Segment Horizon Partition Analyzer — Problem Statement & Solution Guide

Binary SearchHardBinary Search on Answer Matrix

TimeO(n)

SpaceO(1)

Problem Description

You are given an array of integers nums. Your task is to determine the maximum integer X that satisfies the inequality 2 * X <= sum(nums). This problem can be framed as finding the upper bound of a feasible integer range where the value X is constrained by half the total sum of the array elements. The solution requires identifying the largest integer that does not exceed the floor of half the sum.

The input consists of a single array nums of length n. The output should be a single integer representing the maximum valid X. Note that X must be an integer, and the condition 2 * X <= sum(nums) must hold strictly. If the sum of the array is negative, the maximum integer X will also be negative, specifically the floor of sum(nums) / 2.

This problem is a classic application of binary search on the answer space. The search space for X is bounded by the minimum and maximum possible values derived from the array sum. By leveraging the monotonic property of the condition 2 * X <= sum(nums), we can efficiently narrow down the search space to find the optimal X in logarithmic time relative to the value range.

DSA Pattern Breakdown

"Segment Horizon Partition Analyzer"

hard

WHY DOES IT MATTER?

This pattern demonstrates that sometimes the optimal solution is a direct mathematical reduction rather than a search or DP. Recognizing such reductions saves time and avoids overengineering.

OPTIMIZATION CHALLENGE

The key insight is that the inequality 2*X <= sum(nums) is equivalent to X <= sum(nums)/2, so the maximum integer X is the floor of that division. This eliminates the need for any search or iterative refinement.

REAL-WORLD CONNECTION

Think of budgeting: to split a total budget evenly between two departments, you simply divide the total by two. No need to iterate over all possible allocations.

When explaining this to an interviewer, emphasize the mathematical equivalence and point out that the only required operation is a single sum and a division, which is O(n) and O(1).

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to finding the largest integer X such that 2*X does not exceed the total sum of the array. Mathematically this is equivalent to computing the floor of half the sum, i.e., X = ⌊sum(nums)/2⌋. A naive approach might iterate over all possible X values or use a binary search over a range of potential X values, but both would add unnecessary overhead when the answer can be derived in a single pass. The optimal paradigm is a linear scan to accumulate the sum, followed by a constant‑time integer division. This yields an O(n) time complexity and O(1) auxiliary space, which is optimal because every element must be examined at least once to compute the sum.

Interview Questions on This Problem

Q1At Google, how would you explain the time complexity of this problem and why a binary search is unnecessary?

I would state that the answer is simply floor(total_sum/2), so the algorithm runs in O(n) time to compute the sum and O(1) space. A binary search would add log(n) factor and is redundant because the solution is deterministic from the sum.

Q2What edge case should a candidate consider when the array contains negative numbers?

Even with negative numbers, the inequality 2*X <= sum(nums) still holds, so X can be negative. The algorithm must correctly compute floor division for negative sums, which in Python is achieved by integer division //.

Q3In a fintech platform interview, how would you justify using a single pass sum over a more complex DP approach?

I would argue that the problem is a direct arithmetic property; DP would introduce unnecessary state and memory. A single pass sum is both simpler to implement and guarantees correctness for any input size.

Examples

Example 1

Input

nums = [1, 2, 3, 4, 5]

Output

Explanation: The sum of `nums` is 1 + 2 + 3 + 4 + 5 = 15. We need the maximum integer `X` such that `2 * X <= 15`. Testing `X = 7`: `2 * 7 = 14 <= 15` (valid). Testing `X = 8`: `2 * 8 = 16 > 15` (invalid). Thus, the maximum valid `X` is 7? Wait, 15/2 = 7.5. Floor is 7. Let me re-check. 2*7=14<=15. 2*8=16>15. So X=7. My previous thought said 5, that was wrong. Let's correct the example output to 7.

Example 2

Input

nums = [10, 20, 30]

Output

Explanation: The sum of `nums` is 10 + 20 + 30 = 60. We need the maximum integer `X` such that `2 * X <= 60`. This simplifies to `X <= 30`. The maximum integer satisfying this is 30. Verification: `2 * 30 = 60 <= 60` (valid). `2 * 31 = 62 > 60` (invalid). Thus, the answer is 30.

Example 3

Input

nums = [-5, -10, -15]

Output

-15

Explanation: The sum of `nums` is -5 + (-10) + (-15) = -30. We need the maximum integer `X` such that `2 * X <= -30`. This simplifies to `X <= -15`. The maximum integer satisfying this is -15. Verification: `2 * (-15) = -30 <= -30` (valid). `2 * (-14) = -28 > -30` (invalid). Thus, the answer is -15.

Example 4

Input

nums = [1, 1, 1, 1, 1, 1, 1]

Output

Explanation: The sum of `nums` is 7. We need the maximum integer `X` such that `2 * X <= 7`. This simplifies to `X <= 3.5`. The maximum integer satisfying this is 3. Verification: `2 * 3 = 6 <= 7` (valid). `2 * 4 = 8 > 7` (invalid). Thus, the answer is 3.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The sum of nums can be as large as 10^14 in absolute value
The answer X must be an integer

Optimal Approach & Strategy

Compute the sum of the array in O(n) time, then return sum(nums)//2, which is O(1) time and O(1) space.

Brute Force Approach

A brute force method would try every integer X from 0 up to sum(nums) and check if 2*X <= sum(nums), which takes O(sum(nums)) time and is infeasible for large sums.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums) {
    if (!nums || nums.length === 0) return 0;
    const sum = nums.reduce((acc, val) => acc + val, 0);
    return Math.floor(sum / 2);
}

#include <vector>

class Solution {
public:
    long long solution(const std::vector<int>& nums) {
        if (nums.empty()) return 0;
        long long sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return sum / 2;
    }
};

class Solution {
    public long solution(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        long sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return sum / 2;
    }
}

def solution(nums):
    if not nums:
        return 0
    return sum(nums) // 2

function solution(nums) {
    if (!nums || nums.length === 0) return 0;
    const sum = nums.reduce((acc, val) => acc + val, 0);
    return Math.floor(sum / 2);
}

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Segment Horizon Partition Analyzer Solution

Problem Statement

Input

Output

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 — 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