2762. Continuous Subarrays

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

class Solution {
public:
long long continuousSubarrays(vector<int>& nums) {
long ans = 1; // [nums[0]]
int left = nums[0] - 2;
int right = nums[0] + 2;
int l = 0;

// nums[l..r] is a valid window with range in [left, right].
for (int r = 1; r < nums.size(); r++) {
if (left <= nums[r] && nums[r] <= right) {
left = max(left, nums[r] - 2);
right = min(right, nums[r] + 2);
} else {
// nums[r] is out-of-bounds, so reconstruct the window.
left = nums[r] - 2;
right = nums[r] + 2;
l = r;
// If we consistently move leftward in each iteration, it implies that
// the entire left subarray satisfies the given condition. For every
// subarray with l in the range [0, r], the condition is met, preventing
// the code from reaching the final "else" condition. Instead, it stops
// at the "if" condition.
while (nums[r] - 2 <= nums[l] && nums[l] <= nums[r] + 2) {
left = max(left, nums[l] - 2);
right = min(right, nums[l] + 2);
--l;
}
++l;
}
// nums[l..r], nums[l + 1..r], ..., nums[r]
ans += r - l + 1;
}

return ans;
}
};