LeetCode in Kotlin

3101. Count Alternating Subarrays

Medium

You are given a binary array nums.

We call a subarray alternating if no two adjacent elements in the subarray have the same value.

Return the number of alternating subarrays in nums.

Example 1:

Input: nums = [0,1,1,1]

Output: 5

Explanation:

The following subarrays are alternating: [0], [1], [1], [1], and [0,1].

Example 2:

Input: nums = [1,0,1,0]

Output: 10

Explanation:

Every subarray of the array is alternating. There are 10 possible subarrays that we can choose.

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Solution

class Solution {
    fun countAlternatingSubarrays(nums: IntArray): Long {
        var count: Long = 0
        var length: Long
        var start = 0
        var end = 1
        while (end < nums.size) {
            if (nums[end] != nums[end - 1]) {
                end++
            } else {
                length = end - start.toLong()
                count += (length * (length + 1)) / 2
                start = end
                end++
            }
        }
        length = end - start.toLong()
        count += (length * (length + 1)) / 2
        return count
    }
}

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