LeetCode #2488 — HARD

Count Subarrays With Median K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array nums of size n consisting of distinct integers from 1 to n and a positive integer k.

Return the number of non-empty subarrays in nums that have a median equal to k.

Note:

The median of an array is the middle element after sorting the array in ascending order. If the array is of even length, the median is the left middle element.
- For example, the median of [2,3,1,4] is 2, and the median of [8,4,3,5,1] is 4.
A subarray is a contiguous part of an array.

Example 1:

Input: nums = [3,2,1,4,5], k = 4
Output: 3
Explanation: The subarrays that have a median equal to 4 are: [4], [4,5] and [1,4,5].

Example 2:

Input: nums = [2,3,1], k = 3
Output: 1
Explanation: [3] is the only subarray that has a median equal to 3.

Constraints:

n == nums.length
1 <= n <= 10⁵
1 <= nums[i], k <= n
The integers in nums are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of size n consisting of distinct integers from 1 to n and a positive integer k. Return the number of non-empty subarrays in nums that have a median equal to k. Note: The median of an array is the middle element after sorting the array in ascending order. If the array is of even length, the median is the left middle element. For example, the median of [2,3,1,4] is 2, and the median of [8,4,3,5,1] is 4. A subarray is a contiguous part of an array.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[3,2,1,4,5]
4

Example 2

[2,3,1]
3

Core Insight

What unlocks the optimal approach

Consider changing the numbers that are strictly greater than k in the array to 1, the numbers that are strictly smaller than k to -1, and k to 0.
After the change, what property does a subarray with median k have in the new array?
An array with median k should have a sum equal to either 0 or 1 in the new array and should contain the element k. How do you count such 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

Largest constraint values

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 #2488: Count Subarrays With Median K
class Solution {
    public int countSubarrays(int[] nums, int k) {
        int n = nums.length;
        int i = 0;
        for (; nums[i] != k; ++i) {
        }
        int[] cnt = new int[n << 1 | 1];
        int ans = 1;
        int x = 0;
        for (int j = i + 1; j < n; ++j) {
            x += nums[j] > k ? 1 : -1;
            if (x >= 0 && x <= 1) {
                ++ans;
            }
            ++cnt[x + n];
        }
        x = 0;
        for (int j = i - 1; j >= 0; --j) {
            x += nums[j] > k ? 1 : -1;
            if (x >= 0 && x <= 1) {
                ++ans;
            }
            ans += cnt[-x + n] + cnt[-x + 1 + n];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2488: Count Subarrays With Median K
func countSubarrays(nums []int, k int) int {
	i, n := 0, len(nums)
	for nums[i] != k {
		i++
	}
	ans := 1
	cnt := make([]int, n<<1|1)
	x := 0
	for j := i + 1; j < n; j++ {
		if nums[j] > k {
			x++
		} else {
			x--
		}
		if x >= 0 && x <= 1 {
			ans++
		}
		cnt[x+n]++
	}
	x = 0
	for j := i - 1; j >= 0; j-- {
		if nums[j] > k {
			x++
		} else {
			x--
		}
		if x >= 0 && x <= 1 {
			ans++
		}
		ans += cnt[-x+n] + cnt[-x+1+n]
	}
	return ans
}

# Accepted solution for LeetCode #2488: Count Subarrays With Median K
class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        i = nums.index(k)
        cnt = Counter()
        ans = 1
        x = 0
        for v in nums[i + 1 :]:
            x += 1 if v > k else -1
            ans += 0 <= x <= 1
            cnt[x] += 1
        x = 0
        for j in range(i - 1, -1, -1):
            x += 1 if nums[j] > k else -1
            ans += 0 <= x <= 1
            ans += cnt[-x] + cnt[-x + 1]
        return ans

// Accepted solution for LeetCode #2488: Count Subarrays With Median K
// 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 #2488: Count Subarrays With Median K
// class Solution {
//     public int countSubarrays(int[] nums, int k) {
//         int n = nums.length;
//         int i = 0;
//         for (; nums[i] != k; ++i) {
//         }
//         int[] cnt = new int[n << 1 | 1];
//         int ans = 1;
//         int x = 0;
//         for (int j = i + 1; j < n; ++j) {
//             x += nums[j] > k ? 1 : -1;
//             if (x >= 0 && x <= 1) {
//                 ++ans;
//             }
//             ++cnt[x + n];
//         }
//         x = 0;
//         for (int j = i - 1; j >= 0; --j) {
//             x += nums[j] > k ? 1 : -1;
//             if (x >= 0 && x <= 1) {
//                 ++ans;
//             }
//             ans += cnt[-x + n] + cnt[-x + 1 + n];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2488: Count Subarrays With Median K
function countSubarrays(nums: number[], k: number): number {
    const i = nums.indexOf(k);
    const n = nums.length;
    const cnt = new Array((n << 1) | 1).fill(0);
    let ans = 1;
    let x = 0;
    for (let j = i + 1; j < n; ++j) {
        x += nums[j] > k ? 1 : -1;
        ans += x >= 0 && x <= 1 ? 1 : 0;
        ++cnt[x + n];
    }
    x = 0;
    for (let j = i - 1; ~j; --j) {
        x += nums[j] > k ? 1 : -1;
        ans += x >= 0 && x <= 1 ? 1 : 0;
        ans += cnt[-x + n] + cnt[-x + 1 + n];
    }
    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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.