LeetCode in Kotlin
3741. Minimum Distance Between Three Equal Elements II
Medium
You are given an integer array nums.
A tuple (i, j, k) of 3 distinct indices is good if nums[i] == nums[j] == nums[k].
The distance of a good tuple is abs(i - j) + abs(j - k) + abs(k - i), where abs(x) denotes the absolute value of x.
Return an integer denoting the minimum possible distance of a good tuple. If no good tuples exist, return -1.
Example 1:
Input: nums = [1,2,1,1,3]
Output: 6
Explanation:
The minimum distance is achieved by the good tuple (0, 2, 3).
(0, 2, 3) is a good tuple because nums[0] == nums[2] == nums[3] == 1. Its distance is abs(0 - 2) + abs(2 - 3) + abs(3 - 0) = 2 + 1 + 3 = 6.
Example 2:
Input: nums = [1,1,2,3,2,1,2]
Output: 8
Explanation:
The minimum distance is achieved by the good tuple (2, 4, 6).
(2, 4, 6) is a good tuple because nums[2] == nums[4] == nums[6] == 2. Its distance is abs(2 - 4) + abs(4 - 6) + abs(6 - 2) = 2 + 2 + 4 = 8.
Example 3:
Input: nums = [1]
Output: -1
Explanation:
There are no good tuples. Therefore, the answer is -1.
Constraints:
1 <= n == nums.length <= 1051 <= nums[i] <= n
Solution
import kotlin.math.min
class Solution {
fun minimumDistance(nums: IntArray): Int {
val n = nums.size
var ans = Int.MAX_VALUE
val prev1 = IntArray(n + 1)
val prev2 = IntArray(n + 1)
for (i in 0..<n + 1) {
prev2[i] = -1
prev1[i] = prev2[i]
}
for (i in 0..<n) {
val value = nums[i]
if (prev2[value] != -1) {
ans = min(ans, (i - prev2[value]))
}
prev2[value] = prev1[value]
prev1[value] = i
}
if (ans < 100002) {
return ans * 2
}
return -1
}
}