363. Max Sum of Rectangle No Larger Than K

Hard

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Given an m x n matrix matrix and an integer k, return the max sum of a rectangle in the matrix such that its sum is no larger than k.

It is guaranteed that there will be a rectangle with a sum no larger than k.

Example 1:

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

Example 2:

Input: matrix = [[2,2,-1]], k = 3
Output:
 3

Constraints:

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 100

• -100 <= matrix[i][j] <= 100

• -105 <= k <= 105

Follow up: What if the number of rows is much larger than the number of columns?

from sortedcontainers import SortedList
class Solution:
    #.
    def maxSumSubmatrix(self, matrix: List[List[int]], k: int) -> int:
        m, n = len(matrix), len(matrix[0])
        ans = -math.inf
        for r1 in range(m):
            arr = [0] * n  # arr[i] is sum(matrix[r1][c]...matrix[r2][c])
            for r2 in range(r1, m):
                for c in range(n): arr[c] += matrix[r2][c]
                ans = max(ans, self.maxSumSubAarray(arr, n, k))
        return ans

    def maxSumSubAarray(self, arr, n, k):
        right = 0  # PrefixSum so far
        seen = SortedList([0])
        ans = -math.inf
        for i in range(n):
            right += arr[i]
            left = self.ceiling(seen, right - k)  # right - left <= k -> left >= right - k
            if left != None:
                ans = max(ans, right - left)
            seen.add(right)
        return ans

    def ceiling(self, sortedList, key):  # O(logN)
        idx = sortedList.bisect_left(key)
        if idx < len(sortedList): return sortedList[idx]
        return None