Optimal Interval Partition — Problem Statement & Solution Guide
Problem Description
You are given an array of integers. Your task is to find the maximum possible sum that can be obtained by selecting a contiguous subarray (an interval) of the array. The chosen interval may contain only one element, and it must be a non‑empty segment of the original array. The result is the sum of the elements in that optimal interval.
Input format: The first line contains a single integer n (1 ≤ n ≤ 10^5), the length of the array. The second line contains n space‑separated integers a_1, a_2, …, a_n, each satisfying -10^9 ≤ a_i ≤ 10^9.
Output format: Output a single integer – the maximum subarray sum. The answer is guaranteed to fit in a 64‑bit signed integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Interval Partition"
WHY DOES IT MATTER?
Kadane’s pattern demonstrates how a global optimum can be found through local, greedy decisions, a concept that appears in many algorithmic problems such as maximum subarray, longest increasing subsequence, and many dynamic programming challenges. Understanding this pattern equips engineers to recognize when a problem can be solved in linear time rather than quadratic, which is critical for performance‑sensitive applications.
OPTIMIZATION CHALLENGE
The key insight is that the maximum subarray ending at position i depends only on the maximum ending at i‑1 and the current element. By storing just two numbers—current best ending here and global best—we avoid recomputing sums for overlapping intervals, reducing time from O(n^2) to O(n) and space from O(n) to O(1).
REAL-WORLD CONNECTION
Consider a financial trading system that monitors a stream of price changes. The system needs to identify the most profitable contiguous period in real time. Kadane’s algorithm is analogous to a sliding window that continuously updates the best trade window as new data arrives, enabling instant decision‑making without reprocessing the entire history.
When interviewing, emphasize the algorithm’s linear scan and the decision rule: if the running sum becomes negative, reset it to the current element. This simple rule is the crux of Kadane’s efficiency and is often the moment where candidates either shine or falter.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum sum of any contiguous subarray within a given integer array. A naive solution would examine every possible interval, computing its sum and keeping track of the maximum; this requires O(n^2) time and is infeasible for large n. The optimal solution is Kadane’s algorithm, which processes the array in a single pass, maintaining two values: the best sum ending at the current position and the global maximum. By deciding at each step whether to extend the previous subarray or start a new one, Kadane achieves O(n) time and O(1) space, making it ideal for real‑time analytics and large data streams.
Kadane’s insight is that the maximum subarray ending at index i either includes the element at i and the best subarray ending at i‑1, or it starts fresh at i. This local decision property allows dynamic programming to collapse the exponential search space into a linear scan. The algorithm’s simplicity also makes it a staple in interview questions, as it tests understanding of prefix sums, greedy decisions, and edge‑case handling.
Because the array can contain negative numbers, the algorithm must correctly handle the case where all values are negative by initializing the current sum to the first element and updating the maximum accordingly. This nuance is often missed by candidates who default to resetting the sum to zero, leading to incorrect results for arrays with all negative values.
Interview Questions on This Problem
Q1At Google, how would you explain the difference between Kadane’s algorithm and a brute‑force approach for maximum subarray sum?
Kadane’s algorithm runs in O(n) time by maintaining a running maximum ending at each index, whereas a brute‑force approach would examine all O(n^2) intervals, computing sums via nested loops. The key difference is that Kadane uses a greedy, dynamic‑programming step that decides locally whether to extend or restart the subarray, eliminating the need to recompute sums for overlapping intervals.
Q2A fintech startup asks: if the array represents daily profit/loss, how can Kadane’s algorithm help in real‑time risk monitoring?
By maintaining the current best subarray sum as new daily values stream in, the system can instantly detect the most profitable contiguous period up to the current day. This allows the startup to trigger alerts or adjust strategies when a significant negative streak is detected, all in O(1) per update.
Q3At a high‑growth engineering startup, you’re asked to modify Kadane’s algorithm to also return the start and end indices of the optimal subarray. How would you implement this?
Track a temporary start index when resetting the current sum to the current element. Whenever the current sum exceeds the global maximum, update the global start and end indices to the temporary start and current index. This adds O(1) extra bookkeeping without affecting time or space complexity.
Examples
Input
5 -2 1 -3 4 -1
Output
4
Explanation: The subarray consisting of the single element 4 yields the largest sum. All other contiguous segments produce sums of 3, 2, 1, 0, -1, -2, -3, -4, or -5, which are smaller.
Input
8 1 -2 3 4 -5 8 -1 2
Output
11
Explanation: Starting at the third element, the cumulative sums are 3, 7, 2, 10, 9, 11. The maximum of these is 11, achieved by the subarray [3, 4, -5, 8, -1, 2].
Input
4 -5 -1 -3 -2
Output
-1
Explanation: All numbers are negative, so the best choice is the least negative single element, -1.
Input
6 2 3 -1 2 4 -5
Output
10
Explanation: The cumulative sums from the first element are 2, 5, 4, 6, 10, 5. The maximum is 10, obtained by the subarray [2, 3, -1, 2, 4].
Constraints
- 1 <= n <= 10^5
- -10^9 <= a_i <= 10^9
- The answer fits in a 64‑bit signed integer
- The array may contain all negative numbers
- The array may contain all positive numbers
Optimal Approach & Strategy
Traverse the array once, maintaining a running sum that resets to the current element if it becomes negative, and update the global maximum accordingly. This yields O(n) time and O(1) space.
Brute Force Approach
Check every possible subarray by using two nested loops, compute its sum, and track the maximum. This requires O(n^2) time and O(1) additional space.
Verified Code Solutions
function solution(nums) {
let maxSum = nums[0];
let currentSum = nums[0];
for (let i = 1; i < nums.length; i++) {
currentSum = Math.max(nums[i], currentSum + nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
int maxSum = nums[0];
int currentSum = nums[0];
for (int i = 1; i < nums.size(); i++) {
currentSum = max(nums[i], currentSum + nums[i]);
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = nums[0];
int currentSum = nums[0];
for (int i = 1; i < nums.length; i++) {
currentSum = Math.max(nums[i], currentSum + nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
max_sum = nums[0]
current_sum = nums[0]
for i in range(1, len(nums)):
current_sum = max(nums[i], current_sum + nums[i])
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
let maxSum = nums[0];
let currentSum = nums[0];
for (let i = 1; i < nums.length; i++) {
currentSum = Math.max(nums[i], currentSum + nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}Asked in Top Tech Interviews
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