LeetCode in Kotlin
3880. Minimum Absolute Difference Between Two Values
Easy
You are given an integer array nums consisting only of 0, 1, and 2.
A pair of indices (i, j) is called valid if nums[i] == 1 and nums[j] == 2.
Return the minimum absolute difference between i and j among all valid pairs. If no valid pair exists, return -1.
The absolute difference between indices i and j is defined as abs(i - j).
Example 1:
Input: nums = [1,0,0,2,0,1]
Output: 2
Explanation:
The valid pairs are:
- (0, 3) which has absolute difference of
abs(0 - 3) = 3. - (5, 3) which has absolute difference of
abs(5 - 3) = 2.
Thus, the answer is 2.
Example 2:
Input: nums = [1,0,1,0]
Output: -1
Explanation:
There are no valid pairs in the array, thus the answer is -1.
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 2
Solution
import kotlin.math.min
class Solution {
fun minAbsoluteDifference(nums: IntArray): Int {
var min = Int.MAX_VALUE
val n = nums.size
var prev = -1
var last = -1
for (i in 0..<n) {
if (prev == -1) {
if (nums[i] == 1) {
prev = 1
last = i
} else if (nums[i] == 2) {
prev = 2
last = i
}
} else {
if (nums[i] == 1) {
if (prev == 2) {
min = min(min, i - last)
prev = 1
}
last = i
} else if (nums[i] == 2) {
if (prev == 1) {
min = min(min, i - last)
prev = 2
}
last = i
}
}
}
return if (min != Int.MAX_VALUE) min else -1
}
}