410-split-array-largest-sum¶
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Description¶
Given an array nums
which consists of non-negative integers and an integer m
, you can split the array into m
non-empty continuous subarrays.
Write an algorithm to minimize the largest sum among these m
subarrays.
Example 1:
Input: nums = [7,2,5,10,8], m = 2 Output: 18 Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [7,2,5] and [10,8], where the largest sum among the two subarrays is only 18.
Example 2:
Input: nums = [1,2,3,4,5], m = 2 Output: 9
Example 3:
Input: nums = [1,4,4], m = 3 Output: 4
Constraints:
1 <= nums.length <= 1000
0 <= nums[i] <= 106
1 <= m <= min(50, nums.length)
Solution(Python)¶
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
return self.binarySearch(nums, m)
# Time Complexity: O(n^2*m)
# Space Complexity: O(n*m)
def topDownDp(self, nums: List[int], m: int) -> int:
n = len(nums)
prefix_sum = [0] + list(itertools.accumulate(nums))
@functools.lru_cache(None)
def get_min_largest_split_sum(cur_index, sub_arr_cnt):
if sub_arr_cnt == 1:
return prefix_sum[n] - prefix_sum[cur_index]
min_largst_sum = prefix_sum[n]
for i in range(cur_index, n - sub_arr_cnt + 1):
first_split_sum = prefix_sum[i + 1] - prefix_sum[cur_index]
largest_split_sum = max(
first_split_sum, get_min_largest_split_sum(
i + 1, sub_arr_cnt - 1)
)
min_largst_sum = min(largest_split_sum, min_largst_sum)
if first_split_sum >= min_largst_sum:
break
return min_largst_sum
return get_min_largest_split_sum(0, m)
# Time Complexity: O(nlog(S))
# Space Complexity: O(1)
def binarySearch(self, nums: List[int], m: int) -> int:
def min_subarrays_required(max_sum_allowed):
cur_sum = 0
split_req = 0
for ele in nums:
if cur_sum + ele <= max_sum_allowed:
cur_sum += ele
else:
cur_sum = ele
split_req += 1
return split_req + 1
left = max(nums)
right = sum(nums)
while left <= right:
max_sum_allowed = (left + right) >> 1
if min_subarrays_required(max_sum_allowed) <= m:
right = max_sum_allowed - 1
minimum_largest_split_sum = max_sum_allowed
else:
left = max_sum_allowed + 1
return minimum_largest_split_sum