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# Subarray With Elements Greater Than Varying Threshold LeetCode Solution

## Problem – Subarray With Elements Greater Than Varying Threshold LeetCode Solution

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`.

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`

### Subarray With Elements Greater Than Varying Threshold LeetCode Solution in C++

``````class Solution {
public:
int validSubarraySize(vector<int>& nums, int threshold) {
int n = nums.size();
vector<long long> lr(n, n), rl(n, -1);

vector<int> s;
for(int i = 0; i < n; ++i) {
while(!s.empty() and nums[i] < nums[s.back()]) {
lr[s.back()] = i;
s.pop_back();
}
s.push_back(i);
}
s.clear();
for(int i = n - 1; ~i; --i) {
while(!s.empty() and nums[i] < nums[s.back()]) {
rl[s.back()] = i;
s.pop_back();
}
s.push_back(i);
}

for(int i = 0; i < n; ++i) {
long long length = lr[i] - rl[i] - 1;
if(1LL * nums[i] * length > threshold) return length;
}

return -1;
}
};
``````

### Subarray With Elements Greater Than Varying Threshold LeetCode Solution in Python

``````def validSubarraySize(self, nums: List[int], threshold: int) -> int:
nums = [0] + nums + [0]
stack = [0]
for i in range(1,len(nums)):
while nums[i] < nums[stack[-1]]:
tmp = nums[stack.pop()]
if tmp > threshold / (i - stack[-1] - 1):
return i - stack[-1] - 1
stack.append(i)
return -1
``````

### Subarray With Elements Greater Than Varying Threshold LeetCode Solution in Java

``````public int validSubarraySize(int[] nums, int threshold) {
int n = nums.length;
return dfs(nums, 0, n - 1, threshold);
}

private int dfs(int[] nums, int l, int r, int t) {
if(l > r) {
return -1;
}
if(l == r) {
return nums[l] > t ? 1 : -1;
}

int n = r - l + 1;
for(int i = l; i <= r; ) {
if(nums[i] <= t / n) {
i++;
continue;
}
int j = i;
int min = Integer.MAX_VALUE;
while(i <= r && nums[i] > t / n) {
min = Math.min(min, nums[i]);
i++;
}

if(min > t / (i - j)) {
return i - j;
}

int res = dfs(nums, j, i - 1, t);

if(res != -1) {
return res;
}
}

return -1;
}
``````
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