Optimal Range Extent — Problem Statement & Solution Guide
Problem Description
Given a linear array of integers, determine the maximum possible sum of any contiguous subarray. A contiguous subarray is defined as a sequence of elements that are adjacent in the original array, preserving their relative order. The goal is to identify the specific range within the array where the cumulative sum of its elements is maximized.
You are provided with a single array of integers. Your task is to compute the highest aggregate value achievable by selecting a non-empty contiguous segment. If all elements are negative, the result should be the largest (least negative) single element, as the subarray must contain at least one element.
Return the maximum sum as an integer. The solution must efficiently process the array to find this optimal extent without resorting to brute-force enumeration of all possible subarrays.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Range Extent"
WHY DOES IT MATTER?
Maximum subarray is a foundational pattern for any scenario where you need to identify the most profitable contiguous segment, such as profit windows, signal processing peaks, or latency spikes. Mastering this pattern equips engineers to convert quadratic brute‑force scans into linear‑time solutions, a skill that directly impacts performance in large‑scale data pipelines.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that a negative cumulative sum can never contribute to a future optimal subarray, allowing the algorithm to reset the running total instantly. This eliminates the need for nested loops or explicit prefix‑sum arrays, collapsing the problem to constant‑space state updates.
REAL-WORLD CONNECTION
Think of a streaming stock ticker: you want to know the best time window to have bought and sold to maximize profit. Kadane's algorithm mirrors a real‑time monitor that continuously discards losing periods and keeps track of the best observed profit window without storing the entire history.
During an interview, compute the running sum and global max in the same loop, and always keep a temporary start index. If the running sum drops below zero, reset it and move the temporary start to the next element – this tiny detail often differentiates a polished solution from a buggy one.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem of finding the maximum sum of any contiguous subarray is a classic example of a one‑dimensional dynamic programming challenge. A naive solution enumerates every possible start and end index, computing the sum for each subarray, which leads to O(N^2) time and quickly becomes infeasible for large N. The optimal paradigm, known as Kadane's algorithm, leverages the insight that the best subarray ending at position i either extends the best subarray ending at i‑1 or starts fresh at i; this local optimality property enables a single linear pass while maintaining only two scalar variables – the current running sum and the global maximum observed so far. By continuously updating these values, the algorithm implicitly discards sub‑optimal prefixes, guaranteeing that the final global maximum corresponds to the optimal contiguous range without ever revisiting elements.
Kadane's algorithm exemplifies the broader "prefix‑sum reduction" pattern, where the problem is transformed from examining O(N^2) intervals to tracking a running aggregate that captures the essential state. The key mathematical justification is that if the cumulative sum up to index i becomes negative, any future subarray that includes this prefix would be worse than starting after i, so the prefix can be safely dropped. This greedy decision is provably optimal because the subarray sum function is linear and additive, allowing the algorithm to make irrevocable choices without missing the global optimum. Consequently, the solution runs in O(N) time with O(1) auxiliary space, making it suitable for real‑time systems and massive data streams.
Interview Questions on This Problem
Q1How does Kadane's algorithm handle an array where all numbers are negative?
Kadane's algorithm can be initialized with the first element as both the current sum and global max. As it iterates, it updates the current sum to the maximum of the current element and current sum + element, which ensures that the largest (least negative) single element is retained as the answer.
Q2Can you modify the algorithm to also return the start and end indices of the maximum subarray?
Yes. Maintain additional variables for a temporary start index when resetting the current sum, and update the best start/end indices whenever the global max is updated. This yields the exact range in O(N) time.
Q3What is the time‑space trade‑off if you were to solve the problem using a divide‑and‑conquer approach?
The divide‑and‑conquer method runs in O(N log N) time and O(log N) recursion stack space, which is slower than Kadane's O(N) but still linearithmic; it is useful when you need to answer range‑maximum queries on static arrays after preprocessing.
Examples
Input
[-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output
6
Explanation: The contiguous subarray [4, -1, 2, 1] has the maximum sum of 6. Step-by-step: Start with -2 (current= -2, max=-2). Next 1 (current=1, max=1). Next -3 (current=-2, max=1). Next 4 (current=4, max=4). Next -1 (current=3, max=4). Next 2 (current=5, max=5). Next 1 (current=6, max=6). Next -5 (current=1, max=6). Next 4 (current=5, max=6). The global maximum is 6.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: The entire array is the optimal subarray. Step-by-step: Start with 1 (current=1, max=1). Next 2 (current=3, max=3). Next 3 (current=6, max=6). Next 4 (current=10, max=10). Next 5 (current=15, max=15). The global maximum is 15.
Input
[-5, -2, -8, -1]
Output
-1
Explanation: All elements are negative. The optimal subarray is the single element with the highest value. Step-by-step: Start with -5 (current=-5, max=-5). Next -2 (current=-2, max=-2). Next -8 (current=-10, max=-2). Next -1 (current=-1, max=-1). The global maximum is -1.
Input
[3, -1, 2, -1, 5]
Output
8
Explanation: The contiguous subarray [3, -1, 2, -1, 5] has the maximum sum of 8. Step-by-step: Start with 3 (current=3, max=3). Next -1 (current=2, max=3). Next 2 (current=4, max=4). Next -1 (current=3, max=4). Next 5 (current=8, max=8). The global maximum is 8.
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, updating a running sum and a global maximum, resetting the running sum when it becomes negative. This achieves O(N) time and O(1) space.
Brute Force Approach
Enumerate every possible start index, then for each start compute sums for all end indices, tracking the maximum. This double loop yields O(N^2) time and O(1) extra space.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return -1;
let max_sum = nums[0];
let current_sum = nums[0];
for (let i = 1; i < nums.length; i++) {
current_sum = Math.max(nums[i], current_sum + nums[i]);
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}class Solution {
public:
int solution(vector<int> nums) {
if (nums.size() == 0) return -1;
int max_sum = nums[0];
int current_sum = nums[0];
for (int i = 1; i < nums.size(); i++) {
current_sum = max(nums[i], current_sum + nums[i]);
max_sum = max(max_sum, current_sum);
}
return max_sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return -1;
int max_sum = nums[0];
int current_sum = nums[0];
for (int i = 1; i < nums.length; i++) {
current_sum = Math.max(nums[i], current_sum + nums[i]);
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}
}def solution(nums):
if not nums:
return -1
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) {
if (nums.length === 0) return -1;
let max_sum = nums[0];
let current_sum = nums[0];
for (let i = 1; i < nums.length; i++) {
current_sum = Math.max(nums[i], current_sum + nums[i]);
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}Asked in Top Tech Interviews
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