What is the time complexity of Custom Sort String?

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

What data structures are used to solve Custom Sort String?

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

What are the key insights for solving Custom Sort String?

Insight 1: The key is to respect the custom order specified in the `order` string. We can iterate through `order` and extract/append the respective characters from `s` based on their counts. Insight 2: A hash table (or similar data structure like a counter) is useful to efficiently count the occurrences of each character in `s`.

Which companies ask LeetCode 791. Custom Sort String in interviews?

This problem is frequently asked at Snap.

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Custom Sort String - LeetCode 791 Solution

Custom Sort String - Complete Solution Guide

Custom Sort String is LeetCode problem 791, 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 two strings order and s . All the characters of order are unique and were sorted in some custom order previously. Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order , then x should occur before y in the permuted string. Return any permutation of s that satisfies this property . Example 1: Input: order = "cba", s = "abcd" Output: "cbad" Explanation: "a" , "b" , "c" appear in order,

Detailed Explanation

The problem asks us to rearrange the characters of string `s` according to the order specified by string `order`. The string `order` contains unique characters, defining a custom order. We need to permute `s` such that if a character `x` appears before `y` in `order`, then `x` must also appear before `y` in the rearranged `s`. Characters in `s` that are *not* present in `order` can be placed anywhere in the result. The goal is to return *any* valid permutation of `s` adhering to this rule.

Solution Approach

The solution uses a counting approach. First, it counts the occurrences of each character in string `s` using a hash map or a counter (depending on the language). Then, it iterates through the characters in `order`. For each character in `order`, if it exists in the character counts of `s`, it appends that character to the result string as many times as it appears in `s`, updating the count accordingly (or removing it from the count). Finally, it iterates through the remaining characters in `s` (those not in `order`) and appends them to the result string based on their counts.

Step-by-Step Algorithm

Step 1: Create a hash table `s_count` to store the character counts from the string `s`.
Step 2: Iterate through the string `s` and populate `s_count` with each character's frequency.
Step 3: Initialize an empty string `res` to store the result.
Step 4: Iterate through the string `order`. For each character `char` in `order`:
Step 5: Check if `char` exists as a key in `s_count`.
Step 6: If `char` exists in `s_count`, append `char` to `res` `s_count[char]` times.
Step 7: Remove `char` from `s_count` to avoid processing it again.
Step 8: Iterate through the remaining entries in `s_count`. For each character `char` and its count in `s_count`:
Step 9: Append `char` to `res` the number of times specified by its count.
Step 10: Return the resulting string `res`.

Key Insights

Insight 1: The key is to respect the custom order specified in the `order` string. We can iterate through `order` and extract/append the respective characters from `s` based on their counts.
Insight 2: A hash table (or similar data structure like a counter) is useful to efficiently count the occurrences of each character in `s`.
Insight 3: After processing characters present in `order`, we need to append the remaining characters from `s` (those not present in `order`) to the result. The order of these remaining characters doesn't matter.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Sorting.

Companies

Asked at: Snap.

Custom Sort String | LeetCode 791

Medium

Hash Table String Sorting

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public String customSortString(String order, String s) {
        Map<Character, Integer> s_count = new HashMap<>();
        for (char c : s.toCharArray()) {
            s_count.put(c, s_count.getOrDefault(c, 0) + 1);
        }
        
        StringBuilder res = new StringBuilder();
        
        for (char char_ : order.toCharArray()) {
            if (s_count.containsKey(char_)) {
                for (int i = 0; i < s_count.get(char_); i++) {
                    res.append(char_);
                }
                s_count.remove(char_);
            }
        }
        
        for (Map.Entry<Character, Integer> entry : s_count.entrySet()) {
            char char_ = entry.getKey();
            int count = entry.getValue();
            for (int i = 0; i < count; i++) {
                res.append(char_);
            }
        }
        
        return res.toString();
    }
}

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Problem Info

Description

Problem 791: Custom Sort String - Detailed explanation and solution approaches.

Topics

Hash Table String Sorting

Companies

Snap

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:791

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

10 steps

1
Step 1: Create a hash table `s_count` to store the character counts from the string `s`.
2
Step 2: Iterate through the string `s` and populate `s_count` with each character's frequency.
3
Step 3: Initialize an empty string `res` to store the result.
4
Step 4: Iterate through the string `order`. For each character `char` in `order`:
5
Step 5: Check if `char` exists as a key in `s_count`.
6
Step 6: If `char` exists in `s_count`, append `char` to `res` `s_count[char]` times.
7
Step 7: Remove `char` from `s_count` to avoid processing it again.
8
Step 8: Iterate through the remaining entries in `s_count`. For each character `char` and its count in `s_count`:
9
Step 9: Append `char` to `res` the number of times specified by its count.
10
Step 10: Return the resulting string `res`.

💡

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 iterations through `order` and `s`. Building the character counts takes O(m) time. Iterating through `order` takes O(n) time. The final iteration through the hash map (or the remaining characters) takes at most O(m) time. The space complexity is determined by the size of the `s_count` hash table. In the worst case, all characters in `s` are unique, so the hash table will store at most the number of unique possible characters, which for lowercase English letters is a constant.

📖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

Sorting with Custom Comparator

Create a custom comparator that prioritizes characters present in `order` based on their index in `order`. Sort `s` using this comparator. Characters not present in `order` can be assigned a default large index, effectively placing them at the end.

Time: O(m log m), where m is the length of `s` (due to sorting)Space: O(m) or O(log m) depending on the sorting algorithm's space complexity.

Trade-offs: This approach is conceptually simpler to some, but the sorting overhead generally makes it less efficient than the hash table approach, especially when the length of `order` is much smaller than `s`. However, if one needs to sort according to multiple criteria simultaneously, a custom comparator approach might be easier to adapt.

✨

Key Insights

1Insight 1: The key is to respect the custom order specified in the `order` string. We can iterate through `order` and extract/append the respective characters from `s` based on their counts.
2Insight 2: A hash table (or similar data structure like a counter) is useful to efficiently count the occurrences of each character in `s`.
3Insight 3: After processing characters present in `order`, we need to append the remaining characters from `s` (those not present in `order`) to the result. The order of these remaining characters doesn't matter.

📚

Key Terms Explained

For Students

Hash Table

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

String

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `order` is empty. The solution will skip the main loop and directly append all characters from `s` to the result.
Edge case 2: `s` is empty. The solution will return an empty string.
Edge case 3: `order` and `s` have no characters in common. The solution will append all characters of `s` without any particular order change.
Edge case 4: `s` contains characters not present in `order`. The algorithm correctly handles those cases by putting all remaining characters at the end with no ordering constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle characters in `s` that are not present in `order`. These characters must still be included in the final result.

✗Common mistake 2: Modifying the `s` string directly instead of creating a new string. This can lead to incorrect results and is generally not recommended.

✗Common mistake 3: Assuming that the `order` string contains all the characters that appear in `s`. The problem statement clearly states that the `order` string has unique characters, but not necessarily all the characters in the input string `s`.

✗Common mistake 4: Not allocating enough space for the resulting string in C. Forgetting to null terminate the string in C, leading to buffer overflow.

✅

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: Configuration file parsing where the order of fields is preferred but not strictly enforced. The `order` string could represent preferred field order, while `s` represents the actual fields present in a configuration.

🏦Real-world scenario 2: Customizing the order of columns in a spreadsheet or report. You have a preferred order of columns, but some spreadsheets may contain only a subset of those columns.

🎮Real-world scenario 3: Prioritizing search results. The 'order' string could be a user's profile of preferred topics, and search results (string 's') could be sorted to surface results most related to the user's profile.

💡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

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

⚠️ 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: August 11, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Custom Sort 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.

🤖 Need help with this problem?