Monotonic Path Weight — Problem Statement & Solution Guide
Problem Description
Given an array of non-negative integers, compute the monotonic path weight by summing all elements in the array, or return 0 if the array contains a negative number or is empty.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Monotonic Path Weight"
WHY DOES IT MATTER?
Linear scans with early exit are essential for problems that require validation of a property (e.g., non‑negativity) while aggregating a value. They guarantee that you never perform unnecessary work once the answer is determined, which is critical for performance-sensitive applications.
OPTIMIZATION CHALLENGE
The key insight is that you can combine the validation (checking for negatives) and aggregation (summing) into a single pass, eliminating the need for a second traversal or additional data structures.
REAL-WORLD CONNECTION
Consider a real‑time monitoring system that aggregates sensor readings. If any reading falls below a safety threshold, the system must immediately flag an alert and stop further processing to avoid false positives. The same early‑exit logic applies here.
When implementing, always use a 64‑bit accumulator to guard against overflow, and handle the empty array case explicitly before entering the loop.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan of the input array. In the optimal solution we maintain a running sum and simultaneously check for the presence of a negative value. If a negative is encountered we immediately return 0, otherwise after the loop we return the accumulated sum. This approach is O(n) time and O(1) space, which is optimal because every element must be examined at least once to guarantee correctness.
Naive approaches often involve nested loops or repeated summation of sub‑arrays, leading to O(n^2) time. For example, one might recompute the sum after each element or use a recursive divide‑and‑conquer that still ends up touching each element multiple times. Such methods quickly become infeasible for large inputs (e.g., arrays with millions of elements) because the number of operations grows quadratically.
The optimal paradigm here is a single-pass linear scan with early exit. By checking for a negative value on the fly we avoid unnecessary work. This pattern is a classic example of “scan with a guard” and is widely applicable in streaming, real‑time analytics, and data validation scenarios.
Interview Questions on This Problem
Q1How would you modify this algorithm to handle an input that arrives as a stream of integers rather than a pre‑loaded array?
You would process each integer as it arrives, adding it to a running total and checking for negativity. If a negative is seen, you can immediately output 0 and stop processing further elements, which keeps the algorithm O(1) space and O(n) time for the stream length.
Q2What changes would you make if the array could contain very large integers that might overflow a 32‑bit integer type?
Use a 64‑bit integer type (e.g., long in Java, long long in C++) to store the sum. Additionally, you could perform a check for overflow after each addition, or use arbitrary‑precision arithmetic if the language supports it.
Q3In a distributed system where the array is partitioned across multiple nodes, how would you compute the monotonic path weight efficiently?
Each node can compute a local sum and a flag indicating whether it saw a negative number. The results are then reduced: if any node reports a negative, the final answer is 0; otherwise, the global sum is the sum of all local sums. This requires only a single aggregation step and minimal communication.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: We iterate through the array from left to right. We start with the first element, 1. Then we add the second element, 2. Then we add the third element, 3. Then we add the fourth element, 4. Then we add the fifth element, 5. The sum of these elements is 15.
Input
[]
Output
0
Explanation: Step-by-step: The array is empty, so we return 0.
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
Traverse the array once, maintaining a running sum and checking for negatives on the fly. If a negative is found, return 0 immediately; otherwise, return the sum after the loop.
Brute Force Approach
A naive solution might iterate over the array, and for each element, recompute the sum of all elements seen so far, leading to O(n^2) time. It also might ignore the negative check until after summing, which wastes work.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
if (num >= 0) {
sum += num;
} else {
return 0;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
if (num >= 0) {
sum += num;
} else {
return 0;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
if (num >= 0) {
sum += num;
} else {
return 0;
}
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
if num >= 0:
sum += num
else:
return 0
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
if (num >= 0) {
sum += num;
} else {
return 0;
}
}
return sum;
}Asked in Top Tech Interviews
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