1529. Max Difference You Can Get From Changing An Integer¶

Difficulty: Medium

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1529. Max Difference You Can Get From Changing an Integer

Medium

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

Pick a digit x (0 <= x <= 9).
Pick another digit y (0 <= y <= 9). Note y can be equal to x.
Replace all the occurrences of x in the decimal representation of num by y.

Let a and b be the two results from applying the operation to num independently.

Return the max difference between a and b.

Note that neither a nor b may have any leading zeros, and must not be 0.

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

Constraints:

1 <= num <= 10⁸

Solution¶

class Solution {
    public int maxDiff(int num) {
        String  s = Integer.toString(num);
        int maxi = 0, mini = 0;
        int maxi_ind = -1, mini_ind = -1;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) != '9') {
                maxi_ind = i;
                break;
            }
        }
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (maxi_ind == -1)
                maxi = maxi * 10 + (ch - '0');
            else if (ch == s.charAt(maxi_ind))
                maxi = maxi * 10 + 9;
            else
                maxi = maxi * 10 + ch - '0';
        }
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) != '1' && s.charAt(i) != '0') {
                mini_ind = i;
                break;
            }
        }
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (mini_ind == -1)
                mini = mini * 10 + (ch - '0');
            else if (ch == s.charAt(mini_ind) && mini_ind == 0)
                mini = mini * 10 + 1;
            else if (ch == s.charAt(mini_ind) && mini_ind != 0)
                mini = mini * 10 + 0;
            else
                mini = mini * 10 + ch - '0';
        }
        return maxi - mini;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here