater than `c`, and removing the last element of `res` still allows forming a subsequence of length `k`, pop the last element. When popping, carefully check to see if the element being popped is equal to `letter`. If it is, we need to check if we can still meet the repetition criteria. If popping would reduce the number of letters we've taken such that we can no longer meet the repetition criteria, we must stop popping.
5
Step 5: Enter the pushing phase: If `res` has length less than `k`, check the type of the charcter `c`. If `c` is equal to `letter`, push `c` to `res` and increment the `letters_in_res` count. If `c` is not equal to `letter`, check if adding `c` will leave enough space to add the minimum needed `letter` characters to reach the repetition constraint. If it does, add `c` to `res`.
6
Step 6: After processing each character, decrement `letters_left` if `c` was equal to `letter`.
7
Step 7: After iterating through all characters in `s`, convert `res` to a string and return.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(k)

📝

Why?

The time complexity is O(n) because each character in the input string `s` is processed at most twice (once for pushing and at most once for popping). The space complexity is O(k) because the stack `res` will never contain more than `k` characters.

📖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

Brute-force generation and checking

Generate all possible subsequences of length `k` from `s` and check which ones contain `letter` at least `repetition` times. Then, select the lexicographically smallest one.

Time: O(C(n, k) * k), where C(n, k) represents the binomial coefficient (n choose k), which is the number of subsequences of length k. Checking each subsequence for the letter counts takes O(k) time.Space: O(k)

Trade-offs: The brute-force approach is simple to understand but highly inefficient due to the exponential number of subsequences. It's suitable for very small input sizes but quickly becomes impractical.

✨

Key Insights

1Insight 1: A monotonic stack can be used to efficiently build the lexicographically smallest subsequence. The stack helps in maintaining a sequence that always prioritizes smaller characters at the beginning.
2Insight 2: A greedy approach is suitable because at each step, we try to keep the smallest possible character while still being able to satisfy the length (`k`) and repetition requirements.
3Insight 3: Maintaining a count of the occurrences of the `letter` remaining in the input string and in the resulting subsequence is critical to ensuring that the repetition constraint is met.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: `k` is equal to the length of `s`, and the `letter` appears `repetition` times. The solution will simply return `s` because no smaller subsequence of length `k` is possible and there is only one.
Edge case 2: `repetition` is equal to `k` and all characters in `s` are `letter`. The algorithm will return a string containing `k` repetitions of `letter`.
Edge case 3: When the current char is equal to the target `letter`, we increment `letter_in_res` but we should only increment it when we actually add it to result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to check if there are enough letters remaining to satisfy the repetition requirement before popping a letter from the stack. This could lead to subsequences that don't contain enough `letter` characters.

✗Common mistake 2: Not considering the length constraint `k` when adding characters to the subsequence. It is crucial to check that the current character can be added without exceeding the desired length and while leaving sufficient spaces for letters to achieve repetition.

✅

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: Finding the most relevant keywords in a user query to optimize search results, where relevance is defined by lexicographical order and keyword repetition requirements.

🏦Real-world scenario 2: Compressing a string by selecting a representative subsequence that preserves essential information while minimizing the length, subject to certain character constraints.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 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 21, 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 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Smallest K-Length Subsequence With Occurrences of a Letter"?

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

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