Medium
You are given an integer num
. You will apply the following steps exactly two times:
x (0 <= x <= 9)
.y (0 <= y <= 9)
. The digit y
can be equal to x
.x
in the decimal representation of num
by y
.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
Constraints:
1 <= num <= 10
8import java.util.ArrayDeque;
import java.util.Deque;
public class Solution {
public int maxDiff(int num) {
Deque<Integer> stack = new ArrayDeque<>();
int xMax = 9;
int yMax = 9;
int xMin = 0;
int yMin = 0;
int min = 0;
int max = 0;
boolean areDigitsUnique = true;
while (num != 0) {
if (!stack.isEmpty() && num % 10 != stack.peek()) {
areDigitsUnique = false;
}
stack.push(num % 10);
num /= 10;
if (stack.peek() != 9) {
xMax = stack.peek();
}
if (stack.peek() > 1) {
xMin = stack.peek();
}
}
if (areDigitsUnique || stack.peek() == xMin) {
// Handles no leading zeros/non zero constraints.
yMin = 1;
}
while (!stack.isEmpty()) {
min = min * 10 + (stack.peek() == xMin ? yMin : stack.peek());
max = max * 10 + (stack.peek() == xMax ? yMax : stack.peek());
stack.pop();
}
return max - min;
}
}