Leetcode 2334 Subarray With Elements Greater Than Varying Threshold Solution in c++ | Hindi Coding Community

 

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

C++ Code :

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;

}

};