What is the time complexity of Construct String With Repeat Limit?

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

What data structures are used to solve Construct String With Repeat Limit?

This problem uses Hash Table, String, Greedy, Heap (Priority Queue), Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Construct String With Repeat Limit?

Insight 1: Since we want the lexicographically largest string, we should start by using the largest character ('z') as much as possible until either we exhaust all 'z's or we reach the repeat limit. Insight 2: If we reach the repeat limit for a character, we need to use a smaller character to break the repetition, but we should use the next largest available character to maintain the lexicographical order.

Which companies ask LeetCode 2182. Construct String With Repeat Limit in interviews?

This problem is frequently asked at Arista Networks, Fortinet.

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

Construct String With Repeat Limit - LeetCode 2182 Solution

Construct String With Repeat Limit - Complete Solution Guide

Construct String With Repeat Limit is LeetCode problem 2182, 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 string s and an integer repeatLimit . Construct a new string repeatLimitedString using the characters of s such that no letter appears more than repeatLimit times in a row . You do not have to use all characters from s . Return the lexicographically largest repeatLimitedString possible . A string a is lexicographically larger than a string b if in the first position where a and b differ, string a has a letter that appears later in the alphabet than the corresponding letter in b .

Detailed Explanation

The problem asks us to construct the lexicographically largest string from the given string 's' such that no character repeats more than 'repeatLimit' times consecutively. We don't necessarily have to use all the characters from 's'. The inputs are the string 's' and the integer 'repeatLimit'. The output is the lexicographically largest string satisfying the constraint.

Solution Approach

The solution uses a greedy approach. We start with the largest character ('z') and try to append it to the result as many times as possible, up to the repeat limit. If we exhaust all occurrences of 'z', we move on to the next largest character. If we hit the repeat limit for a character before exhausting all occurrences, we find the next largest available character and append it once to break the repetition. We continue this process until we run out of characters to append.

Step-by-Step Algorithm

Step 1: Create a frequency count array (counts) of size 26 to store the number of occurrences of each lowercase English letter in the input string 's'.
Step 2: Initialize two pointers, 'i' and 'j', to 25 and 24 respectively. 'i' represents the current largest character we're considering, and 'j' represents the next largest character to use as a breaker when 'i' hits repeatLimit.
Step 3: Iterate while 'i' is greater than or equal to 0. In each iteration, check if counts[i] is greater than 0. If not, decrement 'i' and continue to the next iteration.
Step 4: Calculate the number of times (num_to_add) we can add the current character 'i' to the result. It's the minimum of counts[i] and repeatLimit.
Step 5: Append the character 'i' num_to_add times to the result string. Update counts[i] by subtracting num_to_add.
Step 6: If counts[i] is still greater than 0, it means we hit the repeat limit. We need to find a breaker character. Set j = min(j, i - 1) and decrement j until we find a character with a positive count.
Step 7: If we find a breaker character (j >= 0), append it once to the result and decrement counts[j]. Otherwise (j < 0), it means there are no smaller characters available, so we break the loop as we can't satisfy the repeatLimit constraint for character i.
Step 8: If counts[i] is 0, it means we exhausted all occurrences of the current character 'i', so decrement 'i' to consider the next smaller character.
Step 9: Finally, return the result string.

Key Insights

Insight 1: Since we want the lexicographically largest string, we should start by using the largest character ('z') as much as possible until either we exhaust all 'z's or we reach the repeat limit.
Insight 2: If we reach the repeat limit for a character, we need to use a smaller character to break the repetition, but we should use the next largest available character to maintain the lexicographical order.
Insight 3: We can use a frequency count array to keep track of the number of occurrences of each character in the input string. This allows us to efficiently find the available characters.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Greedy, Heap (Priority Queue), Counting.

Companies

Asked at: Arista Networks, Fortinet.

Construct String With Repeat Limit | LeetCode 2182

Medium

Hash Table String Greedy Heap (Priority Queue)Counting

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

Solution Code

Available in:

class Solution {
    public String repeatLimitedString(String s, int repeatLimit) {
        int[] counts = new int[26];
        for (char char_ : s.toCharArray()) {
            counts[char_ - 'a']++;
        }

        StringBuilder res = new StringBuilder();
        int i = 25;
        int j = 24;

        while (i >= 0) {
            if (counts[i] == 0) {
                i--;
                continue;
            }

            int numToAdd = Math.min(counts[i], repeatLimit);

            char charI = (char) ('a' + i);
            for (int k = 0; k < numToAdd; k++) {
                res.append(charI);
            }
            counts[i] -= numToAdd;

            if (counts[i] > 0) {
                j = Math.min(j, i - 1);
                while (j >= 0 && counts[j] == 0) {
                    j--;
                }

                if (j < 0) {
                    break;
                }

                char charJ = (char) ('a' + j);
                res.append(charJ);
                counts[j]--;
            } else {
                i--;
            }
        }

        return res.toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2182: Construct String With Repeat Limit - Detailed explanation and solution approaches.

Topics

Hash Table String Greedy Heap (Priority Queue)Counting

Companies

Arista NetworksFortinet

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2182

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 frequency count array (counts) of size 26 to store the number of occurrences of each lowercase English letter in the input string 's'.
2
Step 2: Initialize two pointers, 'i' and 'j', to 25 and 24 respectively. 'i' represents the current largest character we're considering, and 'j' represents the next largest character to use as a breaker when 'i' hits repeatLimit.
3
Step 3: Iterate while 'i' is greater than or equal to 0. In each iteration, check if counts[i] is greater than 0. If not, decrement 'i' and continue to the next iteration.
4
Step 4: Calculate the number of times (num_to_add) we can add the current character 'i' to the result. It's the minimum of counts[i] and repeatLimit.
5
Step 5: Append the character 'i' num_to_add times to the result string. Update counts[i] by subtracting num_to_add.
6
Step 6: If counts[i] is still greater than 0, it means we hit the repeat limit. We need to find a breaker character. Set j = min(j, i - 1) and decrement j until we find a character with a positive count.
7
Step 7: If we find a breaker character (j >= 0), append it once to the result and decrement counts[j]. Otherwise (j < 0), it means there are no smaller characters available, so we break the loop as we can't satisfy the repeatLimit constraint for character i.
8
Step 8: If counts[i] is 0, it means we exhausted all occurrences of the current character 'i', so decrement 'i' to consider the next smaller character.
9
Step 9: Finally, 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 time complexity is linear because we iterate through the input string once to count the characters, and then the while loop processes each character at most once in the worst case. The space complexity comes from the `counts` array (fixed size of 26) and the resultant string which, in the worst case, will be a modified version of the original string (size 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

Priority Queue (Heap)

Use a priority queue (max heap) to store the characters and their frequencies. Repeatedly extract the character with the highest frequency, add it to the result (up to repeatLimit), and then push it back into the queue if there are remaining occurrences. If adding it to the result would violate the repeatLimit, poll the next highest frequency character, add it to the result, and then push both characters back onto the queue.

Time: O(n log 26) which simplifies to O(n log n). Each insertion/deletion operation into the heap takes O(log k) time, where k is the number of distinct characters (at most 26). Since we perform up to n such operations, the overall complexity becomes O(n log k).Space: O(26), which is O(1). The heap stores the character counts, so it needs space proportional to the number of unique characters in 's' which is bounded by 26.

Trade-offs: While the time complexity might seem similar to the main solution, the constant factor associated with heap operations can make it slower in practice, especially for larger input strings. The main advantage is that the heap-based approach is more adaptable to variations of the problem where the character set is much larger than just lowercase English letters.

✨

Key Insights

1Insight 1: Since we want the lexicographically largest string, we should start by using the largest character ('z') as much as possible until either we exhaust all 'z's or we reach the repeat limit.
2Insight 2: If we reach the repeat limit for a character, we need to use a smaller character to break the repetition, but we should use the next largest available character to maintain the lexicographical order.
3Insight 3: We can use a frequency count array to keep track of the number of occurrences of each character in the input string. This allows us to efficiently find the available characters.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

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: Empty input string. The code handles it correctly by returning an empty string.
Edge case 2: repeatLimit is greater than the occurrences of any character. The code correctly adds all characters up to their occurrences without violating the repeatLimit.
Edge case 3: repeatLimit is 1. The code correctly interleaves the characters to produce the lexicographically largest result with no consecutive repetitions.
Edge case 4: String containing only one character. The code correctly adds the character up to the repeatLimit or the number of occurrences of the character, whichever is smaller.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case when there are no smaller characters available to break the repetition.

✗Common mistake 2: Not ensuring the lexicographical order when appending characters to the result.

✗Common mistake 3: Failing to account for the repeatLimit constraint correctly.

✗Common mistake 4: Integer overflow when calculating num_to_add.

✅

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: Data compression algorithms where you want to limit the number of consecutive repetitions of a symbol to improve compression efficiency.

🏦Real-world scenario 2: Scheduling tasks where you want to prioritize tasks with higher importance (lexicographically larger) but also prevent any single task from monopolizing the resources for too long (repeatLimit).

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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 23, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 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 "Construct String With Repeat Limit"?

For this medium-level Hash Table 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?