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impact: "High" nr: false confidence: 4

📦 Binary Search: Split Array Largest Sum

📝 Problem Description

Given an array nums consisting of non-negative integers and an integer k, split the array into k or fewer contiguous subarrays such that the largest sum among these subarrays is minimized.

Return the minimized largest sum.

LeetCode 410

Real-World Application

Used in Workload Partitioning and Distributed Systems. For example, dividing tasks among k servers such that the maximum load on any server is minimized, ensuring balanced performance.

🛠️ Constraints & Edge Cases

  • \(1 \le nums.length \le 10^3\)
  • \(0 \le nums[i] \le 10^6\)
  • \(1 \le k \le \min(50, nums.length)\)

  • Edge Cases to Watch:

  • \(k = 1\) → entire array is one subarray → answer = \(\sum nums\)

  • \(k = nums.length\) → each element is its own subarray → answer = \(\max(nums)\)
  • Large values in nums → must avoid brute force

🧠 Approach & Intuition

The Aha! Moment

This is a Binary Search on the Answer problem. The maximum subarray sum is monotonic: if we can split the array with max sum ≤ x, then we can also do it for any x' > x.

🐢 Brute Force (Naive)

Try all possible ways to split the array into k parts and compute the largest sum.

  • Time Complexity: Exponential
  • Why it fails: Too many possible partitions → infeasible

🐇 Optimal Approach

Use Binary Search on the answer:

  1. Search space:

  2. low = max(nums) (minimum possible largest sum)

  3. high = sum(nums) (maximum possible largest sum)

  4. While low < high:

  5. Compute mid

  6. Check if we can split into ≤ k subarrays such that no subarray sum exceeds mid
  7. If possible → try smaller (high = mid)
  8. Else → increase (low = mid + 1)
  9. Return low

🧩 Visual Tracing

graph TD
    A["Range: max(nums) to sum(nums)"] --> B{"Check max sum mid"}
    B -->|Can split into ≤ k| C["Search Left: minimize further"]
    B -->|Need > k splits| D["Search Right: increase limit"]
    style C fill:#ccffcc,stroke:#00aa00
    style D fill:#ffcccc,stroke:#aa0000

💻 Solution Implementation

(Implementation details need to be added...)

⏱️ Complexity Analysis

  • Time Complexity: \(\mathcal{O}(N \log(\sum nums))\) — Binary search over range, each check is linear
  • Space Complexity: \(\mathcal{O}(1)\) — Constant extra space

🎤 Interview Toolkit

  • Greedy Insight: Always pack elements into current subarray until exceeding limit, then split
  • Why max(nums)? The largest element must be in some subarray → lower bound
  • Why sum(nums)? No split case → upper bound
  • Pattern Recognition: Classic minimize maximum → think binary search