Balanced Path Weight — Problem Statement & Solution Guide
Problem Description
In the Balanced Path Weight problem you are given a sequence of integers that represent the weights of consecutive segments along a path. Your task is to find the maximum possible total weight that can be obtained by selecting a contiguous portion of the path. In other words, you must determine the largest sum of any subarray of the given sequence. The input consists of the number of segments followed by the list of weights. The output is a single integer: the maximum subarray sum. If all weights are negative, the answer is the largest (least negative) single weight, since an empty subarray is not allowed.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Balanced Path Weight"
WHY DOES IT MATTER?
Kadane’s algorithm exemplifies the "prefix sum with greedy update" pattern, a cornerstone for solving linear‑time subarray problems. Mastering this pattern equips engineers to tackle a wide range of real‑world challenges, from financial time‑series analysis to network traffic monitoring, where contiguous segments with optimal properties must be identified quickly.
OPTIMIZATION CHALLENGE
The core insight is that the maximum subarray ending at index i depends only on the maximum ending at i‑1 and the current element. This local dependency allows a single pass and constant‑size state, reducing both time from O(n^2) to O(n) and space from O(n) to O(1).
REAL-WORLD CONNECTION
Consider a streaming service that monitors user engagement over time. The platform needs to detect the longest streak of high activity to trigger marketing campaigns. By treating engagement scores as a sequence, Kadane’s algorithm can identify the optimal streak in real time, enabling instant decision‑making without storing the entire history.
When explaining Kadane in an interview, emphasize the "current best ending here" concept and show how resetting to the current element when the running sum becomes negative preserves optimality. Practicing the mental model of "extend or restart" helps candidates articulate the algorithm clearly under time pressure.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Balanced Path Weight problem is a classic example of the maximum subarray problem, solvable in linear time using Kadane’s algorithm. The naive approach examines every possible contiguous segment, leading to an O(n^2) time complexity and quadratic memory usage when storing intermediate sums. This quickly becomes infeasible for large input sizes, as the number of subarrays grows quadratically with the sequence length.
Kadane’s algorithm transforms the problem into a single pass through the array by maintaining two values: the maximum sum ending at the current position and the global maximum found so far. At each step, the algorithm decides whether to extend the previous subarray or start a new one at the current element, based on whether the current element alone yields a larger sum. This greedy decision guarantees that the optimal subarray is found without revisiting any element, achieving O(n) time and O(1) additional space.
The key insight is that the maximum subarray ending at position i can be computed from the maximum subarray ending at i-1 plus the current element, or by starting fresh at i. By iteratively updating these two values, we avoid the combinatorial explosion of the brute‑force method and obtain a solution that scales linearly with input size.
Interview Questions on This Problem
Q1How would you explain the difference between Kadane’s algorithm and a dynamic programming approach for maximum subarray sum to a candidate at a fintech company?
Kadane’s algorithm is a specific dynamic programming solution that uses only two variables to track the best subarray ending at each index and the global best. It runs in O(n) time and O(1) space, making it ideal for real‑time analytics on transaction streams where memory and latency are critical. A generic DP approach might store an array of intermediate results, leading to O(n) space, which is unnecessary for this problem.
Q2A senior engineer at a high‑growth startup asks: "What would happen if all weights are negative? How does Kadane handle that?"
Kadane’s algorithm initializes the current sum to the first element and updates the global maximum accordingly. If all numbers are negative, the algorithm will correctly return the largest (least negative) value, because it never resets the current sum to zero unless a positive sum is achieved. This ensures the algorithm works for both all‑positive and all‑negative sequences.
Q3During an interview at a global product company, a candidate proposes a divide‑and‑conquer solution. What are the trade‑offs compared to Kadane’s algorithm?
Divide‑and‑conquer runs in O(n log n) time and requires O(log n) stack space, which is slower and uses more memory than Kadane’s O(n) time and O(1) space. While the divide‑and‑conquer approach is elegant and can be extended to related problems (e.g., maximum subarray with constraints), Kadane’s simplicity and optimal performance make it the preferred choice for interview scenarios.
Examples
Input
5 1 -2 3 4 -5
Output
7
Explanation: The subarray [3, 4] yields the maximum sum 3 + 4 = 7. All other contiguous subarrays produce sums of 5, 4, 3, 2, 1, -1, -2, -3, -4, -5, or 0, none of which exceed 7.
Input
4 -1 -2 -3 -4
Output
-1
Explanation: All weights are negative, so the best choice is to take the single element with the largest value, which is -1. Any longer subarray would produce a smaller sum.
Input
6 2 3 -1 2 4 -5
Output
10
Explanation: The subarray [2, 3, -1, 2, 4] has sum 2 + 3 + (-1) + 2 + 4 = 10, which is the highest achievable. Adding the final -5 would reduce the sum to 5.
Input
3 0 0 0
Output
0
Explanation: All elements are zero, so any non‑empty subarray has sum 0. The maximum sum is therefore 0.
Constraints
- 1 <= N <= 100000
- -1000000000 <= weight[i] <= 1000000000
- The answer fits within a 64‑bit signed integer.
Optimal Approach & Strategy
Kadane’s algorithm scans the array once, maintaining the maximum subarray ending at the current position and the overall maximum. It updates these in constant time per element, achieving O(n) time and O(1) space.
Brute Force Approach
The naive solution iterates over all start indices and, for each, sums all possible end indices, storing the maximum. This requires O(n^2) time and O(1) space, but becomes impractical for large arrays.
Verified Code Solutions
function solution(nums) { let sum = 0; let maxSum = 0; for (let i = 0; i < nums.length; i++) { sum += nums[i]; maxSum = Math.max(maxSum, sum); } return maxSum; }class Solution { public: int solution(vector<int>& nums) { int sum = 0; int maxSum = 0; for (int num : nums) { sum += num; maxSum = max(maxSum, sum); } return maxSum; } };class Solution { public int solution(int[] nums) { int sum = 0; int maxSum = 0; for (int num : nums) { sum += num; maxSum = Math.max(maxSum, sum); } return maxSum; } }def solution(nums): sum = 0; max_sum = 0; for num in nums: sum += num; max_sum = max(max_sum, sum); return max_sumfunction solution(nums) { let sum = 0; let maxSum = 0; for (let i = 0; i < nums.length; i++) { sum += nums[i]; maxSum = Math.max(maxSum, sum); } return maxSum; }Asked in Top Tech Interviews
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