Leetcode #2762: Continuous Subarrays

In this guide, we solve Leetcode #2762 Continuous Subarrays in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Queue, Array, Ordered Set, Sliding Window, Monotonic Queue, Heap (Priority Queue)

Intuition

We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.

Expanding and shrinking the window lets us maintain validity without restarting the scan.

Approach

Grow the window with a right pointer, and shrink from the left only when the constraint is violated.

Track the best window as you go to keep the solution linear.

Steps:

Expand the right end of the window.
While invalid, move the left end to restore constraints.
Update the best window found.

Example

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].
There are no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.

Python Solution

class Solution:
    def continuousSubarrays(self, nums: List[int]) -> int:
        ans = i = 0
        sl = SortedList()
        for x in nums:
            sl.add(x)
            while sl[-1] - sl[0] > 2:
                sl.remove(nums[i])
                i += 1
            ans += len(sl)
        return ans

Complexity

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$ , where $n$ is the length of the array $nums$ . The space complexity is $O(n)$ , where $n$ is the length of the array $nums$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Example

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].
There are no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.

Python Solution

class Solution:
    def continuousSubarrays(self, nums: List[int]) -> int:
        ans = i = 0
        sl = SortedList()
        for x in nums:
            sl.add(x)
            while sl[-1] - sl[0] > 2:
                sl.remove(nums[i])
                i += 1
            ans += len(sl)
        return ans

Leetcode #2762: Continuous Subarrays

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2762: Continuous Subarrays

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview