LeetCode in Kotlin
2170. Minimum Operations to Make the Array Alternating
Medium
You are given a 0-indexed array nums consisting of n positive integers.
The array nums is called alternating if:
nums[i - 2] == nums[i], where2 <= i <= n - 1.nums[i - 1] != nums[i], where1 <= i <= n - 1.
In one operation, you can choose an index i and change nums[i] into any positive integer.
Return the minimum number of operations required to make the array alternating.
Example 1:
Input: nums = [3,1,3,2,4,3]
Output: 3
Explanation:
One way to make the array alternating is by converting it to [3,1,3,1,3,1].
The number of operations required in this case is 3.
It can be proven that it is not possible to make the array alternating in less than 3 operations.
Example 2:
Input: nums = [1,2,2,2,2]
Output: 2
Explanation:
One way to make the array alternating is by converting it to [1,2,1,2,1].
The number of operations required in this case is 2.
Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105
Solution
class Solution {
fun minimumOperations(nums: IntArray): Int {
var maxOdd = 0
var maxEven = 0
var max = 0
val n = nums.size
for (num in nums) {
max = Math.max(max, num)
}
val even = IntArray(max + 1)
val odd = IntArray(max + 1)
for (i in 0 until n) {
if (i % 2 == 0) {
even[nums[i]]++
} else {
odd[nums[i]]++
}
}
var t1 = 0
var t2 = 0
for (i in 0 until max + 1) {
if (even[i] > maxEven) {
maxEven = even[i]
t1 = i
}
if (odd[i] > maxOdd) {
maxOdd = odd[i]
t2 = i
}
}
val ans: Int
if (t1 == t2) {
var secondEven = 0
var secondOdd = 0
for (i in 0 until max + 1) {
if (i != t1 && even[i] > secondEven) {
secondEven = even[i]
}
if (i != t2 && odd[i] > secondOdd) {
secondOdd = odd[i]
}
}
ans = Math.min(
n / 2 + n % 2 - maxEven + (n / 2 - secondOdd),
n / 2 + n % 2 - secondEven + (n / 2 - maxOdd)
)
} else {
ans = n / 2 + n % 2 - maxEven + n / 2 - maxOdd
}
return ans
}
}