split-array-largest-sum.py

"""

410. Split Array Largest Sum
Hard

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)

"""

# V0

# V1
# IDEA : BINARY SEARCH
# https://leetcode.com/problems/split-array-largest-sum/solution/
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

# V1'
# IDEA : TOP DOWN DP
# https://leetcode.com/problems/split-array-largest-sum/solution/
class Solution:
    def splitArray(self, nums: List[int], m: int) -> int:
        n = len(nums)
        
        # Create a prefix sum array of nums.
        prefix_sum = [0] + list(itertools.accumulate(nums))
        
        @functools.lru_cache(None)
        def get_min_largest_split_sum(curr_index: int, subarray_count: int):
            # Base Case: If there is only one subarray left, then all of the remaining numbers
            # must go in the current subarray. So return the sum of the remaining numbers.
            if subarray_count == 1:
                return prefix_sum[n] - prefix_sum[curr_index]
        
            # Otherwise, use the recurrence relation to determine the minimum largest subarray sum
            # between curr_index and the end of the array with subarray_count subarrays remaining.
            minimum_largest_split_sum = prefix_sum[n]
            for i in range(curr_index, n - subarray_count + 1):
                # Store the sum of the first subarray.
                first_split_sum = prefix_sum[i + 1] - prefix_sum[curr_index]

                # Find the maximum subarray sum for the current first split.
                largest_split_sum = max(first_split_sum, 
                                        get_min_largest_split_sum(i + 1, subarray_count - 1))

                # Find the minimum among all possible combinations.
                minimum_largest_split_sum = min(minimum_largest_split_sum, largest_split_sum)

                if first_split_sum >= minimum_largest_split_sum:
                    break
            
            return minimum_largest_split_sum
        
        return get_min_largest_split_sum(0, m)

# V1''
# IDEA : BOTTOM UP DP
# https://leetcode.com/problems/split-array-largest-sum/solution/
class Solution:
    def splitArray(self, nums: List[int], m: int) -> int:
        n = len(nums)
        memo = [[0] * (m + 1) for _ in range(n)]
        
        # Create a prefix sum array of nums.
        prefix_sum = [0] + list(itertools.accumulate(nums))
        
        for subarray_count in range(1, m + 1):
            for curr_index in range(n):
                # Base Case: If there is only one subarray left, then all of the remaining numbers
                # must go in the current subarray. So return the sum of the remaining numbers.
                if subarray_count == 1:
                    memo[curr_index][subarray_count] = prefix_sum[n] - prefix_sum[curr_index]
                    continue

                # Otherwise, use the recurrence relation to determine the minimum largest subarray sum
                # between curr_index and the end of the array with subarray_count subarrays remaining.
                minimum_largest_split_sum = prefix_sum[n]
                for i in range(curr_index, n - subarray_count + 1):
                    # Store the sum of the first subarray.
                    first_split_sum = prefix_sum[i + 1] - prefix_sum[curr_index]

                    # Find the maximum subarray sum for the current first split.
                    largest_split_sum = max(first_split_sum, memo[i + 1][subarray_count - 1])

                    # Find the minimum among all possible combinations.
                    minimum_largest_split_sum = min(minimum_largest_split_sum, largest_split_sum)

                    if first_split_sum >= minimum_largest_split_sum:
                        break
            
                memo[curr_index][subarray_count] = minimum_largest_split_sum
        
        return memo[0][m]

# V2