LeetCode in Kotlin

2369. Check if There is a Valid Partition For The Array

Medium

You are given a 0-indexed integer array nums. You have to partition the array into one or more contiguous subarrays.

We call a partition of the array valid if each of the obtained subarrays satisfies one of the following conditions:

1. The subarray consists of exactly 2 equal elements. For example, the subarray [2,2] is good.
2. The subarray consists of exactly 3 equal elements. For example, the subarray [4,4,4] is good.
3. The subarray consists of exactly 3 consecutive increasing elements, that is, the difference between adjacent elements is 1. For example, the subarray [3,4,5] is good, but the subarray [1,3,5] is not.

Return true if the array has at least one valid partition. Otherwise, return false.

Example 1:

Input: nums = [4,4,4,5,6]

Output: true

Explanation: The array can be partitioned into the subarrays [4,4] and [4,5,6]. This partition is valid, so we return true.

Example 2:

Input: nums = [1,1,1,2]

Output: false

Explanation: There is no valid partition for this array.

Constraints:

• 2 <= nums.length <= 105
• 1 <= nums[i] <= 106

Solution

class Solution {
    fun validPartition(nums: IntArray): Boolean {
        val canPartition = BooleanArray(nums.size + 1)
        canPartition[0] = true
        var diff = nums[1] - nums[0]
        var equal = diff == 0
        var incOne = diff == 1
        canPartition[2] = equal
        for (i in 3 until canPartition.size) {
            diff = nums[i - 1] - nums[i - 2]
            if (diff == 0) {
                canPartition[i] = canPartition[i - 2] || equal && canPartition[i - 3]
                equal = true
                incOne = false
            } else if (diff == 1) {
                canPartition[i] = incOne && canPartition[i - 3]
                equal = false
                incOne = true
            }
        }
        return canPartition[nums.size]
    }
}

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