How do you solve Split Array Largest Sum on LeetCode?

Split Array Largest Sum (LeetCode #410) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(sum)) time complexity and O(1) space complexity. Key insight: The largest sum of any subarray must be at least the maximum element in the array.

What is the brute force solution for Split Array Largest Sum?

The brute force approach for Split Array Largest Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to split the array into k subarrays and calculates the largest sum for each split. It then selects the minimum of these largest sums. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log(sum)).

What are the interview tips for Split Array Largest Sum?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process clearly while coding; it shows your understanding. Practice writing both brute force and optimal solutions to strengthen your problem-solving skills.

What are common mistakes when solving Split Array Largest Sum?

Students often forget to check if the number of splits exceeds k during the greedy check. Not considering edge cases where k equals the length of the array.

#410

Split Array Largest Sum

Hard

ArrayBinary SearchDynamic ProgrammingGreedyPrefix SumBinary SearchGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(sum))
Space	O(1)	O(1)

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Intuition

Time O(n log(sum))Space O(1)

The optimal approach uses binary search combined with a greedy strategy. We search for the minimum possible largest sum that can be achieved by splitting the array into k subarrays. This is efficient because it narrows down the possible sums quickly.

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Algorithm

4 steps

1Step 1: Set the left boundary of the binary search to the maximum element in nums and the right boundary to the sum of all elements in nums.
2Step 2: While left is less than right, calculate the mid-point and check if it's possible to split the array into k subarrays with this mid-point as the largest sum.
3Step 3: If it's possible, move the right boundary to mid; otherwise, move the left boundary to mid + 1.
4Step 4: Return the left boundary, which will be the minimized largest sum.

solution.py22 lines

1# Full working Python code
2def canSplit(nums, maxSum, k):
3    currentSum = 0
4    count = 1
5    for num in nums:
6        currentSum += num
7        if currentSum > maxSum:
8            currentSum = num
9            count += 1
10            if count > k:
11                return False
12    return True
13
14def splitArray(nums, k):
15    left, right = max(nums), sum(nums)
16    while left < right:
17        mid = (left + right) // 2
18        if canSplit(nums, mid, k):
19            right = mid
20        else:
21            left = mid + 1
22    return left

ℹ

Complexity note: The time complexity is O(n log(sum)), where n is the length of the array and sum is the total sum of the array. The log(sum) comes from the binary search over possible sums, and O(n) is for checking if a split is possible.

1The largest sum of any subarray must be at least the maximum element in the array.
2The minimized largest sum can never be less than the average of the total sum divided by k.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.