What is the time complexity of Decoded String at Index?

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

What data structures are used to solve Decoded String at Index?

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

What are the key insights for solving Decoded String at Index?

Insight 1: Instead of building the entire decoded string (which could be huge and cause memory issues), we can work backward to find the character at the k-th position. Insight 2: The crucial idea is to keep track of the size of the decoded string and use the modulo operator (%) to simulate the repetitions and find the original position that contributes to the k-th character.

Which companies ask LeetCode 880. Decoded String at Index in interviews?

This problem is frequently asked at National Instruments, PhonePe.

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Decoded String at Index - LeetCode 880 Solution

Decoded String at Index - Complete Solution Guide

Decoded String at Index is LeetCode problem 880, 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 an encoded string s . To decode the string to a tape, the encoded string is read one character at a time and the following steps are taken: If the character read is a letter, that letter is written onto the tape. If the character read is a digit d , the entire current tape is repeatedly written d - 1 more times in total. Given an integer k , return the k th letter ( 1-indexed) in the decoded string . Example 1: Input: s = "leet2code3", k = 10 Output: "o" Explanation: The decoded st

Detailed Explanation

The problem asks us to decode a string that is encoded with letters and digits. Letters are added directly to the decoded string. Digits represent repeating the current decoded string a number of times equal to the digit. Given an encoded string `s` and an index `k`, we must find the character at the `k`-th position in the fully decoded string (1-indexed). The problem ensures the decoded string's length is within reasonable bounds (less than 2^63) and `k` is always a valid index within the decoded string.

Solution Approach

The provided solution uses a reverse iteration approach. First, it calculates the total size of the decoded string by iterating through the encoded string. Then, it iterates backward through the encoded string. If a digit is encountered, it implies a repetition. We update `size` and `k` based on the repetition. If a letter is encountered, we check if `k` is equal to the current `size`. If they're equal, it means the current letter is at the `k`-th position, so we return it.

Step-by-Step Algorithm

Step 1: Calculate the size of the decoded string. Iterate through the encoded string `s`. If the character is a digit, multiply `size` by the digit. Otherwise, increment `size` by 1.
Step 2: Iterate through the encoded string `s` in reverse order.
Step 3: If the current character is a digit, divide `size` by the digit, and update `k` to `k % size`. If `k` becomes 0, set it to `size` to account for the edge case where `k` is a multiple of the current string length.
Step 4: If the current character is a letter, check if `k` is equal to `size`. If it is, then the current character is the answer and return it.
Step 5: If the current character is a letter and `k` is not equal to `size`, decrement `size`.

Key Insights

Insight 1: Instead of building the entire decoded string (which could be huge and cause memory issues), we can work backward to find the character at the k-th position.
Insight 2: The crucial idea is to keep track of the size of the decoded string and use the modulo operator (%) to simulate the repetitions and find the original position that contributes to the k-th character.
Insight 3: If we encounter a digit while traversing backwards, we divide the `size` by that digit and take `k % size` to simulate moving back within the repeated section. If `k % size == 0`, it means `k` is actually `size` in this repetition.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: String, Stack.

Companies

Asked at: National Instruments, PhonePe.

Decoded String at Index | LeetCode 880

Medium

String Stack

Time: O(N)Space: O(1)

Solution Code

Available in:

class Solution {
    public String decodeAtIndex(String s, int k) {
        long size = 0;
        for (char charr : s.toCharArray()) {
            if (Character.isDigit(charr)) {
                size *= (charr - '0');
            } else {
                size += 1;
            }
        }

        for (int i = s.length() - 1; i >= 0; i--) {
            char charr = s.charAt(i);

            if (Character.isDigit(charr)) {
                int digit = charr - '0';
                size /= digit;
                k %= size;

                if (k == 0) {
                    k = (int)size;
                }
            } else {
                if (k == size) {
                    return String.valueOf(charr);
                }
                size -= 1;
            }
        }
        return null;
    }
}

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Problem Info

Description

Problem 880: Decoded String at Index - Detailed explanation and solution approaches.

Topics

String Stack

Companies

National InstrumentsPhonePe

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:880

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

5 steps

1
Step 1: Calculate the size of the decoded string. Iterate through the encoded string `s`. If the character is a digit, multiply `size` by the digit. Otherwise, increment `size` by 1.
2
Step 2: Iterate through the encoded string `s` in reverse order.
3
Step 3: If the current character is a digit, divide `size` by the digit, and update `k` to `k % size`. If `k` becomes 0, set it to `size` to account for the edge case where `k` is a multiple of the current string length.
4
Step 4: If the current character is a letter, check if `k` is equal to `size`. If it is, then the current character is the answer and return it.
5
Step 5: If the current character is a letter and `k` is not equal to `size`, decrement `size`.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the two linear traversals of the input string. The space complexity is constant because we use a fixed number of variables regardless of the input size.

📖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

Building the string incrementally

One could attempt to build the decoded string iteratively, repeating the string as necessary. This would require string concatenation and checks against the length of the string to stop building once it surpasses k.

Time: O(k), in the worst case. The concatenated string might grow until its length equals `k`.Space: O(k), due to the space used to store the growing decoded string.

Trade-offs: This approach is simple to understand, but very inefficient and impractical for larger values of `k`, as it would require a large amount of memory to store the decoded string.

✨

Key Insights

1Insight 1: Instead of building the entire decoded string (which could be huge and cause memory issues), we can work backward to find the character at the k-th position.
2Insight 2: The crucial idea is to keep track of the size of the decoded string and use the modulo operator (%) to simulate the repetitions and find the original position that contributes to the k-th character.
3Insight 3: If we encounter a digit while traversing backwards, we divide the `size` by that digit and take `k % size` to simulate moving back within the repeated section. If `k % size == 0`, it means `k` is actually `size` in this repetition.

📚

Key Terms Explained

For Students

Stack

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

String

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

Edge Cases

Edge case 1: `k` is a multiple of the length of the decoded substring. The solution correctly handles this by setting `k = size` when `k % size == 0` in the digit case. This is necessary because the modulo operation will return 0 when `k` is a multiple of `size`, but we need `k` to represent the length of the repeated substring in that case.
Edge case 2: Input like 'a2345678999999999999999', k=1. It shows that you should use long to store size. The initial approach of calculating the full string is impractical as it can become extremely large quickly. This approach efficiently addresses this limitation by calculating the length and working backwards using the modulo operator.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using `int` instead of `long` for the `size` variable. The decoded string length can easily exceed the maximum value of an integer, causing incorrect results.

✗Common mistake 2: Not handling the case when `k % size == 0`. In this scenario, `k` should be set to `size` before proceeding.

✅

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 that use repetition to compress data. Understanding how to decode these compressed strings is useful.

🏦Real-world scenario 2: Text editors or IDEs with features like code folding or macro expansion, where repetitive patterns are used to represent code blocks or commands.

💡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

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

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 8, 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.

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 "Decoded String at Index"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?