363. Max Sum of Rectangle No Larger Than K

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2 
Explanation: Because the sum of rectangle [[0, 1], [-2, 3]] is 2,
             and 2 is the max number no larger than k (k = 2).

Note:

  1. The rectangle inside the matrix must have an area > 0.
  2. What if the number of rows is much larger than the number of columns?

363. Max Sum of Rectangle No Larger Than K
struct Solution;
use std::collections::BTreeSet;

impl Solution {
    fn max_sum_submatrix(matrix: Vec<Vec<i32>>, k: i32) -> i32 {
        let n = matrix.len();
        let m = matrix[0].len();
        let mut prefix = vec![vec![0; m + 1]; n];
        for i in 0..n {
            for j in 0..m {
                prefix[i][j + 1] = prefix[i][j] + matrix[i][j];
            }
        }
        let mut res = std::i32::MIN;
        for start in 0..m {
            for end in start + 1..=m {
                let mut bts: BTreeSet<i32> = BTreeSet::new();
                bts.insert(0);
                let mut sum = 0;
                for i in 0..n {
                    sum += prefix[i][end] - prefix[i][start];
                    if let Some(prev) = bts.range(sum - k..).take(1).next() {
                        res = res.max(sum - prev);
                    }
                    bts.insert(sum);
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[1, 0, 1], [0, -2, 3]];
    let k = 2;
    let res = 2;
    assert_eq!(Solution::max_sum_submatrix(matrix, k), res);
}