What is the time complexity of Shifting Letters II?

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

What data structures are used to solve Shifting Letters II?

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

What are the key insights for solving Shifting Letters II?

Insight 1: We need to apply multiple shifts to potentially overlapping ranges of the string, so a naive approach of directly shifting characters for each shift in `shifts` would be inefficient (O(n*m) where n is the length of string and m is length of shifts). Insight 2: We can use the concept of prefix sums (or difference arrays) to efficiently calculate the net shift for each character in the string after applying all shift operations.

Which companies ask LeetCode 2381. Shifting Letters II in interviews?

This problem is frequently asked at Veritas.

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Shifting Letters II - LeetCode 2381 Solution

Shifting Letters II - Complete Solution Guide

Shifting Letters II is LeetCode problem 2381, 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 of lowercase English letters and a 2D integer array shifts where shifts[i] = [start i , end i , direction i ] . For every i , shift the characters in s from the index start i to the index end i ( inclusive ) forward if direction i = 1 , or shift the characters backward if direction i = 0 . Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a' ). Similarly, shifting a character backward means replacin

Detailed Explanation

The problem requires us to shift characters in a given string `s` based on a series of shifts defined in the `shifts` array. Each shift operation `shifts[i] = [start_i, end_i, direction_i]` dictates that characters from index `start_i` to `end_i` (inclusive) should be shifted either forward (direction = 1) or backward (direction = 0) in the alphabet. The shifting operation wraps around, so 'z' becomes 'a' when shifted forward, and 'a' becomes 'z' when shifted backward. The goal is to apply all shifts sequentially and return the final modified string.

Solution Approach

The solution uses a difference array (or prefix sum) technique to efficiently compute the net shift applied to each character in the string. First, a difference array `diff` is created with a size of `n+1` (where `n` is the length of the string) to store the changes in shift values. We iterate through the `shifts` array, and for each shift, we increment `diff[start]` and decrement `diff[end + 1]`. This represents the start and end of the shifted region. Then, we iterate through the `diff` array, calculating the prefix sum to determine the actual shift value at each index. Finally, we apply the calculated shift to each character in the original string, handling the wraparound behavior correctly by using the modulo operator. The updated characters are then combined to form the final shifted string.

Step-by-Step Algorithm

Step 1: Initialize a difference array `diff` of size `n + 1` with all elements set to 0.
Step 2: Iterate through the `shifts` array.
Step 3: For each shift `[start, end, direction]`, determine the shift value `val` (1 for forward, -1 for backward).
Step 4: Update the difference array: `diff[start] += val` and `diff[end + 1] -= val`.
Step 5: Initialize `current_shift` to 0.
Step 6: Iterate through the string `s` from index 0 to `n - 1`.
Step 7: Update `current_shift` by adding `diff[i]` to it (`current_shift += diff[i]`). This represents the total shift amount at the current index.
Step 8: Calculate the new character code: `new_code = (ord(s[i]) - ord('a') + current_shift) % 26`. Important to add 26 and modulo 26 to ensure positive values for all languages for negative current_shift
Step 9: Convert the `new_code` back to a character and update the string `s[i]` with the new character.
Step 10: After processing all characters, return the modified string `s`.
Step 11: Handle potential negative `current_shift` values correctly in languages like C++ where the modulo operator can return negative results for negative inputs. Add 26 to the `new_code` before taking the modulo if needed to ensure a positive remainder.

Key Insights

Insight 1: We need to apply multiple shifts to potentially overlapping ranges of the string, so a naive approach of directly shifting characters for each shift in `shifts` would be inefficient (O(n*m) where n is the length of string and m is length of shifts).
Insight 2: We can use the concept of prefix sums (or difference arrays) to efficiently calculate the net shift for each character in the string after applying all shift operations.
Insight 3: When calculating the new character code, it's crucial to handle the wrapping around from 'z' to 'a' and vice-versa correctly, using the modulo operator (%) to ensure the resulting character code stays within the range of 0-25.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Prefix Sum.

Companies

Asked at: Veritas.

Shifting Letters II | LeetCode 2381

Medium

Array String Prefix Sum

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

Solution Code

Available in:

class Solution {
    public String shiftingLetters(String s, int[][] shifts) {
        int n = s.length();
        int[] diff = new int[n + 1];

        for (int[] shift : shifts) {
            int start = shift[0];
            int end = shift[1];
            int direction = shift[2];
            
            int val = (direction == 1) ? 1 : -1;
            diff[start] += val;
            diff[end + 1] -= val;
        }

        char[] res = s.toCharArray();
        int current_shift = 0;
        for (int i = 0; i < n; i++) {
            current_shift += diff[i];

            int char_code = res[i] - 'a';
            int new_code = (char_code + current_shift % 26 + 26) % 26; // Ensure non-negative modulo
            
            res[i] = (char) ('a' + new_code);
        }

        return new String(res);
    }
}

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Problem Info

Description

Problem 2381: Shifting Letters II - Detailed explanation and solution approaches.

Topics

Array String Prefix Sum

Companies

Veritas

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2381

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

11 steps

1
Step 1: Initialize a difference array `diff` of size `n + 1` with all elements set to 0.
2
Step 2: Iterate through the `shifts` array.
3
Step 3: For each shift `[start, end, direction]`, determine the shift value `val` (1 for forward, -1 for backward).
4
Step 4: Update the difference array: `diff[start] += val` and `diff[end + 1] -= val`.
5
Step 5: Initialize `current_shift` to 0.
6
Step 6: Iterate through the string `s` from index 0 to `n - 1`.
7
Step 7: Update `current_shift` by adding `diff[i]` to it (`current_shift += diff[i]`). This represents the total shift amount at the current index.
8
Step 8: Calculate the new character code: `new_code = (ord(s[i]) - ord('a') + current_shift) % 26`. Important to add 26 and modulo 26 to ensure positive values for all languages for negative current_shift
9
Step 9: Convert the `new_code` back to a character and update the string `s[i]` with the new character.
10
Step 10: After processing all characters, return the modified string `s`.
11
Step 11: Handle potential negative `current_shift` values correctly in languages like C++ where the modulo operator can return negative results for negative inputs. Add 26 to the `new_code` before taking the modulo if needed to ensure a positive remainder.

💡

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 + m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the two loops: one iterating through the `shifts` array and the other iterating through the string. The space complexity is dominated by the `diff` array, which has a size proportional to the length of the input string.

📖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

Iterate through each shift in `shifts` and directly shift the characters within the specified range. For each character in the range, calculate the shifted character based on the direction. Perform modular arithmetic to handle wrapping.

Time: O(n*m), where n is the string length and m is the number of shifts.Space: O(1) or O(n) depending on if the string is mutable or not (in-place modification).

Trade-offs: Simpler to understand but significantly less efficient for a large number of shifts because you modify the string potentially multiple times. Suitable for smaller input sizes, but not recommended for large `n` and `m` due to its quadratic time complexity. The main solution is superior due to its linear time complexity.

✨

Key Insights

1Insight 1: We need to apply multiple shifts to potentially overlapping ranges of the string, so a naive approach of directly shifting characters for each shift in `shifts` would be inefficient (O(n*m) where n is the length of string and m is length of shifts).
2Insight 2: We can use the concept of prefix sums (or difference arrays) to efficiently calculate the net shift for each character in the string after applying all shift operations.
3Insight 3: When calculating the new character code, it's crucial to handle the wrapping around from 'z' to 'a' and vice-versa correctly, using the modulo operator (%) to ensure the resulting character code stays within the range of 0-25.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

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

Edge Cases

Edge case 1: Empty string `s`. The algorithm correctly handles this case as the loops will not execute, and the empty string will be returned.
Edge case 2: Empty `shifts` array. The algorithm correctly handles this case because the loop that processes the shifts won't execute and no changes will be made to the prefix sum array.
Edge case 3: Large number of shifts or large shift values, leading to a large `current_shift`. The modulo operator (%) ensures that the `new_code` always stays within the valid range (0-25), preventing out-of-bounds character codes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the wraparound from 'z' to 'a' (or vice-versa). The modulo operator (%) is essential to ensure the shifted character stays within the alphabet.

✗Common mistake 2: Incorrectly updating the `diff` array. It's important to add the shift value at the start index and subtract it at the `end + 1` index. Missing either step leads to incorrect shift calculations.

✗Common mistake 3: Not considering negative values when using the modulo operator, especially in languages like C++. A common fix is adding 26 to the result of the modulo operation and then modulo again.

✗Common mistake 4: Not allocating enough memory for the result string in C, and not null-terminating the result string which can lead to undefined behaviour.

✅

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 encryption. Shifting characters based on a series of keys is a basic form of encryption (Caesar cipher).

🏦Real-world scenario 2: Text transformations. Applying a series of transformations to a text document based on certain rules or patterns. Can be helpful in string processing tasks, log analysis and data anonymization.

🎮Real-world scenario 3: Game development. Encoding and decoding messages or in manipulating in-game text in creative ways.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 12, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Shifting Letters II"?

For this medium-level Array 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.

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