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max-sum-of-sub-matrix-no-larger-than-k.py
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max-sum-of-sub-matrix-no-larger-than-k.py
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# Time: O(min(m, n)^2 * max(m, n) * log(max(m, n)))
# Space: O(max(m, n))
# 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:
# Given matrix = [
# [1, 0, 1],
# [0, -2, 3]
# ]
# k = 2
# The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]]
# is 2 and 2 is the max number no larger than k (k = 2).
#
# Note:
# The rectangle inside the matrix must have an area > 0.
# What if the number of rows is much larger than the number of columns?
# Time: O(min(m, n)^2 * max(m, n)^2)
# Space: O(max(m, n))
# 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:
# Given matrix = [
# [1, 0, 1],
# [0, -2, 3]
# ]
# k = 2
# The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]]
# is 2 and 2 is the max number no larger than k (k = 2).
#
# Note:
# The rectangle inside the matrix must have an area > 0.
# What if the number of rows is much larger than the number of columns?
# Time: O(min(m, n)^2 * max(m, n)^2)
# Space: O(max(m, n))
from bisect import bisect_left, insort
class Solution(object):
def maxSumSubmatrix(self, matrix, k):
"""
:type matrix: List[List[int]]
:type k: int
:rtype: int
"""
if not matrix:
return 0
m = min(len(matrix), len(matrix[0]))
n = max(len(matrix), len(matrix[0]))
result = float("-inf")
for i in xrange(m):
sums = [0] * n
for j in xrange(i, m):
for l in xrange(n):
sums[l] += matrix[j][l] if m == len(matrix) else matrix[l][j]
# Find the max subarray no more than K.
accu_sum_set, accu_sum = [0], 0
for sum in sums:
accu_sum += sum
it = bisect_left(accu_sum_set, accu_sum - k) # Time: O(logn)
if it != len(accu_sum_set):
result = max(result, accu_sum - accu_sum_set[it])
insort(accu_sum_set, accu_sum) # Time: O(n)
return result
# Time: O(min(m, n)^2 * max(m, n) * log(max(m, n))) ~ O(min(m, n)^2 * max(m, n)^2)
# Space: O(max(m, n))
class Solution_TLE(object):
def maxSumSubmatrix(self, matrix, k):
"""
:type matrix: List[List[int]]
:type k: int
:rtype: int
"""
class BST(object): # not avl, rbtree
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def insert(self, val): # Time: O(h) = O(logn) ~ O(n)
curr = self
while curr:
if curr.val >= val:
if curr.left:
curr = curr.left
else:
curr.left = BST(val)
return
else:
if curr.right:
curr = curr.right
else:
curr.right = BST(val)
return
def lower_bound(self, val): # Time: O(h) = O(logn) ~ O(n)
result, curr = None, self
while curr:
if curr.val >= val:
result, curr = curr, curr.left
else:
curr = curr.right
return result
if not matrix:
return 0
m = min(len(matrix), len(matrix[0]))
n = max(len(matrix), len(matrix[0]))
result = float("-inf")
for i in xrange(m):
sums = [0] * n
for j in xrange(i, m):
for l in xrange(n):
sums[l] += matrix[j][l] if m == len(matrix) else matrix[l][j]
# Find the max subarray no more than K.
accu_sum_set = BST(0)
accu_sum = 0
for sum in sums:
accu_sum += sum
node = accu_sum_set.lower_bound(accu_sum - k);
if node:
result = max(result, accu_sum - node.val)
accu_sum_set.insert(accu_sum)
return result