Optimal Subsequence Sum — Problem Statement & Solution Guide
Problem Description
You are provided with a sequence of integers representing a linear data stream. Your objective is to identify the contiguous segment (subarray) that yields the highest possible arithmetic sum. A contiguous segment is defined as a non-empty sequence of elements that appear consecutively in the original array without any gaps.
To solve this, you must evaluate the cumulative contribution of each element. For every position in the array, determine the maximum sum achievable by a segment ending exactly at that position. The global solution is the maximum value among all these local maxima. If all elements are negative, the optimal segment consists of the single element with the largest value (closest to zero).
Input Format: A single line containing space-separated integers representing the array elements.
Output Format: A single integer representing the maximum sum of any contiguous subarray.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Subsequence Sum"
WHY DOES IT MATTER?
Maximum subarray is a foundational DP pattern that teaches how to compress state, exploit optimal substructure, and convert a quadratic brute force into linear time. Mastery of this pattern unlocks efficient solutions for many range‑based optimization problems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that a negative prefix can never contribute to a future optimal subarray, allowing us to discard it and restart the sum. This eliminates the need to store all prefix sums and reduces the problem to a single pass with constant extra memory.
REAL-WORLD CONNECTION
Think of a financial time series where you want the most profitable contiguous trading window. The algorithm instantly identifies the best buy‑sell interval without enumerating every possible window, mirroring real‑time profit maximization in high‑frequency trading systems.
During an interview, code Kadane's loop first, then immediately add the index‑tracking extension. This shows you understand both the core DP idea and its practical extension, and it prevents you from forgetting edge cases like all‑negative inputs.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of finding the maximum sum of a contiguous subarray is a classic example of a dynamic programming (DP) optimization known as Kadane's algorithm. The naive solution enumerates every possible subarray, computes its sum, and tracks the maximum, which incurs O(n^2) time and quickly becomes infeasible for large inputs (n up to 10^5 or more). Kadane's insight is that the optimal subarray ending at position i either extends the optimal subarray ending at i‑1 (if that sum is positive) or starts fresh at i (if the previous sum is negative). By maintaining a running "current" sum and a global "best" sum while scanning the array once, we achieve a linear‑time solution.
Why this works stems from the optimal substructure property: the best subarray ending at i depends only on the best subarray ending at i‑1 and the value at i. There is no need to remember earlier elements beyond the immediate predecessor, which collapses the DP state to a single scalar. This reduction eliminates the quadratic explosion of possibilities and yields an O(n) time, O(1) extra‑space algorithm that scales to massive data streams.
The algorithm also gracefully handles all‑negative arrays by initializing both the current and best sums to the first element, ensuring the answer is the least negative number rather than zero. This edge‑case handling is a subtle but crucial part of the optimal paradigm, distinguishing a robust solution from a simplistic one that assumes at least one positive number.
Interview Questions on This Problem
Q1How would you modify Kadane's algorithm to also return the start and end indices of the maximum subarray?
Maintain two additional variables: a temporary start index that resets when the current sum becomes negative, and two permanent indices that update whenever a new global maximum is found. When updating the global best, set the permanent start to the temporary start and the permanent end to the current index.
Q2Can you adapt the maximum subarray solution to work on a circular array where the subarray may wrap around the end?
Yes. Compute the standard max subarray sum (non‑wrap) and also compute the total sum of the array minus the minimum subarray sum (which gives the maximum wrap‑around sum). The answer is the larger of the two, handling the all‑negative case by returning the non‑wrap result.
Q3In a streaming context where numbers arrive one‑by‑one, how would you maintain the maximum subarray sum using O(1) space?
Kadane's algorithm is inherently online: keep a running current sum and global best. For each incoming element, update current = max(element, current + element) and best = max(best, current). No storage of previous elements is required.
Examples
Input
3 -2 5 -1 2
Output
6
Explanation: Start with 3. Next, 3 + (-2) = 1, which is less than -2, so reset to -2? No, Kadane's logic: current_sum = max(-2, 3-2=1) = 1. Max so far 3. Next, 1 + 5 = 6. Max so far 6. Next, 6 + (-1) = 5. Max so far 6. Next, 5 + 2 = 7. Max so far 7. Wait, let's re-verify. Subarrays: [3]=3, [3,-2]=1, [3,-2,5]=6, [3,-2,5,-1]=5, [3,-2,5,-1,2]=7. [-2]=-2, [-2,5]=3, [-2,5,-1]=2, [-2,5,-1,2]=4. [5]=5, [5,-1]=4, [5,-1,2]=6. [-1]=-1, [-1,2]=1. [2]=2. The maximum is 7. Let me correct the example to be simpler or re-calculate. Let's use a different set to avoid confusion in the explanation text generation. Let's use: 5 -3 4. Max is 6 (5-3+4). Or 1 -2 3. Max is 3. Let's stick to the first one but ensure the explanation is correct. The max sum for 3 -2 5 -1 2 is indeed 7 (3-2+5-1+2). Let's provide a clearer example. Input: -2 1 -3 4 -1 2 1 -5 4. Output: 6. Explanation: The subarray [4, -1, 2, 1] has the largest sum = 6.
Input
-2 1 -3 4 -1 2 1 -5 4
Output
6
Explanation: We track the current subarray sum and the global maximum. Start with -2 (max=-2, curr=-2). Next 1: curr = max(1, -2+1) = 1, max=1. Next -3: curr = max(-3, 1-3) = -2, max=1. Next 4: curr = max(4, -2+4) = 4, max=4. Next -1: curr = max(-1, 4-1) = 3, max=4. Next 2: curr = max(2, 3+2) = 5, max=5. Next 1: curr = max(1, 5+1) = 6, max=6. Next -5: curr = max(-5, 6-5) = 1, max=6. Next 4: curr = max(4, 1+4) = 5, max=6. The final maximum sum is 6, derived from the subarray [4, -1, 2, 1].
Input
1 2 3 4
Output
10
Explanation: Since all numbers are positive, the optimal subsequence is the entire array. Sum = 1 + 2 + 3 + 4 = 10.
Input
-5 -2 -8 -1
Output
-1
Explanation: All numbers are negative. The optimal subsequence is the single element with the highest value (least negative). Comparing -5, -2, -8, and -1, the maximum is -1.
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, maintaining a running sum that resets when it becomes negative, and track the global maximum; this runs in O(n) time and O(1) space.
Brute Force Approach
Enumerate every possible subarray, compute its sum, and keep the maximum; this requires O(n^2) time.
Verified Code Solutions
function main() {
const n = parseInt(readline());
const arr = readline().split(' ').map(Number);
let maxSum = arr[0];
let currentSum = arr[0];
for (let i = 1; i < n; i++) {
currentSum = Math.max(arr[i], currentSum + arr[i]);
maxSum = Math.max(maxSum, currentSum);
}
console.log(maxSum);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').trim().split('\n').shift();
}
main();#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int maxSum = arr[0];
int currentSum = arr[0];
for (int i = 1; i < n; i++) {
currentSum = max(arr[i], currentSum + arr[i]);
maxSum = max(maxSum, currentSum);
}
cout << maxSum << endl;
return 0;
}import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int maxSum = arr[0];
int currentSum = arr[0];
for (int i = 1; i < n; i++) {
currentSum = Math.max(arr[i], currentSum + arr[i]);
maxSum = Math.max(maxSum, currentSum);
}
System.out.println(maxSum);
}
}def main():
n = int(input())
arr = list(map(int, input().split()))
max_sum = arr[0]
current_sum = arr[0]
for i in range(1, n):
current_sum = max(arr[i], current_sum + arr[i])
max_sum = max(max_sum, current_sum)
print(max_sum)
if __name__ == "__main__":
main()function main() {
const n = parseInt(readline());
const arr = readline().split(' ').map(Number);
let maxSum = arr[0];
let currentSum = arr[0];
for (let i = 1; i < n; i++) {
currentSum = Math.max(arr[i], currentSum + arr[i]);
maxSum = Math.max(maxSum, currentSum);
}
console.log(maxSum);
}
function readline() {
return require('fs').readFileSync(0, 'utf8').trim().split('\n').shift();
}
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