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Problem solution · C++

Range Sum of Sorted Subarray Sums

Range Sum of Sorted Subarray Sums: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
50 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

Sliding window or two pointers

For Range Sum of Sorted Subarray Sums, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 50 lines of C++ from the credited upstream file 1508.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeRange Sum of Sorted Subarray Sums · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int rangeSum(vector<int>& nums, int n, int left, int right) {    constexpr int kMod = 1'000'000'007;     auto subarraysAndSumNoGreaterThan = [&](int m) -> pair<int, long> {      int count = 0;   // the number of subarrays <= m      long total = 0;  // sum(subarrays)      int sum = 0;     // the current sum      int window = 0;  // the window sum       for (int i = 0, j = 0; j < n; ++j) {        sum += nums[j] * (j - i + 1);        window += nums[j];  // Extend each subarray that ends in j.        while (window > m) {          sum -= window;          window -= nums[i++];  // Shrink the window.        }        count += j - i + 1;        total += sum;      }       return {count, total};    };     // [L, R] is the possible range of the sum of any subarray.    const int L = ranges::min(nums);    const int R = accumulate(nums.begin(), nums.end(), 0);     auto firstKSubarraysSum = [&](int k) -> long {      int l = L;      int r = R;       while (l < r) {        const int m = l + (r - l) / 2;        if (subarraysAndSumNoGreaterThan(m).first < k)          l = m + 1;        else          r = m;      }       const auto& [count, total] = subarraysAndSumNoGreaterThan(l);      // If count != k, there're subarray(s) have the same sum as l.      return total - l * (count - k);    };     return (firstKSubarraysSum(right) - firstKSubarraysSum(left - 1)) % kMod;  }};

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