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Problem solution · C++

Max Sum of Sub Matrix No Larger Than K

Max Sum of Sub Matrix No Larger Than 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
40 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 Max Sum of Sub Matrix No Larger Than 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

  • 40 lines of C++ from the credited upstream file max-sum-of-sub-matrix-no-larger-than-k.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 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 codeMax Sum of Sub Matrix No Larger Than K · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(min(m, n)^2 * max(m, n) * log(max(m, n)))// Space: O(max(m, n)) class Solution {public:    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {        if (matrix.empty()) {            return 0;        }         const int m = min(matrix.size(), matrix[0].size());        const int n = max(matrix.size(), matrix[0].size());        int result = numeric_limits<int>::min();         for (int i = 0; i < m; ++i) {            vector<int> sums(n, 0);            for (int j = i; j < m; ++j) {                for (int l = 0; l < n; ++l) {                    sums[l] += (m == matrix.size()) ? matrix[j][l] : matrix[l][j];                }                    // Find the max subarray no more than K.                set<int> accu_sum_set;                accu_sum_set.emplace(0);                int accu_sum = 0;                for (int sum : sums) {                    accu_sum += sum;                    auto it = accu_sum_set.lower_bound(accu_sum - k);                    if (it != accu_sum_set.end()) {                        result = max(result, accu_sum - *it);                    }                    accu_sum_set.emplace(accu_sum);                }            }        }         return result;    }};

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