How do you solve Maximum Value after Insertion on LeetCode?

Maximum Value after Insertion (LeetCode #1881) 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: Inserting 'x' before a smaller digit maximizes the value for positive numbers.

What is the brute force solution for Maximum Value after Insertion?

The brute force approach for Maximum Value after Insertion is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to insert the digit 'x' at every possible position in the string representation of 'n' and checking which resulting number is the largest. This is simple but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Value after Insertion?

Maximum Value after Insertion uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, String Manipulation.

What are the interview tips for Maximum Value after Insertion?

Always clarify how to handle negative numbers before coding. Think about edge cases, such as inserting at the beginning or end. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Maximum Value after Insertion?

Not considering the negative sign when inserting for negative numbers. Failing to check all possible insertion points efficiently.

#1881

Maximum Value after Insertion

Medium

StringGreedyGreedyString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we can directly determine where to insert 'x' by iterating through the string once. We check if inserting 'x' before a digit will yield a larger number, and we stop as soon as we find the correct position.

⚙️

Algorithm

4 steps

1Step 1: Check if the number is negative. If so, handle it differently by looking for the first digit greater than 'x'.
2Step 2: Loop through the digits of 'n'.
3Step 3: If 'x' is greater than the current digit, insert 'x' before it and break the loop.
4Step 4: If no insertion has been made, append 'x' at the end.

solution.py13 lines

1def insert_max(n, x):
2    if n[0] == '-':
3        for i in range(1, len(n)):
4            if x < n[i]:
5                return n[:i] + str(x) + n[i:]
6        return n + str(x)
7    else:
8        for i in range(len(n)):
9            if x > n[i]:
10                return n[:i] + str(x) + n[i:]
11        return n + str(x)
12
13print(insert_max("99", 9))

ℹ

Complexity note: This complexity is linear because we only make a single pass through the string to find the correct insertion point, and the space complexity is O(n) due to the new string created during insertion.

1Inserting 'x' before a smaller digit maximizes the value for positive numbers.
2For negative numbers, inserting 'x' before a larger digit minimizes the value, which is optimal.

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