LeetCode #2762 — MEDIUM

Continuous Subarrays

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Let i, i + 1, ..., jbe 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].
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.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window · Segment Tree · Monotonic Queue

Example 1

[5,4,2,4]

Example 2

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

Try using the sliding window technique.
Use a set or map to keep track of the maximum and minimum of subarrays.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2762: Continuous Subarrays
class Solution {
    public long continuousSubarrays(int[] nums) {
        long ans = 0;
        int i = 0, n = nums.length;
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        for (int j = 0; j < n; ++j) {
            tm.merge(nums[j], 1, Integer::sum);
            while (tm.lastEntry().getKey() - tm.firstEntry().getKey() > 2) {
                tm.merge(nums[i], -1, Integer::sum);
                if (tm.get(nums[i]) == 0) {
                    tm.remove(nums[i]);
                }
                ++i;
            }
            ans += j - i + 1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2762: Continuous Subarrays
func continuousSubarrays(nums []int) (ans int64) {
	i := 0
	tm := treemap.NewWithIntComparator()
	for j, x := range nums {
		if v, ok := tm.Get(x); ok {
			tm.Put(x, v.(int)+1)
		} else {
			tm.Put(x, 1)
		}
		for {
			a, _ := tm.Min()
			b, _ := tm.Max()
			if b.(int)-a.(int) > 2 {
				if v, _ := tm.Get(nums[i]); v.(int) == 1 {
					tm.Remove(nums[i])
				} else {
					tm.Put(nums[i], v.(int)-1)
				}
				i++
			} else {
				break
			}
		}
		ans += int64(j - i + 1)
	}
	return
}

# Accepted solution for LeetCode #2762: Continuous Subarrays
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

// Accepted solution for LeetCode #2762: Continuous Subarrays
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2762: Continuous Subarrays
// class Solution {
//     public long continuousSubarrays(int[] nums) {
//         long ans = 0;
//         int i = 0, n = nums.length;
//         TreeMap<Integer, Integer> tm = new TreeMap<>();
//         for (int j = 0; j < n; ++j) {
//             tm.merge(nums[j], 1, Integer::sum);
//             while (tm.lastEntry().getKey() - tm.firstEntry().getKey() > 2) {
//                 tm.merge(nums[i], -1, Integer::sum);
//                 if (tm.get(nums[i]) == 0) {
//                     tm.remove(nums[i]);
//                 }
//                 ++i;
//             }
//             ans += j - i + 1;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2762: Continuous Subarrays
function continuousSubarrays(nums: number[]): number {
    const [minQ, maxQ]: [number[], number[]] = [[], []];
    const n = nums.length;
    let res = 0;

    for (let r = 0, l = 0; r < n; r++) {
        const x = nums[r];
        while (minQ.length && nums[minQ.at(-1)!] > x) minQ.pop();
        while (maxQ.length && nums[maxQ.at(-1)!] < x) maxQ.pop();
        minQ.push(r);
        maxQ.push(r);

        while (minQ.length && maxQ.length && nums[maxQ[0]] - nums[minQ[0]] > 2) {
            if (maxQ[0] < minQ[0]) {
                l = maxQ[0] + 1;
                maxQ.shift();
            } else {
                l = minQ[0] + 1;
                minQ.shift();
            }
        }

        res += r - l + 1;
    }

    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.