Leetcode #1432: Max Difference You Can Get From Changing an Integer
In this guide, we solve Leetcode #1432 Max Difference You Can Get From Changing an Integer in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer num. You will apply the following steps to num two separate times: Pick a digit x (0 <= x <= 9).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Math
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: num = 555
Output: 888
Explanation: The first time pick x = 5 and y = 9 and store the new integer in a.
The second time pick x = 5 and y = 1 and store the new integer in b.
We have now a = 999 and b = 111 and max difference = 888
Python Solution
class Solution:
def maxDiff(self, num: int) -> int:
a, b = str(num), str(num)
for c in a:
if c != "9":
a = a.replace(c, "9")
break
if b[0] != "1":
b = b.replace(b[0], "1")
else:
for c in b[1:]:
if c not in "01":
b = b.replace(c, "0")
break
return int(a) - int(b)
Complexity
The time complexity is , and the space complexity is , where is the given integer. The space complexity is , where is the given integer.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.