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 <= 10000 <= nums[i] <= 1061 <= m <= min(50, nums.length)
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class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
def min_subarrays_required(max_sum_allowed: int) -> int:
current_sum = 0
splits_required = 0
for element in nums:
# Add element only if the sum doesn't exceed max_sum_allowed
if current_sum + element <= max_sum_allowed:
current_sum += element
else:
# If the element addition makes sum more than max_sum_allowed
# Increment the splits required and reset sum
current_sum = element
splits_required += 1
# Return the number of subarrays, which is the number of splits + 1
return splits_required + 1
# Define the left and right boundary of binary search
left = max(nums)
right = sum(nums)
while left <= right:
# Find the mid value
max_sum_allowed = (left + right) // 2
# Find the minimum splits. If splits_required is less than
# or equal to m move towards left i.e., smaller values
if min_subarrays_required(max_sum_allowed) <= m:
right = max_sum_allowed - 1
minimum_largest_split_sum = max_sum_allowed
else:
# Move towards right if splits_required is more than m
left = max_sum_allowed + 1
return minimum_largest_split_sum
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