What is the time complexity of Largest Number After Mutating Substring?

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

What data structures are used to solve Largest Number After Mutating Substring?

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

What are the key insights for solving Largest Number After Mutating Substring?

Insight 1: The problem can be solved using a greedy approach. We iterate through the number from left to right and start mutating the substring as soon as we find a digit that can be increased by applying the change array. Insight 2: Once we start mutating, we continue as long as the mutated digit is greater than or equal to the original digit. As soon as the mutated digit becomes smaller, we stop the mutation.

Which companies ask LeetCode 1946. Largest Number After Mutating Substring in interviews?

This problem is frequently asked at Infosys.

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Largest Number After Mutating Substring - LeetCode 1946 Solution

Largest Number After Mutating Substring - Complete Solution Guide

Largest Number After Mutating Substring is LeetCode problem 1946, 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 num , which represents a large integer. You are also given a 0-indexed integer array change of length 10 that maps each digit 0-9 to another digit. More formally, digit d maps to digit change[d] . You may choose to mutate a single substring of num . To mutate a substring, replace each digit num[i] with the digit it maps to in change (i.e. replace num[i] with change[num[i]] ). Return a string representing the largest possible integer after mutating (or choosing not to) a si

Detailed Explanation

The problem asks us to find the largest possible number that can be obtained by mutating at most one substring of a given number `num`. We are given a `change` array that maps each digit from 0 to 9 to another digit. To mutate a substring, we replace each digit in the substring with the digit it maps to in the `change` array. If no mutation results in a larger number, we can return the original number. The input `num` is a string representing a large integer, and `change` is an array of integers representing the mapping of digits.

Solution Approach

The solution iterates through the input number string. For each digit, it checks if the changed value (using the `change` array) is greater than the original digit. If it is, it starts a mutation. During the mutation, it replaces the original digit with the changed one. It continues the mutation as long as the changed digit is greater than or equal to the original. Once the changed digit is less than the original, the mutation stops. If no mutation occurs, the original string is returned.

Step-by-Step Algorithm

Step 1: Convert the input number string into a list/array of characters for easier manipulation.
Step 2: Initialize a boolean variable `mutated` to `False` to track if a mutation has started.
Step 3: Iterate through the list/array of characters representing the number.
Step 4: For each digit, find its corresponding changed value using the `change` array.
Step 5: If the changed value is greater than the original digit, start/continue the mutation by replacing the original digit with the changed value and set `mutated` to `True`.
Step 6: If the changed value is equal to the original digit, continue the mutation by replacing the original digit with the changed value, but do not start a new mutation.
Step 7: If the changed value is smaller than the original digit and `mutated` is `True`, stop the mutation.
Step 8: If `mutated` is `True` and the changed value is smaller, break the loop. If `mutated` is `False` and the changed value is smaller, continue the iteration.
Step 9: After the iteration, convert the list/array of characters back into a string and return it.

Key Insights

Insight 1: The problem can be solved using a greedy approach. We iterate through the number from left to right and start mutating the substring as soon as we find a digit that can be increased by applying the change array.
Insight 2: Once we start mutating, we continue as long as the mutated digit is greater than or equal to the original digit. As soon as the mutated digit becomes smaller, we stop the mutation.
Insight 3: There may be no profitable mutation, in which case we return the original string.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Greedy.

Companies

Asked at: Infosys.

Largest Number After Mutating Substring | LeetCode 1946

Medium

Array String Greedy

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

Solution Code

Available in:

class Solution {
    public String maximumNumber(String num, int[] change) {
        char[] numList = num.toCharArray();
        boolean mutated = false;

        for (int i = 0; i < numList.length; i++) {
            int digit = Character.getNumericValue(numList[i]);

            if (change[digit] > digit) {
                numList[i] = Character.forDigit(change[digit], 10);
                mutated = true;
            } else if (change[digit] < digit) {
                if (mutated) {
                    break;
                }
            }
        }

        return new String(numList);
    }
}

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Problem Info

Description

Problem 1946: Largest Number After Mutating Substring - Detailed explanation and solution approaches.

Topics

Array String Greedy

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1946

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: Convert the input number string into a list/array of characters for easier manipulation.
2
Step 2: Initialize a boolean variable `mutated` to `False` to track if a mutation has started.
3
Step 3: Iterate through the list/array of characters representing the number.
4
Step 4: For each digit, find its corresponding changed value using the `change` array.
5
Step 5: If the changed value is greater than the original digit, start/continue the mutation by replacing the original digit with the changed value and set `mutated` to `True`.
6
Step 6: If the changed value is equal to the original digit, continue the mutation by replacing the original digit with the changed value, but do not start a new mutation.
7
Step 7: If the changed value is smaller than the original digit and `mutated` is `True`, stop the mutation.
8
Step 8: If `mutated` is `True` and the changed value is smaller, break the loop. If `mutated` is `False` and the changed value is smaller, continue the iteration.
9
Step 9: After the iteration, convert the list/array of characters back into a string and return it.

💡

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 traverse the input string at most once. The space complexity is linear because in each of the provided solutions (python, java, cpp, c), a new array/list is constructed from the original string to hold the modified digits.

📖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 all possible substrings and mutate them. Compare the resulting numbers and return the largest.

Time: O(n^3), where n is the length of the number string. O(n^2) for generating all substrings, and O(n) for mutating and comparing each substring.Space: O(n).

Trade-offs: This approach is much less efficient than the greedy approach and is not suitable for large input strings. Use the greedy approach instead.

✨

Key Insights

1Insight 1: The problem can be solved using a greedy approach. We iterate through the number from left to right and start mutating the substring as soon as we find a digit that can be increased by applying the change array.
2Insight 2: Once we start mutating, we continue as long as the mutated digit is greater than or equal to the original digit. As soon as the mutated digit becomes smaller, we stop the mutation.
3Insight 3: There may be no profitable mutation, in which case we return the original string.

📚

Key Terms Explained

For Students

Array

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

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: When the change array results in no mutations. The algorithm correctly returns the original number string.
Edge case 2: When the number string consists of a single digit. The algorithm correctly handles this case by checking if the mutation results in a larger number and mutating only if necessary.
Edge case 3: When the change array maps every digit to the same digit or a smaller digit, in which case the algorithm returns the original number.
Edge case 4: A large string, the algorithm handles it linearly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not stopping the mutation when the mutated digit becomes smaller than the original digit.

✗Common mistake 2: Trying to mutate multiple substrings instead of just one.

✗Common mistake 3: Incorrectly converting between string and integer representations of digits.

✗Common mistake 4: Forgetting to consider the case where no mutation is necessary.

✅

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: Optimizing financial data representation. For example, replacing certain digits in a stock price to represent a more favorable scenario within defined constraints.

🏦Real-world scenario 2: Image processing: Applying transformations on pixel values within a specific region (substring) of an image to enhance or modify the image.

🎮Real-world scenario 3: Data masking or sanitization: Replacing parts of sensitive data (like credit card numbers or social security numbers) while trying to maximize the numerical value within allowed changes.

💡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: October 25, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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 "Largest Number After Mutating Substring"?

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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