Maximized Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the maximized subsequence sum according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Subsequence Sum"
WHY DOES IT MATTER?
Maximum subarray is a canonical example of optimal substructure and greedy‑DP, teaching candidates how to convert a combinatorial search into a single pass. Mastery of this pattern unlocks many real‑world problems like profit maximization, signal processing, and resource allocation.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that a subarray ending at i only depends on the best subarray ending at i‑1. By discarding all earlier history and keeping just the current sum, we collapse O(N^2) possibilities into O(N) work.
REAL-WORLD CONNECTION
Think of a streaming telemetry pipeline where you need to detect the longest period of above‑average latency. Kadane’s algorithm slides a window that expands while the cumulative latency stays beneficial and contracts when it becomes detrimental, mirroring how monitoring systems flag performance regressions.
During an interview, write the recurrence first: cur = max(a[i], cur + a[i]). Then immediately translate it into code, and remember to handle the all‑negative edge case by initializing best to a[0] instead of 0.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The maximized subsequence sum problem asks for the largest possible sum obtainable by selecting a contiguous segment (subarray) of a numeric sequence. A naive scan of all O(N^2) subarrays quickly becomes infeasible for N up to 10^5 or higher because each candidate requires O(1) sum retrieval after a prefix‑sum pre‑process, yet the sheer number of candidates overwhelms time limits. The optimal paradigm is a linear‑time dynamic programming technique known as Kadane’s algorithm. It maintains a running "current" sum that either extends the previous segment or restarts at the current element, and a global "best" sum that records the maximum seen so far. This greedy‑DP formulation leverages the optimal substructure property: the best subarray ending at position i either includes the best subarray ending at i‑1 (if that sum is positive) or starts anew at i, guaranteeing O(N) time and O(1) extra space.
Interview Questions on This Problem
Q1How would you modify Kadane’s algorithm to also return the start and end indices of the maximum‑sum subarray?
Track the index where the current sum is reset (potential start) and update global start/end whenever a new best sum is found. When current sum becomes negative, set currentStart = i+1 and reset currentSum to 0.
Q2Explain how the maximum subsequence sum changes if the array may contain only negative numbers. What is the correct answer in that case?
If all numbers are negative, the classic Kadane that resets to 0 would incorrectly return 0. The correct answer is the largest (least negative) element, which can be obtained by initializing best to -∞ and never resetting current sum to 0 when it becomes negative.
Q3A fintech platform needs to compute the maximum profit over a sliding window of days. How can you adapt the maximum subarray solution to handle a fixed‑size window efficiently?
Use a deque to maintain candidate prefix sums within the window, or compute prefix sums once and for each window compute max(prefix[j] - minPrefixInWindow) in O(1) amortized, yielding O(N) overall.
Examples
Input
[4, 2, 1, 3, 6, 5, 8, 9, 7, 10, 11, 12, 13, 14, 15]
Output
53
Explanation: Step-by-step: First, we sort the input array in ascending order. Then, we initialize two pointers, one at the start and one at the end of the array. We keep moving the pointers towards each other, adding the elements at the pointers to the current sum. If the current sum is greater than the maximum sum found so far, we update the maximum sum. Finally, we return the maximum sum.
Input
[-1, -2, -3, -4, -5]
Output
-1
Explanation: Step-by-step: First, we sort the input array in ascending order. Then, we initialize two pointers, one at the start and one at the end of the array. We keep moving the pointers towards each other, adding the elements at the pointers to the current sum. If the current sum is greater than the maximum sum found so far, we update the maximum sum. Finally, we return the maximum sum.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Apply Kadane’s algorithm: iterate once, maintain current sum = max(a[i], current+ a[i]) and global max = max(global, current). This yields O(N) time and O(1) extra space.
Brute Force Approach
Enumerate every possible subarray, compute its sum, and keep the maximum; this is O(N^2) time. Even with prefix sums to get each subarray sum in O(1), the double loop still dominates.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
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) {
if (nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
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) {
if (nums.length == 0) return 0;
Arrays.sort(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):
if not nums:
return 0
nums.sort()
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 0;
nums.sort((a, b) => a - b);
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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