Split Array Largest Sum

To solve this problem, you can apply binary search on the result, which would be in the range from the maximum single element to the sum of all elements in the array. This approach will help you find the minimum largest sum for the given integer array and ( k ) subarrays.

Here’s the step-by-step approach:

  1. Initialize Binary Search Bounds: The lower bound will be the maximum single element in the array, and the upper bound will be the sum of all elements in the array.
  2. Apply Binary Search: In each step, you can check whether it’s possible to split the array into ( k ) subarrays with a maximum sum less than or equal to the current mid value. If it is possible, you know that the actual result will be less than or equal to mid, so you update the upper bound. Otherwise, you update the lower bound.
  3. Check Feasibility: Write a helper function to check if it’s possible to split the array into ( k ) subarrays with a given maximum sum.

Here’s the code:

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class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        # Function to check feasibility
        def is_feasible(mid):
            subarrays = 1
            current_sum = 0
            for num in nums:
                current_sum += num
                if current_sum > mid:
                    subarrays += 1
                    current_sum = num
            return subarrays <= k

        # Initialize binary search bounds
        low, high = max(nums), sum(nums)

        # Apply binary search
        while low < high:
            mid = (low + high) // 2
            if is_feasible(mid):
                high = mid
            else:
                low = mid + 1

        return low

Explanation:

  • The is_feasible function checks whether it’s possible to split the array into ( k ) subarrays with the given maximum sum.
  • The binary search helps find the minimum largest sum by iteratively narrowing down the search range.

The time complexity of this solution is ( O(n \log m) ), where ( n ) is the length of the array, and ( m ) is the sum of the array elements, making it efficient for the given constraints.

10 Prerequisite LeetCode Problems

“Split Array Largest Sum” involves binary search and dynamic programming. Here are 10 problems to prepare for it:

  1. “Maximum Subarray” (LeetCode Problem #53): A classic problem that introduces the concept of finding subarray with maximum sum.

  2. “Best Time to Buy and Sell Stock” (LeetCode Problem #121): It helps to understand the notion of tracking maximum and minimum values for optimization.

  3. “Sum of Subarray Minimums” (LeetCode Problem #907): This problem helps in understanding handling of subarray sums.

  4. “Find Minimum in Rotated Sorted Array” (LeetCode Problem #153): It helps understand the binary search in a rotated sorted array.

  5. “Binary Search” (LeetCode Problem #704): This is a foundational binary search problem.

  6. “Guess Number Higher or Lower” (LeetCode Problem #374): A good problem to understand the application of binary search.

  7. “House Robber” (LeetCode Problem #198): This problem applies dynamic programming in a context where you need to make decisions to maximize the sum.

  8. “Partition Equal Subset Sum” (LeetCode Problem #416): This problem introduces the concept of partitioning an array with the aim of equalizing the sums of the subsets.

  9. “Minimum Size Subarray Sum” (LeetCode Problem #209): This problem helps you learn how to manage subarrays to reach a target sum.

  10. “Can Partition” (LeetCode Problem #416): This problem is about dividing an array into two parts with the same sum, which will give you practice working with similar constraints.

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    def splitArray(self, nums: List[int], m: int) -> int:
        def cannot_split(max_sum, m):
            cuts, curr_sum  = 0, 0
            for x in nums:
                curr_sum += x
                if curr_sum > max_sum:
                    cuts += 1
                    curr_sum = x
            subs = cuts + 1
            return (subs > m)

        low, high = max(nums), sum(nums)
        while low < high:
            guess = low + (high - low) // 2
            if cannot_split(guess, m):
                low = guess + 1
            else:
                high = guess
        return low

Problem Classification

This problem involves splitting an array into subarrays to minimize the maximum sum among them. This is a Partitioned Sum Minimization Problem.

Language Agnostic Coding Drills

This code solves the problem of splitting an array into m subarrays, such that the maximum sum of the subarrays is minimized. It utilizes binary search to find the minimum possible maximum sum. The code can be broken down into the following parts:

  1. Binary Search: The problem uses binary search to narrow down the range of possible solutions. Binary search is an algorithm that divides the search space in half repeatedly until the solution is found.

  2. Helper function: The code uses a helper function cannot_split(max_sum, m) that checks whether the current array can be divided into m subarrays with a maximum sum of max_sum. If it’s not possible, it returns True, otherwise False.

  3. Subarray Sums: The problem involves computing the sums of subarrays. This is a common task in many array-based problems.

  4. Array Iteration: It iterates over the array to calculate the sum of elements and the number of cuts required to split the array.

  5. Comparing and updating variables: Inside the binary search loop, the function checks if the array can be split with the given guessed maximum sum. Depending on the outcome, it updates the binary search range accordingly.

Now, let’s describe a step by step problem-solving approach:

  1. Identify the problem as a binary search problem since we are trying to minimize the maximum sum of subarrays. This involves finding an optimal point in a sorted list (sums of subsets), which is a classic sign of problems that require binary search.

  2. Establish the bounds for the binary search. In this case, the lower bound (low) is the maximum number in the list, and the upper bound (high) is the sum of all numbers in the list. This is because the maximum sum of a subarray must be within this range.

  3. Implement the helper function cannot_split(max_sum, m). This function tries to split the array into subarrays where each subarray’s sum is less than or equal to max_sum. If it needs more than m subarrays to do so, it returns True.

  4. Implement the binary search. Start a loop that continues until the lower bound is less than the upper bound. Calculate the middle value (guess) and use the helper function to check if the array can be split into m subarrays with a maximum sum of guess. If it can, decrease the upper bound to guess, else increase the lower bound.

  5. When the loop ends, return the lower bound as the minimum possible maximum sum.

This explanation is language-agnostic, details of implementation may vary across different programming languages. For instance, Python supports the // operator for integer division, but other languages like JavaScript use / for all types of division, and results would need to be manually converted to an integer with the Math.floor() function.

Targeted Drills in Python

1. Binary Search

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def binary_search(array, target):
    low, high = 0, len(array) - 1
    while low <= high:
        mid = low + (high - low) // 2
        if array[mid] == target:
            return mid
        elif array[mid] < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

2. Helper Function

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def helper_function(nums, max_sum, m):
    cuts, curr_sum = 0, 0
    for num in nums:
        curr_sum += num
        if curr_sum > max_sum:
            cuts += 1
            curr_sum = num
    return cuts < m

3. Subarray Sums

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def subarray_sums(nums):
    curr_sum = 0
    for num in nums:
        curr_sum += num
        print(curr_sum)
    return curr_sum

4. Array Iteration

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def array_iteration(nums):
    for num in nums:
        print(num)

5. Comparing and updating variables

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def compare_and_update(nums):
    guess = 5
    for num in nums:
        if num > guess:
            guess = num
    return guess

Note: For the problem-specific concept of splitting an array into subarrays based on the sum, I’ve included that within the ‘Helper Function’ drill. This concept is a combination of Array Iteration, Subarray Sums, and Comparing and updating variables concepts, which have been addressed in their own drills. Thus, practicing these drills will indirectly provide practice for this concept too.