LeetCode #2962 — MEDIUM

Count Subarrays Where Max Element Appears at Least K Times

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

The Problem

Problem Statement

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

Return the number of subarrays where the maximum element of nums appears at least k times in that subarray.

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

Example 1:

Input: nums = [1,3,2,3,3], k = 2
Output: 6
Explanation: The subarrays that contain the element 3 at least 2 times 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
Explanation: No subarray contains the element 4 at least 3 times.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶
1 <= k <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and a positive integer k. Return the number of subarrays where the maximum element of nums appears at least k times in that subarray. A subarray is a contiguous 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

Example 1

[1,3,2,3,3]
2

Example 2

[1,4,2,1]
3

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #2962: Count Subarrays Where Max Element Appears at Least K Times
class Solution {
    public long countSubarrays(int[] nums, int k) {
        int mx = Arrays.stream(nums).max().getAsInt();
        int n = nums.length;
        long ans = 0;
        int cnt = 0, j = 0;
        for (int x : nums) {
            while (j < n && cnt < k) {
                cnt += nums[j++] == mx ? 1 : 0;
            }
            if (cnt < k) {
                break;
            }
            ans += n - j + 1;
            cnt -= x == mx ? 1 : 0;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2962: Count Subarrays Where Max Element Appears at Least K Times
func countSubarrays(nums []int, k int) (ans int64) {
	mx := slices.Max(nums)
	n := len(nums)
	cnt, j := 0, 0
	for _, x := range nums {
		for ; j < n && cnt < k; j++ {
			if nums[j] == mx {
				cnt++
			}
		}
		if cnt < k {
			break
		}
		ans += int64(n - j + 1)
		if x == mx {
			cnt--
		}
	}
	return
}

# Accepted solution for LeetCode #2962: Count Subarrays Where Max Element Appears at Least K Times
class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        mx = max(nums)
        n = len(nums)
        ans = cnt = j = 0
        for x in nums:
            while j < n and cnt < k:
                cnt += nums[j] == mx
                j += 1
            if cnt < k:
                break
            ans += n - j + 1
            cnt -= x == mx
        return ans

// Accepted solution for LeetCode #2962: Count Subarrays Where Max Element Appears at Least K Times
/**
 * [2962] Count Subarrays Where Max Element Appears at Least K Times
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn count_subarrays(nums: Vec<i32>, k: i32) -> i64 {
        let max = *nums.iter().max().unwrap();
        let n = nums.len();

        let mut result = 0;
        let mut count = 0;
        let mut left = 0;

        for right in 0..n {
            count += if nums[right] == max { 1 } else { 0 };

            if count >= k {
                result += (n - right) as i64;
            }

            while left <= right && count >= k {
                count -= if nums[left] == max { 1 } else { 0 };

                if count >= k {
                    result += (n - right) as i64;
                }
                left += 1;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2962() {
        assert_eq!(6, Solution::count_subarrays(vec![1, 3, 2, 3, 3], 2));
        assert_eq!(0, Solution::count_subarrays(vec![1, 4, 2, 1], 3));
    }
}

// Accepted solution for LeetCode #2962: Count Subarrays Where Max Element Appears at Least K Times
function countSubarrays(nums: number[], k: number): number {
    const mx = Math.max(...nums);
    const n = nums.length;
    let [cnt, j] = [0, 0];
    let ans = 0;
    for (const x of nums) {
        for (; j < n && cnt < k; ++j) {
            cnt += nums[j] === mx ? 1 : 0;
        }
        if (cnt < k) {
            break;
        }
        ans += n - j + 1;
        cnt -= x === mx ? 1 : 0;
    }
    return ans;
}

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)

Space

O(k)

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.