Leetcode 2566: Maximum Difference by Remapping a Digit
You are given an integer num, and you know that Bob will sneakily remap one of the 10 possible digits (0 to 9) to another digit. Your task is to return the difference between the maximum and minimum values Bob can make by remapping exactly one digit in num. If no change is made, the value of num remains the same.
Problem
Approach
Steps
Complexity
Input: You are given a single integer num, which can be as large as 10^8.
Example: For example, num = 13579.
Constraints:
• 1 <= num <= 10^8
Output: Return the difference between the maximum and minimum possible values that Bob can obtain after remapping one digit.
Example: For example, the output for num = 13579 is 99850.
Constraints:
• The number should be in its integer form and the difference between max and min should be returned.
Goal: To find the largest possible difference between the maximum and minimum values obtained after remapping a digit.
Steps:
• 1. For each digit in num, calculate the value after remapping it to every other possible digit (0 to 9).
• 2. Track the minimum and maximum values obtained.
• 3. Return the difference between the maximum and minimum values.
Goal: num will always be a valid integer and will lie between 1 and 10^8 inclusive.
Steps:
• 1 <= num <= 10^8
Assumptions:
• The input num will always be an integer within the specified range.
• Input: For num = 13579, the largest difference comes from remapping the digit 1 to 9 for maximum value and 9 to 0 for minimum value.
• Explanation: This leads to the maximum value of 99579 and minimum value of 13570, resulting in a difference of 99850.
• Input: For num = 81, no remapping gives a higher or lower value than 81, so the difference is 0.
• Explanation: The result is 81 - 81 = 0.
Approach: To solve this problem, we will iterate through each digit and simulate all possible remapping operations.
Observations:
• We need to compute both the maximum and minimum numbers Bob can create by changing a single digit.
• By testing all digits for remapping and calculating the resulting number, we can identify the max-min difference.
Steps:
• 1. Convert the integer num into a string to facilitate digit access.
• 2. For each digit, test remapping it to every other digit and calculate the new number.
• 3. Track the maximum and minimum numbers during this process.
• 4. Return the difference between the maximum and minimum values.
Empty Inputs:
• There are no empty inputs as the input is always a valid integer.
Large Inputs:
• When num is a large integer (up to 10^8), the solution needs to handle the digits efficiently.
Special Values:
• If num has all identical digits, remapping any digit to another won't change the number.
Constraints:
• The input num will always be a valid positive integer.
int minMaxDifference(int num) {
int min_v = num, max_v = num;
for (int i = 0; i < 10; ++i) {
int lo = 0, hi = 0, mul = 1;
for (int n = num; n; n /= 10) {
bool replace = n % 10 == i;
lo += (replace ? 0 : n % 10) * mul;
hi += (replace ? 9 : n % 10) * mul;
mul *= 10;
}
min_v = min(min_v, lo);
max_v = max(max_v, hi);
}
return max_v - min_v;
}
1 : Function Definition
int minMaxDifference(int num) {
Defines the function `minMaxDifference` which takes an integer `num` as input.
2 : Variable Initialization
int min_v = num, max_v = num;
Initializes two variables `min_v` and `max_v` to the value of `num`, which will track the minimum and maximum values during the calculations.
3 : Looping
for (int i = 0; i < 10; ++i) {
A loop that iterates over each digit (0 to 9) to simulate replacing digits of `num` with each possible value.
4 : Variable Initialization
int lo = 0, hi = 0, mul = 1;
Initializes variables `lo` and `hi` to track the lowest and highest possible numbers, respectively, and `mul` to control the place value during digit manipulation.
5 : Looping and Modulo Operation
for (int n = num; n; n /= 10) {
A nested loop that processes each digit of `num` from right to left, dividing by 10 after each iteration to isolate each digit.
6 : Conditional Statement
bool replace = n % 10 == i;
Checks if the current digit (`n % 10`) matches the digit `i` being considered for replacement in the outer loop.
7 : Mathematical Operations
lo += (replace ? 0 : n % 10) * mul;
Calculates the lower number by replacing the matching digit with 0, and keeps the other digits unchanged by adding them to `lo`.
8 : Mathematical Operations
hi += (replace ? 9 : n % 10) * mul;
Calculates the higher number by replacing the matching digit with 9, and keeps the other digits unchanged by adding them to `hi`.
9 : Mathematical Operations
mul *= 10;
Multiplies `mul` by 10 to move to the next higher place value for the next digit.
10 : Mathematical Operations
min_v = min(min_v, lo);
Updates `min_v` to store the smallest possible value between the current `min_v` and the new `lo` value.
11 : Mathematical Operations
max_v = max(max_v, hi);
Updates `max_v` to store the largest possible value between the current `max_v` and the new `hi` value.
12 : Return Statement
return max_v - min_v;
Returns the difference between the maximum and minimum values calculated during the loops.
Best Case: O(d), where d is the number of digits in num.
Average Case: O(d * 10), where d is the number of digits in num.
Worst Case: O(d * 10), where d is the number of digits in num.
Description: For each digit, we try remapping it to all other digits, so the time complexity is linear with respect to the number of digits.
Best Case: O(d), where d is the number of digits in num.
Worst Case: O(d), where d is the number of digits in num.
Description: We need space to store the digits of num during the remapping process.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus