LeetCode #2444 — HARD

Count Subarrays With Fixed Bounds

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums and two integers minK and maxK.

A fixed-bound subarray of nums is a subarray that satisfies the following conditions:

The minimum value in the subarray is equal to minK.
The maximum value in the subarray is equal to maxK.

Return the number of fixed-bound subarrays.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5
Output: 2
Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].

Example 2:

Input: nums = [1,1,1,1], minK = 1, maxK = 1
Output: 10
Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i], minK, maxK <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers minK and maxK. A fixed-bound subarray of nums is a subarray that satisfies the following conditions: The minimum value in the subarray is equal to minK. The maximum value in the subarray is equal to maxK. Return the number of fixed-bound subarrays. 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 · Sliding Window · Monotonic Queue

Example 1

[1,3,5,2,7,5]
1
5

Example 2

[1,1,1,1]
1
1

Core Insight

What unlocks the optimal approach

Can you solve the problem if all the numbers in the array were between minK and maxK inclusive?
Think of the inclusion-exclusion principle.
Divide the array into multiple subarrays such that each number in each subarray is between minK and maxK inclusive, solve the previous problem for each subarray, and sum all the answers.

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 #2444: Count Subarrays With Fixed Bounds
class Solution {
    public long countSubarrays(int[] nums, int minK, int maxK) {
        long ans = 0;
        int j1 = -1, j2 = -1, k = -1;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] < minK || nums[i] > maxK) {
                k = i;
            }
            if (nums[i] == minK) {
                j1 = i;
            }
            if (nums[i] == maxK) {
                j2 = i;
            }
            ans += Math.max(0, Math.min(j1, j2) - k);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2444: Count Subarrays With Fixed Bounds
func countSubarrays(nums []int, minK int, maxK int) int64 {
	ans := 0
	j1, j2, k := -1, -1, -1
	for i, v := range nums {
		if v < minK || v > maxK {
			k = i
		}
		if v == minK {
			j1 = i
		}
		if v == maxK {
			j2 = i
		}
		ans += max(0, min(j1, j2)-k)
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2444: Count Subarrays With Fixed Bounds
class Solution:
    def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
        j1 = j2 = k = -1
        ans = 0
        for i, v in enumerate(nums):
            if v < minK or v > maxK:
                k = i
            if v == minK:
                j1 = i
            if v == maxK:
                j2 = i
            ans += max(0, min(j1, j2) - k)
        return ans

// Accepted solution for LeetCode #2444: Count Subarrays With Fixed Bounds
impl Solution {
    pub fn count_subarrays(nums: Vec<i32>, min_k: i32, max_k: i32) -> i64 {
        let mut ans: i64 = 0;
        let mut j1: i64 = -1;
        let mut j2: i64 = -1;
        let mut k: i64 = -1;
        for (i, &v) in nums.iter().enumerate() {
            let i = i as i64;
            if v < min_k || v > max_k {
                k = i;
            }
            if v == min_k {
                j1 = i;
            }
            if v == max_k {
                j2 = i;
            }
            let m = j1.min(j2);
            if m > k {
                ans += m - k;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2444: Count Subarrays With Fixed Bounds
function countSubarrays(nums: number[], minK: number, maxK: number): number {
    let ans = 0;
    let [j1, j2, k] = [-1, -1, -1];
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i] < minK || nums[i] > maxK) k = i;
        if (nums[i] === minK) j1 = i;
        if (nums[i] === maxK) j2 = i;
        ans += Math.max(0, Math.min(j1, j2) - k);
    }
    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.