LeetCode 410: Split Array Largest Sum Solution in Python – A Step-by-Step Guide

What Is LeetCode 410: Split Array Largest Sum?

In LeetCode 410: Split Array Largest Sum, you’re given an array nums of non-negative integers (e.g., [7, 2, 5, 10, 8]) and an integer m. Your task is to split the array into m contiguous subarrays and return the smallest possible largest sum among them. For example, with [7, 2, 5, 10, 8] and m = 2, you could split into [7, 2, 5] (sum 14) and [10, 8] (sum 18), where the largest sum is 18. But can we do better? The goal is to minimize that max sum.

Problem Statement

• Input: An integer array nums, an integer m (number of splits).
• Output: An integer—the minimized largest sum after splitting into m subarrays.
• Rules:
• Split into exactly m contiguous subarrays.
• Minimize the maximum sum among them.

Constraints

• 1 <= nums.length <= 1000.
• 0 <= nums[i] <= 10⁶.
• 1 <= m <= min(50, nums.length).

Examples

• Input: nums = [7,2,5,10,8], m = 2
• Output: 18 (e.g., [7,2,5] and [10,8]).
• Input: nums = [1,2,3,4,5], m = 2
• Output: 9 (e.g., [1,2,3] and [4,5]).
• Input: nums = [1,4,4], m = 3
• Output: 4 (e.g., [1], [4], [4]).

Understanding the Problem: Minimizing the Max Load

To solve LeetCode 410: Split Array Largest Sum in Python, we need to divide nums into m chunks so the biggest chunk’s sum is as small as possible. A naive idea might be to try every possible split—but with up to 1000 elements and multiple partitions, that’s a combinatorial nightmare! Instead, we’ll use:

• Best Solution (Binary Search): O(n log(sum(nums))) time, O(1) space—finds the sweet spot fast.
• Alternative Solution (Dynamic Programming): O(n²m) time, O(nm) space—builds it systematically.

Let’s dive into the binary search solution with a clear, step-by-step explanation.

Best Solution: Binary Search

Why This Is the Best Solution

The binary search method is the top pick because it’s fast—O(n log(sum(nums))) time, O(1) space—and cleverly turns a partitioning problem into a guessing game. It searches for the smallest possible max sum by testing midpoints between the array’s max element and total sum, adjusting based on how many splits that guess allows. It’s like guessing the perfect weight limit for your friends’ book piles and tweaking until it fits just right!

How It Works

Think of splitting the array as setting a weight limit:

• Step 1: Define the Search Range:
• Lower bound: max element in nums (e.g., 10 in [7,2,5,10,8])—can’t split below this.
• Upper bound: sum of nums (e.g., 32)—one big pile.
• Step 2: Guess a Max Sum:
• Pick the midpoint (e.g., (10 + 32) // 2 = 21).
• Count how many subarrays fit under this limit (greedy split).
• Step 3: Adjust the Guess:
• Too many splits (> m)? Guess higher (left = mid + 1).
• Too few splits (< m)? Guess lower (right = mid).
• Just right (≤ m)? Try lower to minimize (right = mid).
• Step 4: Why This Works:
• Binary search narrows down to the smallest valid max sum.
• Greedy splitting ensures we use the least splits possible for a given limit.
• It’s like dialing in the perfect load balance with quick guesses!

Step-by-Step Example

Example: nums = [7,2,5,10,8], m = 2

• Range: Left = 10 (max element), Right = 32 (sum).
• Guess 21:
• Split: [7,2,5] (14), [10,8] (18) → 2 subarrays.
• Max sum = 18 ≤ 21, splits = 2 ≤ m, try lower, right = 21.
• Guess 15:
• Split: [7,2] (9), [5] (5), [10] (10), [8] (8) → 4 subarrays.
• Splits > m, left = 16.
• Guess 18:
• Split: [7,2,5] (14), [10,8] (18) → 2 subarrays.
• Max sum = 18 ≤ 18, splits = 2 = m, try lower, right = 18.
• Guess 17:
• Split: [7,2,5] (14), [10] (10), [8] (8) → 3 subarrays.
• Splits > m, left = 18.
• Converge: Left = 18, Right = 18, answer = 18.
• Result: 18.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down so you can follow every step:

Copy codeclass Solution:
    def splitArray(self, nums: List[int], m: int) -> int:
        # Step 1: Define search bounds
        left = max(nums)  # Min possible max sum
        right = sum(nums) # Max possible max sum

        # Step 2: Binary search
        while left < right:
            mid = (left + right) // 2  # Guess a max sum

            # Count subarrays with sum <= mid
            count = 1  # Number of subarrays
            curr_sum = 0  # Current subarray sum
            for num in nums:
                if curr_sum + num <= mid:
                    curr_sum += num
                else:
                    count += 1
                    curr_sum = num

            # Step 3: Adjust bounds
            if count <= m:
                right = mid  # Try a smaller max sum
            else:
                left = mid + 1  # Need a larger max sum

        return left

• Line 4-5: Set bounds:
• left: Max element (e.g., 10)—smallest possible max sum.
• right: Total sum (e.g., 32)—largest possible max sum.
• Line 8-19: Binary search loop:
• Line 9: Guess midpoint (e.g., 21).
• Line 12-17: Count subarrays:
• Start with 1 subarray, sum = 0.
• Add numbers; if exceeds mid, start new subarray (e.g., 14 ≤ 21, 24 > 21 → new).
• Line 20-23: Adjust:
• count ≤ m: Mid works, try lower (right = mid).
• count > m: Need more room, increase (left = mid + 1).
• Line 25: Return left (converged smallest valid sum).
• Time Complexity: O(n log(sum(nums)))—binary search over sum range, n per check.
• Space Complexity: O(1)—just a few variables.

This is like guessing the perfect pile size with a clever tweak each time!

Alternative Solution: Dynamic Programming

Why an Alternative Approach?

The dynamic programming (DP) method builds the solution by computing the minimum max sum for each prefix and number of splits. It’s O(n²m) time and O(nm) space—slower but shows a systematic buildup. It’s like planning every possible split and picking the best one step-by-step!

How It Works

Picture it as a table of options:

• Step 1: DP[i][j] = min max sum for first i elements split into j subarrays.
• Step 2: Fill table by trying all split points.
• Step 3: Final answer in DP[n][m].

Step-by-Step Example

Example: nums = [7,2,5], m = 2

• DP Table:
• DP[1][1] = 7.
• DP[2][1] = 9, DP[2][2] = 7 (split [7],[2]).
• DP[3][1] = 14, DP[3][2] = 9 (split [7,2],[5]).
• Result: 9.

Code for DP Approach

Copy codeclass Solution:
    def splitArray(self, nums: List[int], m: int) -> int:
        n = len(nums)
        # DP[i][j] = min max sum for first i elements in j subarrays
        dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]

        # Prefix sums for efficiency
        prefix = [0] * (n + 1)
        for i in range(n):
            prefix[i + 1] = prefix[i] + nums[i]

        # Base case: 1 subarray
        for i in range(1, n + 1):
            dp[i][1] = prefix[i]

        # Fill DP table
        for i in range(2, n + 1):
            for j in range(2, min(i + 1, m + 1)):
                for k in range(i - 1):
                    curr_max = max(dp[k][j - 1], prefix[i] - prefix[k])
                    dp[i][j] = min(dp[i][j], curr_max)

        return dp[n][m]

• Time Complexity: O(n²m)—n states, n splits, m subarrays.
• Space Complexity: O(nm)—DP table.

It’s a methodical split planner!

Comparing the Two Solutions

• Binary Search (Best):
• Pros: O(n log(sum(nums))), O(1), fast and lean.
• Cons: Less intuitive.
• Dynamic Programming (Alternative):
• Pros: O(n²m), O(nm), systematic.
• Cons: Slower, more space.

Binary search wins for speed.

Additional Examples and Edge Cases

• [1], 1: 1.
• [1,2], 2: 2.
• [10,20], 1: 30.

Binary search handles all.

Complexity Breakdown

• Binary Search: Time O(n log(sum(nums))), Space O(1).
• DP: Time O(n²m), Space O(nm).

Binary’s the champ.

Key Takeaways

• Binary Search: Guess smart!
• DP: Build it up!
• Splits: Balance is key.
• Python Tip: Loops rock—see [Python Basics](/python/basics).

Final Thoughts: Split That Load

LeetCode 410: Split Array Largest Sum in Python is an array-partitioning quest. Binary search nails it fast, while DP builds it steady. Want more array fun? Try LeetCode 53: Maximum Subarray or LeetCode 287: Find the Duplicate Number. Ready to split? Head to Solve LeetCode 410 on LeetCode and lighten that load today!