2762. Continuous Subarrays

Problem

You are given a 0-indexed integer array nums. A subarray of nums is called continuous if:

• Let i, i + 1, …, j be the indices in the subarray. Then, for each pair of indices i <= i1, i2 <= j, 0 <= |nums[i1] - nums[i2]| <= 2.

Return **the total number of *continuous* subarrays.**

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [5,4,2,4]
Output: 8
Explanation:
Continuous subarray of size 1: [5], [4], [2], [4].
Continuous subarray of size 2: [5,4], [4,2], [2,4].
Continuous subarray of size 3: [4,2,4].
Thereare no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.

Example 2:

Input: nums = [1,2,3]
Output: 6
Explanation:
Continuous subarray of size 1: [1], [2], [3].
Continuous subarray of size 2: [1,2], [2,3].
Continuous subarray of size 3: [1,2,3].
Total continuous subarrays = 3 + 2 + 1 = 6.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

Solution (Java)

class Solution {
    public long continuousSubarrays(int[] nums) {
        long ans = 0;
        int n = nums.length;
        TreeMap<Integer, Integer> map = new TreeMap<>();
        int i=0,j=0;
        while(j <n)
        {
            map.put(nums[j], map.getOrDefault(nums[j], 0) + 1);

            while (map.size() > 1 && map.lastKey() - map.firstKey() > 2)
            {
                map.put(nums[i], map.get(nums[i]) - 1);
                if (map.get(nums[i]) == 0) {
                    map.remove(nums[i]);
                }
                i++;
            }

            ans += j - i + 1;
            j++;
        }

        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).