What is the time complexity of Shortest and Lexicographically Smallest Beautiful String?

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

What data structures are used to solve Shortest and Lexicographically Smallest Beautiful String?

This problem uses String, Sliding Window. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shortest and Lexicographically Smallest Beautiful String?

Insight 1: Pre-calculating the indices of '1's in the string significantly improves efficiency. Iterating through all possible substrings to count '1's would be much slower. Insight 2: Finding the shortest beautiful substring first allows us to prune the search space. We only need to compare substrings of the minimum length found so far.

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Shortest and Lexicographically Smallest Beautiful String - LeetCode 2904 Solution

Q: Which companies ask LeetCode 2904. Shortest and Lexicographically Smallest Beautiful String in interviews?

This problem is frequently asked at IBM, Wells Fargo, Yelp.

Shortest and Lexicographically Smallest Beautiful String - Complete Solution Guide

Shortest and Lexicographically Smallest Beautiful String is LeetCode problem 2904, a Medium 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 binary string s and a positive integer k . A substring of s is beautiful if the number of 1 's in it is exactly k . Let len be the length of the shortest beautiful substring. Return the lexicographically smallest beautiful substring of string s with length equal to len . If s doesn't contain a beautiful substring, return an empty string . A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character st

Detailed Explanation

The problem asks us to find the shortest and lexicographically smallest substring of a given binary string 's' that contains exactly 'k' ones. If no such substring exists, we should return an empty string. The problem emphasizes two criteria: length and lexicographical order. We want the shortest substring that has k '1's, and if multiple substrings have that shortest length, we pick the one that comes earliest in lexicographical order (like in a dictionary).

Solution Approach

The solution focuses on efficiency by first identifying the indices of all '1's in the input string 's'. Then, it iterates through these indices, creating substrings that contain exactly 'k' ones. For each substring, it checks if its length is shorter than the current minimum length. If it is, it updates the minimum length and the result substring. If the length is equal to the current minimum length, it compares the current substring lexicographically with the current result and updates the result if the current substring is smaller.

Step-by-Step Algorithm

Step 1: Create a list/vector/array called `one_indices` to store the indices of all characters '1' in the input string `s`.
Step 2: If the number of '1's in `s` is less than `k`, return an empty string because a beautiful substring cannot exist.
Step 3: Initialize `min_len` to infinity (or a very large number) and `result` to an empty string. These will store the length of the shortest beautiful substring found so far and the substring itself.
Step 4: Iterate through the `one_indices` list from `i = 0` to `len(one_indices) - k`. This loop considers each possible starting index for a beautiful substring.
Step 5: Inside the loop, get the starting index (`start_idx`) from `one_indices[i]` and the ending index (`end_idx`) from `one_indices[i + k - 1]`. These two indices define a substring that contains exactly `k` ones.
Step 6: Calculate the length of the current substring: `current_len = end_idx - start_idx + 1`.
Step 7: Compare `current_len` with `min_len`. If `current_len < min_len`, update `min_len = current_len` and set `result = s[start_idx : end_idx + 1]` (or the equivalent substring operation in the respective language).
Step 8: If `current_len == min_len`, compare the current substring `s[start_idx : end_idx + 1]` with `result` lexicographically. If the current substring is lexicographically smaller than `result`, update `result`.
Step 9: After the loop finishes, return the `result` string.

Key Insights

Insight 1: Pre-calculating the indices of '1's in the string significantly improves efficiency. Iterating through all possible substrings to count '1's would be much slower.
Insight 2: Finding the shortest beautiful substring first allows us to prune the search space. We only need to compare substrings of the minimum length found so far.
Insight 3: Lexicographical comparison using the language's built-in string comparison operators is the most efficient way to compare substrings of the same length.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Sliding Window.

Companies

Asked at: IBM, Wells Fargo, Yelp.

Shortest and Lexicographically Smallest Beautiful String | LeetCode 2904

Medium

String Sliding Window

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public String shortestBeautifulSubstring(String s, int k) {
        List<Integer> one_indices = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '1') {
                one_indices.add(i);
            }
        }

        if (one_indices.size() < k) {
            return "";
        }

        int min_len = Integer.MAX_VALUE;
        String result = "";

        for (int i = 0; i <= one_indices.size() - k; i++) {
            int start_idx = one_indices.get(i);
            int end_idx = one_indices.get(i + k - 1);

            int current_len = end_idx - start_idx + 1;

            if (current_len < min_len) {
                min_len = current_len;
                result = s.substring(start_idx, end_idx + 1);
            } else if (current_len == min_len) {
                String candidate = s.substring(start_idx, end_idx + 1);
                if (candidate.compareTo(result) < 0) {
                    result = candidate;
                }
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 2904: Shortest and Lexicographically Smallest Beautiful String - Detailed explanation and solution approaches.

Topics

String Sliding Window

Companies

IBMWells FargoYelp

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:2904

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

9 steps

1
Step 1: Create a list/vector/array called `one_indices` to store the indices of all characters '1' in the input string `s`.
2
Step 2: If the number of '1's in `s` is less than `k`, return an empty string because a beautiful substring cannot exist.
3
Step 3: Initialize `min_len` to infinity (or a very large number) and `result` to an empty string. These will store the length of the shortest beautiful substring found so far and the substring itself.
4
Step 4: Iterate through the `one_indices` list from `i = 0` to `len(one_indices) - k`. This loop considers each possible starting index for a beautiful substring.
5
Step 5: Inside the loop, get the starting index (`start_idx`) from `one_indices[i]` and the ending index (`end_idx`) from `one_indices[i + k - 1]`. These two indices define a substring that contains exactly `k` ones.
6
Step 6: Calculate the length of the current substring: `current_len = end_idx - start_idx + 1`.
7
Step 7: Compare `current_len` with `min_len`. If `current_len < min_len`, update `min_len = current_len` and set `result = s[start_idx : end_idx + 1]` (or the equivalent substring operation in the respective language).
8
Step 8: If `current_len == min_len`, compare the current substring `s[start_idx : end_idx + 1]` with `result` lexicographically. If the current substring is lexicographically smaller than `result`, update `result`.
9
Step 9: After the loop finishes, return the `result` string.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the string 's' once to create the `one_indices` list (O(n)). The main loop iterates at most 'n' times (more precisely, len(one_indices) - k + 1, which is bounded by n). Inside the loop, substring operations and comparisons take at most O(n) in some languages (e.g., Python), but optimized string implementations make these comparisons efficient in practice, often approaching O(1) on average. The space complexity is O(n) because in the worst case, the `one_indices` list can contain all indices of the string (if the string consists only of '1's). The `result` string can also have a maximum length of 'n'.

📖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

Sliding Window

A sliding window approach can also be used. Maintain a window and adjust its start and end to find substrings with exactly k ones. This approach avoids pre-calculating indices of 1s but involves incrementing/decrementing a 'ones_count' variable within the sliding window.

Time: O(n)Space: O(1)

Trade-offs: The sliding window approach might have slightly better space complexity (O(1)) as it doesn't store the indices of 1s. However, the code might be slightly more complex to implement and reason about, especially regarding the sliding window maintenance and ensuring exactly `k` ones are present within the window. The current solution is generally preferred for its clarity.

✨

Key Insights

1Insight 1: Pre-calculating the indices of '1's in the string significantly improves efficiency. Iterating through all possible substrings to count '1's would be much slower.
2Insight 2: Finding the shortest beautiful substring first allows us to prune the search space. We only need to compare substrings of the minimum length found so far.
3Insight 3: Lexicographical comparison using the language's built-in string comparison operators is the most efficient way to compare substrings of the same length.

📚

Key Terms Explained

For Students

Sliding Window

A moving "window" over data to process subsets efficiently.

String

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

Edge Cases

Edge case 1: `k` is greater than the number of '1's in `s`. The solution handles this by checking if `len(one_indices) < k` and returning an empty string in that case.
Edge case 2: `s` contains no '1's and `k > 0`. This is a special case of the first edge case and is also handled correctly.
Edge case 3: `s` is very long. The linear time complexity ensures that the solution remains efficient even for large input strings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where no beautiful substring exists. The algorithm must correctly return an empty string when no substring with exactly 'k' ones can be found.

✗Common mistake 2: Incorrectly calculating the length of the substring or using the wrong indices when extracting the substring.

✗Common mistake 3: Forgetting to perform lexicographical comparison when multiple substrings have the same minimum length.

✗Common mistake 4: Using a brute-force approach that iterates through all possible substrings and counts the number of '1's in each. This results in a significantly higher time complexity (O(n^2)).

✅

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: In bioinformatics, searching for specific DNA sequences (represented as strings) with a certain number of a particular nucleotide (e.g., finding sequences with a certain number of guanine bases).

🏦Real-world scenario 2: In network traffic analysis, identifying network packets (represented as binary strings) with a specific number of '1' bits, which might indicate a particular type of communication or activity.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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: June 25, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

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.

How do I determine the window size in sliding window problems?

For fixed-size windows (like max sum of k elements), window size is given. For variable-size windows (like smallest subarray with sum >= target), expand right pointer to meet condition, then shrink left pointer to find optimal. Track the best result during shrinking.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Shortest and Lexicographically Smallest Beautiful String"?

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

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