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Problem solution · C++

Count Subarrays with Cost Less Than or Equal to K

Count Subarrays with Cost Less Than or Equal to K: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
33 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Count Subarrays with Cost Less Than or Equal to K, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 33 lines of C++ from the credited upstream file count-subarrays-with-cost-less-than-or-equal-to-k.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 4 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Subarrays with Cost Less Than or Equal to K · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(n) // mono deque, two pointersclass Solution {public:    long long countSubarrays(vector<int>& nums, long long k) {        int64_t result = 0;        deque<int> max_dq, min_dq;        for (int right = 0, left = 0; right < size(nums); ++right) {            while (!empty(max_dq) && nums[max_dq.back()] <= nums[right]) {                max_dq.pop_back();            }            max_dq.emplace_back(right);            while (!empty(min_dq) && nums[min_dq.back()] >= nums[right]) {                min_dq.pop_back();            }            min_dq.emplace_back(right);            while (static_cast<int64_t>(right - left + 1) * (nums[max_dq[0]] - nums[min_dq[0]]) > k) {                if (!empty(max_dq) && max_dq[0] == left) {                    max_dq.pop_front();                }                if (!empty(min_dq) && min_dq[0] == left) {                    min_dq.pop_front();                }                ++left;            }            result += right - left + 1;        }        return result;    }};

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