Optimal Node Cluster — Problem Statement & Solution Guide
Problem Description
In a linear array of integer values representing discrete signal intensities, identify the contiguous subsequence that maximizes the cumulative sum. The subsequence must contain at least one element. Given an array nums of length n, compute the maximum sum obtainable from any non-empty contiguous subarray.
Input: An array nums of integers.
Output: A single integer representing the maximum subarray sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Node Cluster"
WHY DOES IT MATTER?
Maximum subarray is a classic example of the “optimal substructure + greedy choice” pattern, teaching candidates how to reduce a combinatorial search to a linear scan, a skill that recurs in many DP and sliding‑window problems.
OPTIMIZATION CHALLENGE
The key insight is that a negative cumulative sum can never contribute to a future optimum, so you can safely discard it and restart the running total, collapsing O(n^2) possibilities into two scalar variables.
REAL-WORLD CONNECTION
In signal processing, the algorithm finds the strongest burst of energy in a noisy waveform, analogous to detecting a peak traffic surge in a time‑series of server requests.
During an interview, write the update as curr = max(num, curr + num) and best = max(best, curr); this one‑liner captures the entire logic and reduces the chance of off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum sum of any non‑empty contiguous subarray. A naive O(n^2) scan enumerates every possible start and end index, accumulating sums, which quickly becomes prohibitive for n up to 10^5 or more. The optimal paradigm is Kadane’s algorithm, a dynamic‑programming technique that processes the array in a single left‑to‑right pass, maintaining the best subarray ending at the current position and the global maximum seen so far. By deciding at each element whether to extend the previous subarray or start a new one, the algorithm collapses the exponential state space into two constant‑size variables, achieving linear time and constant extra space. The correctness follows from the optimal substructure property: any optimal subarray ending at index i either includes the optimal subarray ending at i‑1 (if its sum is positive) or starts fresh at i (if the previous sum is negative). This greedy choice is provably optimal because discarding a negative prefix can never improve the sum of a later subarray, allowing the algorithm to update the global maximum in O(1) per element.
Interview Questions on This Problem
Q1How would you modify Kadane’s algorithm to also return the start and end indices of the maximum subarray?
Track two additional indices: when you start a new subarray at i, set a temporary start = i; when you update the global maximum, record the current temporary start and i as the answer range.
Q2Can Kadane’s algorithm handle arrays that contain only negative numbers? Explain the result.
Yes. The algorithm initializes both current and global max with the first element, so it will correctly return the largest (least negative) single element when all numbers are negative.
Q3What is the time‑space trade‑off if you need to compute the maximum subarray sum for every possible prefix of the array?
You can compute prefix‑wise maximums in O(n) time and O(1) extra space by maintaining the Kadane state while iterating; storing each prefix result requires O(n) additional space for the output array.
Examples
Input
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output
6
Explanation: The subarray [4, -1, 2, 1] yields the maximum sum of 6. Other candidates like [4] sum to 4, and [-1, 2, 1] sum to 2, which are lower.
Input
nums = [1, 2, 3, 4]
Output
10
Explanation: Since all elements are positive, the entire array constitutes the optimal subarray. The sum is 1 + 2 + 3 + 4 = 10.
Input
nums = [-1, -2, -3]
Output
-1
Explanation: All elements are negative. The maximum sum is the least negative element, which is -1. Any longer subarray would result in a smaller (more negative) sum.
Input
nums = [5, -3, 2, -1, 4]
Output
7
Explanation: The subarray [5, -3, 2, -1, 4] sums to 7. Alternatively, [2, -1, 4] sums to 5, and [5] sums to 5. The global maximum is 7.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The answer is guaranteed to fit in a 64-bit integer.
Optimal Approach & Strategy
Use Kadane’s algorithm: iterate once, keep a running sum that resets to the current element when it becomes negative, and track the global maximum. This yields O(n) time and O(1) extra space.
Brute Force Approach
Enumerate every possible start index, then for each start compute sums for all end indices, updating the maximum; this requires two nested loops. The time complexity is O(n^2) and quickly times out for large n.
Verified Code Solutions
function maxSubArray(nums) {
let maxSoFar = nums[0];
let cur = nums[0];
for (let i = 1; i < nums.length; i++) {
cur = Math.max(nums[i], cur + nums[i]);
maxSoFar = Math.max(maxSoFar, cur);
}
return maxSoFar;
}#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int maxSoFar = nums[0];
int cur = nums[0];
for (size_t i = 1; i < nums.size(); ++i) {
cur = max(nums[i], cur + nums[i]);
maxSoFar = max(maxSoFar, cur);
}
return maxSoFar;
}
};
public class Solution {
public int maxSubArray(int[] nums) {
int maxSoFar = nums[0];
int cur = nums[0];
for (int i = 1; i < nums.length; i++) {
cur = Math.max(nums[i], cur + nums[i]);
maxSoFar = Math.max(maxSoFar, cur);
}
return maxSoFar;
}
}
class Solution:
def maxSubArray(self, nums):
"""Kadane's algorithm: O(n) time, O(1) space."""
max_so_far = nums[0]
cur = nums[0]
for num in nums[1:]:
cur = max(num, cur + num)
max_so_far = max(max_so_far, cur)
return max_so_far
function maxSubArray(nums) {
let maxSoFar = nums[0];
let cur = nums[0];
for (let i = 1; i < nums.length; i++) {
cur = Math.max(nums[i], cur + nums[i]);
maxSoFar = Math.max(maxSoFar, cur);
}
return maxSoFar;
}
Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.