What is the brute force approach and complexity for Count Subarrays Where Max Element Appears at Least K Times?

For each start, grow until mx appears k times, then add n − end with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Count Subarrays Where Max Element Appears at Least K Times?

Shrink while the window has k copies of mx; all starts before left work with optimal time complexity O(N) and space complexity O(1).

Count Subarrays Where Max Element Appears at Least K Times

Q: How do you identify the pattern for Count Subarrays Where Max Element Appears at Least K Times?

Look for key signals in the prompt: "count subarrays" indicates Counting problem → add a block of subarrays for each right end.; "maximum element (of nums)" indicates One fixed value for the whole array. Find it first; then you only count its copies. No deque needed.; "appears at least k times" indicates An "at least" rule: growing keeps it true. Good starts form a prefix, so add `left` after shrinking..

Medium·LeetCode #2962 Pro OnlyMedium

LeetCode

Problem Statement

You are given an integer array nums and a positive integer k.

Let mx be the largest value in the whole array. Count how many contiguous subarrays contain mx at least k times.

Input

An integer array nums and an integer k.

Output

The number of subarrays in which the maximum element of nums appears at least k times (as a 64-bit integer).

Example 1

Input: nums = [1, 3, 2, 3, 3], k = 2

Output: 6

The maximum is 3. The subarrays with at least two 3's are [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].

Example 2

Input: nums = [1, 4, 2, 1], k = 3

Output: 0

The maximum 4 appears only once in the whole array, so no subarray has it 3 times.

Constraints

• 1 ≤ nums.length ≤ 10^5
• 1 ≤ nums[i] ≤ 10^6
• 1 ≤ k ≤ 10^5
• The answer can reach n(n+1)/2 ≈ 5·10^9, so it needs long long.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“count subarrays”

→ Counting problem → add a block of subarrays for each right end.

“maximum element (of nums)”

→ One fixed value for the whole array. Find it first; then you only count its copies. No deque needed.

“appears at least k times”

→ An "at least" rule: growing keeps it true. Good starts form a prefix, so add `left` after shrinking.

Approach Complexity Ladder

Brute Force — first good end for every startO(N²)O(1)

For each start, grow until mx appears k times, then add n − end

Optimal — shrink while good, then add leftO(N)O(1)

Shrink while the window has k copies of mx; all starts before left work

Brute Force Intuition

Find mx. For each start, grow the end and count copies of mx. When the count reaches k, every longer subarray from this start also has at least k copies, so add n − end and move to the next start.

Recognition rule for this sub-pattern:

Count subarrays that satisfy a rule. "At most": for each right end, every start from left to right works, so add (right − left + 1). "At least": once the window is good, every longer one is too, so add left (or n − right).

Remember this · Sliding Window

Trigger

Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".

Think

Can I keep a valid range and move it, instead of checking every subarray from scratch?

Key question

When is my current window valid?

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