LeetCode in Kotlin
3982. Sum of Integers with Maximum Digit Range
Easy
You are given an integer array nums.
The digit range of an integer is defined as the difference between its largest digit and smallest digit.
For example, the digit range of 5724 is 7 - 2 = 5.
Return the sum of all integers in nums whose digit range is equal to the maximum digit range among all integers in the array.
Example 1:
Input: nums = [5724,111,350]
Output: 6074
Explanation:
i |
nums[i] |
Largest | Smallest | Digit Range |
|---|---|---|---|---|
| 0 | 5724 | 7 | 2 | 5 |
| 1 | 111 | 1 | 1 | 0 |
| 2 | 350 | 5 | 0 | 5 |
The maximum digit range is 5. The integers with this digit range are 5724 and 350, so the answer is 5724 + 350 = 6074.
Example 2:
Input: nums = [90,900]
Output: 990
Explanation:
i |
nums[i] |
Largest | Smallest | Digit Range |
|---|---|---|---|---|
| 0 | 90 | 9 | 0 | 9 |
| 1 | 900 | 9 | 0 | 9 |
The maximum digit range is 9. Both integers have this digit range, so the answer is 90 + 900 = 990.
Constraints:
1 <= nums.length <= 10010 <= nums[i] <= 105
Solution
import kotlin.math.max
import kotlin.math.min
class Solution {
fun maxDigitRange(nums: IntArray): Int {
val n = nums.size
val range = nums.clone()
var maxRange = Int.MIN_VALUE
for (i in 0..<n) {
var x = range[i]
var mn = Int.MAX_VALUE
var mx = Int.MIN_VALUE
while (x > 0) {
val d = x % 10
mn = min(mn, d)
mx = max(mx, d)
x /= 10
}
range[i] = mx - mn
maxRange = max(maxRange, range[i])
}
var ans = 0
for (i in 0..<n) {
if (range[i] == maxRange) {
ans += nums[i]
}
}
return ans
}
}