How do you solve Largest Number After Mutating Substring on LeetCode?

Largest Number After Mutating Substring (LeetCode #1946) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Changing digits only when the new digit is larger is crucial for maximizing the number.

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

The optimal solution for Largest Number After Mutating Substring runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(n) due to the result string that we build.

What is the brute force solution for Largest Number After Mutating Substring?

The brute force approach for Largest Number After Mutating Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible substring of the given number and mutating it according to the change mapping. We then compare all mutated results to find the largest possible number. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Largest Number After Mutating Substring?

Largest Number After Mutating Substring uses the following concepts: Array, String, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Largest Number After Mutating Substring?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as when no digits can be improved. Discuss your thought process and approach with the interviewer to show your understanding.

What are common mistakes when solving Largest Number After Mutating Substring?

Failing to stop mutating when a smaller digit is encountered after starting. Not considering the case where no mutation is needed at all.

#1946

Largest Number After Mutating Substring

Medium

ArrayStringGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach involves iterating through the string once while deciding when to start and stop mutating the substring. We only mutate when the new digit is greater than the original, and we stop when we encounter a digit that is not beneficial to change.

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Algorithm

4 steps

1Step 1: Initialize a result string to build the final number.
2Step 2: Iterate through each digit of the num string.
3Step 3: If the current digit can be changed to a larger digit, start mutating until the digit is smaller than the original or the end of the string is reached.
4Step 4: Append either the mutated digit or the original digit to the result.

solution.py15 lines

1# Full working Python code
2
3def largestNumberAfterMutatingSubstring(num: str, change: list) -> str:
4    result = []
5    mutating = False
6    for digit in num:
7        d = int(digit)
8        if change[d] > d:
9            mutating = True
10            result.append(str(change[d]))
11        elif change[d] < d and mutating:
12            break
13        else:
14            result.append(digit)
15    return ''.join(result) + num[len(result):]

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(n) due to the result string that we build.

1Changing digits only when the new digit is larger is crucial for maximizing the number.
2Once we start mutating, we should stop as soon as we encounter a digit that cannot be improved.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.