What is the time complexity of Find Mirror Score of a String?

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

What data structures are used to solve Find Mirror Score of a String?

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

What are the key insights for solving Find Mirror Score of a String?

Insight 1: Use a hash map to store the indices of each character to efficiently find potential mirror pairs. Insight 2: Iterate through the string and, for each character, check if the mirror character exists in the hash map and if there are any unmarked indices available.

Which companies ask LeetCode 3412. Find Mirror Score of a String in interviews?

This problem is frequently asked at carwale.

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Find Mirror Score of a String - LeetCode 3412 Solution

Q: How to solve Find Mirror Score of a String?

The provided code uses a hash map (dictionary in Python) called `positions` to store the indices of each character encountered so far. As it iterates through the string, it checks if the mirror character of the current character exists in the `positions` hash map. If it does and there are available unmarked indices for the mirror character, it takes the last (closest) index, calculates the difference, adds it to the score, and removes the index from the hash map to mark it as used. If the mirror character doesn't exist or there are no available indices, the current character and its index are added to the `positions` hash map.

Find Mirror Score of a String - Complete Solution Guide

Find Mirror Score of a String is LeetCode problem 3412, 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 . We define the mirror of a letter in the English alphabet as its corresponding letter when the alphabet is reversed. For example, the mirror of 'a' is 'z' , and the mirror of 'y' is 'b' . Initially, all characters in the string s are unmarked . You start with a score of 0, and you perform the following process on the string s : Iterate through the string from left to right. At each index i , find the closest unmarked index j such that j < i and s[j] is the mirror of s[i

Detailed Explanation

The problem asks us to calculate a 'mirror score' for a given string `s`. The mirror of a character is its corresponding character when the alphabet is reversed (e.g., 'a' becomes 'z', 'b' becomes 'y'). We iterate through the string `s` from left to right. For each character at index `i`, we look for the closest unmarked index `j` (where `j < i`) such that `s[j]` is the mirror of `s[i]`. If we find such a `j`, we mark both indices `i` and `j` and add `i - j` to the total score. The goal is to return the final total score after processing the entire string.

Solution Approach

The provided code uses a hash map (dictionary in Python) called `positions` to store the indices of each character encountered so far. As it iterates through the string, it checks if the mirror character of the current character exists in the `positions` hash map. If it does and there are available unmarked indices for the mirror character, it takes the last (closest) index, calculates the difference, adds it to the score, and removes the index from the hash map to mark it as used. If the mirror character doesn't exist or there are no available indices, the current character and its index are added to the `positions` hash map.

Step-by-Step Algorithm

Step 1: Initialize a hash map `positions` to store the indices of each character and a variable `score` to 0.
Step 2: Iterate through the string `s` from left to right (index `i`).
Step 3: Calculate the mirror character of the current character `s[i]`.
Step 4: Check if the mirror character exists in the `positions` hash map and if its list of indices is not empty.
Step 5: If the mirror character exists and has available indices, retrieve the last (closest) index `j` from the list, remove it (mark it as used), and add `i - j` to the `score`.
Step 6: If the mirror character does not exist or has no available indices, add the current character and its index `i` to the `positions` hash map.
Step 7: After iterating through the entire string, return the final `score`.

Key Insights

Insight 1: Use a hash map to store the indices of each character to efficiently find potential mirror pairs.
Insight 2: Iterate through the string and, for each character, check if the mirror character exists in the hash map and if there are any unmarked indices available.
Insight 3: Use a stack-like structure (in this case, a list where we pop from the end) to keep track of indices, so we can quickly find the closest unmarked index `j`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Stack, Simulation.

Companies

Asked at: carwale.

Find Mirror Score of a String | LeetCode 3412

Medium

Hash Table String Stack Simulation

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public long calculateScore(String s) {
        Map<Character, List<Integer>> positions = new HashMap<>();
        long score = 0;

        for (int i = 0; i < s.length(); i++) {
            char charAtI = s.charAt(i);
            char mirror_char = (char) ('a' + 'z' - charAtI);

            if (positions.containsKey(mirror_char) && !positions.get(mirror_char).isEmpty()) {
                int j = positions.get(mirror_char).remove(positions.get(mirror_char).size() - 1);
                score += i - j;
            } else {
                positions.computeIfAbsent(charAtI, k -> new ArrayList<>()).add(i);
            }
        }

        return score;
    }
}

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Problem Info

Description

Problem 3412: Find Mirror Score of a String - Detailed explanation and solution approaches.

Topics

Hash Table String Stack Simulation

Companies

carwale

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3412

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 a hash map `positions` to store the indices of each character and a variable `score` to 0.
2
Step 2: Iterate through the string `s` from left to right (index `i`).
3
Step 3: Calculate the mirror character of the current character `s[i]`.
4
Step 4: Check if the mirror character exists in the `positions` hash map and if its list of indices is not empty.
5
Step 5: If the mirror character exists and has available indices, retrieve the last (closest) index `j` from the list, remove it (mark it as used), and add `i - j` to the `score`.
6
Step 6: If the mirror character does not exist or has no available indices, add the current character and its index `i` to the `positions` hash map.
7
Step 7: After iterating through the entire string, return the final `score`.

💡

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 determined by the single loop that iterates through the string `s`. The space complexity is determined by the size of the `positions` hash map, which can grow up to the length of the string in the worst-case scenario where all characters are different.

📖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

Two Pointers

While the hashmap-based approach is efficient, a two-pointer approach could be considered. However, directly implementing a two-pointer solution to find *closest* matches would be complex since the closest match must be found, and elements can be marked. It's less suitable than using a hashmap to store indices.

Time: O(n^2) in the worst caseSpace: O(1)

Trade-offs: This approach would avoid the extra space required for the hash map, but the quadratic time complexity makes it much less efficient for longer strings. The hashmap-based solution offers a good balance.

✨

Key Insights

1Insight 1: Use a hash map to store the indices of each character to efficiently find potential mirror pairs.
2Insight 2: Iterate through the string and, for each character, check if the mirror character exists in the hash map and if there are any unmarked indices available.
3Insight 3: Use a stack-like structure (in this case, a list where we pop from the end) to keep track of indices, so we can quickly find the closest unmarked index `j`.

📚

Key Terms Explained

For Students

Hash Table

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

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: Empty string. The code handles this implicitly because the loop won't execute, and the score will remain 0.
Edge case 2: String with no mirror pairs. The code correctly skips adding to the score when no mirror pairs are found, resulting in a score of 0.
Edge case 3: String with duplicate characters. The `positions` hash map correctly stores all indices of duplicate characters, ensuring that the closest mirror is found when it becomes available.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the order of indices when multiple mirror characters exist. The solution needs to select the closest (last) available index.

✗Common mistake 2: Failing to mark an index as used after it has been paired with its mirror, which could lead to the same index being used multiple times.

✗Common mistake 3: Incorrectly calculating the mirror character. Ensuring that the formula `mirror_char = 'a' + 'z' - char_val` is correct is crucial.

✅

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: Analyzing palindromic sequences in DNA. Finding complementary base pairs (A-T, C-G) is similar to finding mirror characters.

🏦Real-world scenario 2: Text processing and pattern matching where finding reversed or mirrored patterns is required.

💡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: November 20, 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 "Find Mirror Score of a String"?

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.

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