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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 <= 1051 <= nums[i], threshold <= 109
Solutions
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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]; } } -
using pii = pair<int, int>; class Solution { public: vector<int> p; vector<int> size; int validSubarraySize(vector<int>& nums, int threshold) { int n = nums.size(); p.resize(n); for (int i = 0; i < n; ++i) p[i] = i; size.assign(n, 1); vector<pii> arr(n); for (int i = 0; i < n; ++i) arr[i] = {nums[i], i}; sort(arr.begin(), arr.end()); vector<bool> vis(n); for (int j = n - 1; ~j; --j) { int v = arr[j].first, i = arr[j].second; if (i && vis[i - 1]) merge(i, i - 1); if (j < n - 1 && vis[i + 1]) merge(i, i + 1); if (v > threshold / size[find(i)]) return size[find(i)]; vis[i] = true; } return -1; } int find(int x) { if (p[x] != x) p[x] = find(p[x]); return p[x]; } void merge(int a, int b) { int pa = find(a), pb = find(b); if (pa == pb) return; p[pa] = pb; size[pb] += size[pa]; } }; -
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 -
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 }