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花花酱 LeetCode 1432. Max Difference You Can Get From Changing an Integer

You are given an integer num. You will apply the following steps exactly two times:

  • Pick a digit x (0 <= x <= 9).
  • Pick another digit y (0 <= y <= 9). The digit y can be equal to x.
  • Replace all the occurrences of x in the decimal representation of num by y.
  • The new integer cannot have any leading zeros, also the new integer cannot be 0.

Let a and b be the results of applying the operations to num the first and second times, respectively.

Return the max difference between a and b.

Example 1:

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

Example 2:

Input: num = 9
Output: 8
Explanation: The first time pick x = 9 and y = 9 and store the new integer in a.
The second time pick x = 9 and y = 1 and store the new integer in b.
We have now a = 9 and b = 1 and max difference = 8

Example 3:

Input: num = 123456
Output: 820000

Example 4:

Input: num = 10000
Output: 80000

Example 5:

Input: num = 9288
Output: 8700

Constraints:

  • 1 <= num <= 10^8

Solution 1: Brute Force

Try all possible pairs of (x, y)

Time complexity: O(10*10*logn)
Space complexity: O(logn)

C++

Solution 2: Greedy

Maximize A and Minimize B

A – B will the answer.

To maximize A, find the left most digit that is not 9, replace that digit with 9.
e.g. 121 => 919
e.g. 9981 => 9991
To minimize B, fin the left most digit that is not 1 or 0, replace that digit with 0 (if not the first one) or 1.
e.g. 1024 -> 1004
e.g. 2024 -> 1014

Time complexity: O(logn)
Space complexity: O(logn)

C++

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