Segment Horizon Partition Analyzer 4 — Problem Statement & Solution Guide

Binary SearchHardBinary Search on Answer Matrix

TimeO(N * log(S))

SpaceO(1)

Problem Description

You are tasked with optimizing the partitioning of a linear sequence of sensor readings to minimize the maximum segment load. Given an array of integers representing discrete data points, you must divide the array into exactly K contiguous non-empty segments. The cost of a segment is defined as the sum of its elements. Your objective is to find the minimum possible value for the maximum segment sum across all valid partitions.

This problem requires determining the optimal threshold for segment capacity. You are provided with an array arr of length N and an integer K. You need to compute the smallest integer M such that it is possible to partition arr into K contiguous subarrays where the sum of elements in each subarray does not exceed M. If no such partition exists for a given M, the partition is invalid. The solution involves finding the minimal feasible M that satisfies the partitioning constraint.

Input consists of an integer array arr and an integer K. Output is a single integer representing the minimum possible maximum segment sum. The partition must cover all elements of the array exactly once, and each segment must be contiguous.

DSA Pattern Breakdown

"Segment Horizon Partition Analyzer 4"

hard

WHY DOES IT MATTER?

This pattern—binary search on answer space combined with a linear feasibility check—is a cornerstone for many partitioning, allocation, and load‑balancing problems. It turns an otherwise exponential combinatorial search into a logarithmic number of cheap scans, enabling solutions that scale to millions of items.

OPTIMIZATION CHALLENGE

The key insight is recognizing the monotonic predicate: as the allowed maximum segment sum grows, the required number of segments never increases. This property permits binary search, collapsing the search space from exponential to O(log(range)).

REAL-WORLD CONNECTION

Think of distributing files across K servers to keep the busiest server's storage low. The greedy feasibility test mimics a real‑world scheduler that fills a server until adding another file would exceed a threshold, then moves to the next server. Binary search finds the smallest threshold that still fits all files.

During an interview, first write the greedy feasibility function and test it manually on edge cases; then wrap it in a binary search loop. Keep the loop invariant clear (low works, high may not) to avoid off‑by‑one bugs.

COMPLEXITY AT A GLANCE

⏱ Time:O(N * log(S))

💾 Space:O(1)

Core Theory — Why This Approach?

The problem is a classic instance of the "split array largest sum" optimization, which can be modeled as a decision problem: given a candidate maximum segment sum X, can we partition the array into at most K contiguous sub‑arrays such that each sub‑array sum does not exceed X? This decision can be answered greedily in linear time by scanning the array and starting a new segment whenever the running sum would surpass X. The monotonic nature of the decision (if X works, any larger X also works) enables a binary search over the answer space, reducing the exponential search space of all possible partitions to a logarithmic number of feasibility checks.

A naive exhaustive search would try every way to place K‑1 cuts among N‑1 possible positions, leading to O( C(N‑1, K‑1) ) possibilities – infeasible for N up to 10^5. Dynamic programming can solve the problem in O(N·K) time, but with K potentially large (e.g., up to N) this still exceeds the required limits for hard constraints. The binary‑search‑plus‑greedy paradigm yields O(N·log(S)) time, where S is the sum of all elements, and O(1) extra space, making it optimal for the typical input limits.

The optimal paradigm therefore combines two fundamental algorithmic ideas: (1) a feasibility test that runs in linear time using a greedy scan, and (2) a binary search over a numeric range that leverages the monotonic predicate. This synergy transforms a combinatorial partitioning problem into a tractable numeric optimization.

Interview Questions on This Problem

Q1How would you determine the search bounds for the binary search when minimizing the maximum segment sum?

The lower bound is the maximum single element because no segment can be smaller than any element it contains. The upper bound is the total sum of the array, representing the case of a single segment. Binary search runs between these two inclusive bounds.

Q2Explain why a greedy scan can correctly answer the feasibility predicate for a given candidate X.

When scanning left to right, the greedy choice of extending the current segment as far as possible without exceeding X never harms feasibility: any optimal partition must also respect the X limit, so cutting earlier would only increase the number of segments, never reduce it. Thus the greedy count of required segments is minimal for that X.

Q3If the array contains negative numbers, does the binary‑search‑plus‑greedy approach still work? Why or why not?

No. Negative values break the monotonicity of the feasibility predicate because adding a negative can reduce a segment sum below X, allowing later large positives to be absorbed. The greedy scan may under‑count segments, and the lower bound is no longer the max element. A different DP or transformation is needed for arrays with negatives.

Examples

Example 1

Input

arr = [3, 1, 4, 1, 5], K = 2

Output

Explanation: We need to split [3, 1, 4, 1, 5] into 2 contiguous parts to minimize the max sum. Possible splits: 1. [3, 1, 4] and [1, 5] -> Sums: 8, 6 -> Max: 8 2. [3, 1] and [4, 1, 5] -> Sums: 4, 10 -> Max: 10 3. [3] and [1, 4, 1, 5] -> Sums: 3, 11 -> Max: 11 4. [3, 1, 4, 1] and [5] -> Sums: 9, 5 -> Max: 9 The minimum of the maximums is 8? Wait, let's re-evaluate. Split 1: [3,1,4] sum=8, [1,5] sum=6. Max=8. Is 8 possible? Yes. Is 7 possible? Total sum is 14. If max is 7, we need two segments summing to <=7. 7+7=14. Can we split into 7 and 7? [3,1,4] is 8, too big. [3,1] is 4, next [4,1,5] is 10. No. [3,1,4,1] is 9. No. So 7 is not possible. Thus, the answer is 8.

Example 2

Input

arr = [1, 2, 3, 4, 5], K = 3

Output

Explanation: Total sum is 15. We need 3 segments. Lower bound is max(arr) = 5. Upper bound is sum(arr) = 15. Check M=5: Can we partition into 3 segments with max sum 5? [1,2] sum=3, [3] sum=3, [4] sum=4, [5] sum=5. This uses 4 segments. We need exactly 3. Since we used more than 3 segments, M=5 is too small (we need larger segments to reduce count). Check M=6: [1,2,3] sum=6, [4] sum=4, [5] sum=5. This uses 3 segments. Valid. Thus, the minimum M is 6.

Example 3

Input

arr = [10, 10, 10], K = 1

Output

Explanation: K=1 means the entire array is one segment. The sum is 10+10+10=30. The maximum segment sum is 30. Thus, the answer is 30.

Constraints

1 <= arr.length <= 10^5
1 <= arr[i] <= 10^9
1 <= K <= arr.length
The sum of all elements in arr will not exceed 10^14

Optimal Approach & Strategy

Perform a binary search on the answer range, using a greedy linear scan to test if a candidate maximum segment sum can be achieved with ≤K segments.

Brute Force Approach

Enumerate all ways to place K‑1 cuts among N‑1 positions and compute the maximum segment sum for each partition.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N * log(S))

function solution(nums) {
   if (nums.length === 0) {
       return 0;
   }
   let prefixSum = new Array(nums.length + 1).fill(0);
   for (let i = 0; i < nums.length; i++) {
       prefixSum[i + 1] = prefixSum[i] + nums[i];
   }
   let maxSum = -Infinity;
   for (let i = 0; i < nums.length; i++) {
       for (let j = i; j < nums.length; j++) {
           let sum = prefixSum[j + 1] - prefixSum[i];
           maxSum = Math.max(maxSum, sum);
       }
   }
   return maxSum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       if (nums.size() == 0) {
           return 0;
       }
       vector<int> prefixSum(nums.size() + 1, 0);
       for (int i = 0; i < nums.size(); i++) {
           prefixSum[i + 1] = prefixSum[i] + nums[i];
       }
       int maxSum = INT_MIN;
       for (int i = 0; i < nums.size(); i++) {
           for (int j = i; j < nums.size(); j++) {
               int sum = prefixSum[j + 1] - prefixSum[i];
               maxSum = max(maxSum, sum);
           }
       }
       return maxSum;
   }
};

class Solution {
   public int solution(int[] nums) {
       if (nums.length == 0) {
           return 0;
       }
       int[] prefixSum = new int[nums.length + 1];
       for (int i = 0; i < nums.length; i++) {
           prefixSum[i + 1] = prefixSum[i] + nums[i];
       }
       int maxSum = Integer.MIN_VALUE;
       for (int i = 0; i < nums.length; i++) {
           for (int j = i; j < nums.length; j++) {
               int sum = prefixSum[j + 1] - prefixSum[i];
               maxSum = Math.max(maxSum, sum);
           }
       }
       return maxSum;
   }
}

def solution(nums):
   if not nums:
       return 0
   prefix_sum = [0] * (len(nums) + 1)
   for i in range(len(nums)):
       prefix_sum[i + 1] = prefix_sum[i] + nums[i]
   max_sum = float('-inf')
   for i in range(len(nums)):
       for j in range(i, len(nums)):
           sum_ = prefix_sum[j + 1] - prefix_sum[i]
           max_sum = max(max_sum, sum_)
   return max_sum

function solution(nums) {
   if (nums.length === 0) {
       return 0;
   }
   let prefixSum = new Array(nums.length + 1).fill(0);
   for (let i = 0; i < nums.length; i++) {
       prefixSum[i + 1] = prefixSum[i] + nums[i];
   }
   let maxSum = -Infinity;
   for (let i = 0; i < nums.length; i++) {
       for (let j = i; j < nums.length; j++) {
           let sum = prefixSum[j + 1] - prefixSum[i];
           maxSum = Math.max(maxSum, sum);
       }
   }
   return maxSum;
}

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Segment Horizon Partition Analyzer 4 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Binary Search

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Segment Horizon Partition Analyzer 4 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor