LeetCode #2334 — HARD

Subarray With Elements Greater Than Varying Threshold

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

The Problem

Problem Statement

You are given an integer array nums and an integer threshold.

Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.

Return the size of any such subarray. If there is no such subarray, return -1.

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

Example 1:

Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.

Example 2:

Input: nums = [6,5,6,5,8], threshold = 7
Output: 1
Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. 
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], threshold <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer threshold. Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k. Return the size of any such subarray. If there is no such subarray, return -1. 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 · Stack · Union-Find

Example 1

[1,3,4,3,1]
6

Example 2

[6,5,6,5,8]
7

Core Insight

What unlocks the optimal approach

For all elements to be greater than the threshold/length, the minimum element in the subarray must be greater than the threshold/length.
For a given index, could you find the largest subarray such that the given index is the minimum element?
Could you use a monotonic stack to get the next and previous smallest element for every index?

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 #2334: Subarray With Elements Greater Than Varying Threshold
class Solution {
    private int[] p;
    private int[] size;

    public int validSubarraySize(int[] nums, int threshold) {
        int n = nums.length;
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i][0] = nums[i];
            arr[i][1] = i;
        }
        Arrays.sort(arr, (a, b) -> b[0] - a[0]);
        boolean[] vis = new boolean[n];
        for (int[] e : arr) {
            int v = e[0], i = e[1];
            if (i > 0 && vis[i - 1]) {
                merge(i, i - 1);
            }
            if (i < n - 1 && vis[i + 1]) {
                merge(i, i + 1);
            }
            if (v > threshold / size[find(i)]) {
                return size[find(i)];
            }
            vis[i] = true;
        }
        return -1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    private void merge(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return;
        }
        p[pa] = pb;
        size[pb] += size[pa];
    }
}

// Accepted solution for LeetCode #2334: Subarray With Elements Greater Than Varying Threshold
func validSubarraySize(nums []int, threshold int) int {
	n := len(nums)
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	merge := func(a, b int) {
		pa, pb := find(a), find(b)
		if pa == pb {
			return
		}
		p[pa] = pb
		size[pb] += size[pa]
	}

	arr := make([][]int, n)
	for i, v := range nums {
		arr[i] = []int{v, i}
	}
	sort.Slice(arr, func(i, j int) bool {
		return arr[i][0] > arr[j][0]
	})
	vis := make([]bool, n)
	for _, e := range arr {
		v, i := e[0], e[1]
		if i > 0 && vis[i-1] {
			merge(i, i-1)
		}
		if i < n-1 && vis[i+1] {
			merge(i, i+1)
		}
		if v > threshold/size[find(i)] {
			return size[find(i)]
		}
		vis[i] = true
	}
	return -1
}

# Accepted solution for LeetCode #2334: Subarray With Elements Greater Than Varying Threshold
class Solution:
    def validSubarraySize(self, nums: List[int], threshold: int) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def merge(a, b):
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            p[pa] = pb
            size[pb] += size[pa]

        n = len(nums)
        p = list(range(n))
        size = [1] * n
        arr = sorted(zip(nums, range(n)), reverse=True)
        vis = [False] * n
        for v, i in arr:
            if i and vis[i - 1]:
                merge(i, i - 1)
            if i < n - 1 and vis[i + 1]:
                merge(i, i + 1)
            if v > threshold // size[find(i)]:
                return size[find(i)]
            vis[i] = True
        return -1

// Accepted solution for LeetCode #2334: Subarray With Elements Greater Than Varying Threshold
/**
 * [2334] Subarray With Elements Greater Than Varying Threshold
 *
 * You are given an integer array nums and an integer threshold.
 * Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.
 * Return the size of any such subarray. If there is no such subarray, return -1.
 * A subarray is a contiguous non-empty sequence of elements within an array.
 *  
 * Example 1:
 *
 * Input: nums = [1,3,4,3,1], threshold = 6
 * Output: 3
 * Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
 * Note that this is the only valid subarray.
 *
 * Example 2:
 *
 * Input: nums = [6,5,6,5,8], threshold = 7
 * Output: 1
 * Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
 * Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5.
 * Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
 * Therefore, 2, 3, 4, or 5 may also be returned.
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 10^5
 * 	1 <= nums[i], threshold <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/subarray-with-elements-greater-than-varying-threshold/
// discuss: https://leetcode.com/problems/subarray-with-elements-greater-than-varying-threshold/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn valid_subarray_size(nums: Vec<i32>, threshold: i32) -> i32 {
        0
    }
}

// submission codes end

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

    #[test]
    #[ignore]
    fn test_2334_example_1() {
        let nums = vec![1, 3, 4, 3, 1];
        let threshold = 6;

        let result = 3;

        assert_eq!(Solution::valid_subarray_size(nums, threshold), result);
    }

    #[test]
    #[ignore]
    fn test_2334_example_2() {
        let nums = vec![6, 5, 6, 5, 8];
        let threshold = 7;

        let result = 1;

        assert_eq!(Solution::valid_subarray_size(nums, threshold), result);
    }
}

// Accepted solution for LeetCode #2334: Subarray With Elements Greater Than Varying Threshold
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2334: Subarray With Elements Greater Than Varying Threshold
// class Solution {
//     private int[] p;
//     private int[] size;
// 
//     public int validSubarraySize(int[] nums, int threshold) {
//         int n = nums.length;
//         p = new int[n];
//         size = new int[n];
//         for (int i = 0; i < n; ++i) {
//             p[i] = i;
//             size[i] = 1;
//         }
//         int[][] arr = new int[n][2];
//         for (int i = 0; i < n; ++i) {
//             arr[i][0] = nums[i];
//             arr[i][1] = i;
//         }
//         Arrays.sort(arr, (a, b) -> b[0] - a[0]);
//         boolean[] vis = new boolean[n];
//         for (int[] e : arr) {
//             int v = e[0], i = e[1];
//             if (i > 0 && vis[i - 1]) {
//                 merge(i, i - 1);
//             }
//             if (i < n - 1 && vis[i + 1]) {
//                 merge(i, i + 1);
//             }
//             if (v > threshold / size[find(i)]) {
//                 return size[find(i)];
//             }
//             vis[i] = true;
//         }
//         return -1;
//     }
// 
//     private int find(int x) {
//         if (p[x] != x) {
//             p[x] = find(p[x]);
//         }
//         return p[x];
//     }
// 
//     private void merge(int a, int b) {
//         int pa = find(a), pb = find(b);
//         if (pa == pb) {
//             return;
//         }
//         p[pa] = pb;
//         size[pb] += size[pa];
//     }
// }

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

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.