LeetCode 2334. Subarray With Elements Greater Than Varying Threshold Solution in Java, C++, Python & Go | Explanation + Code

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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 1

PythonJavaC++Go
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(code-box)

Solution 2

PythonJavaC++Go
class Solution: def validSubarraySize(self, nums: List[int], threshold: int) -> int: n = len(nums) left = [-1] * n right = [n] * n stk = [] for i, v in enumerate(nums): while stk and nums[stk[-1]] >= v: stk.pop() if stk: left[i] = stk[-1] stk.append(i) stk = [] for i in range(n - 1, -1, -1): while stk and nums[stk[-1]] >= nums[i]: stk.pop() if stk: right[i] = stk[-1] stk.append(i) for i, v in enumerate(nums): k = right[i] - left[i] - 1 if v > threshold // k: return k return -1(code-box)

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