2334. Subarray With Elements Greater Than Varying Threshold

Description

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

 

Solutions

Solution: Stack

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} threshold
 * @return {number}
 */
const validSubarraySize = function (nums, threshold) {
  const n = nums.length;
  const stack = [];
  const prev = Array.from({ length: n }, () => -1);
  const next = Array.from({ length: n }, () => n);

  for (let index = 0; index < n; index++) {
    const num = nums[index];

    while (stack.length && nums[stack.at(-1)] > num) {
      const pos = stack.pop();

      next[pos] = index;
    }

    if (stack.length) {
      prev[index] = stack.at(-1);
    }

    stack.push(index);
  }

  for (let index = 0; index < n; index++) {
    const k = index - prev[index] + (next[index] - index) - 1;

    if (nums[index] > threshold / k) return k;
  }

  return -1;
};