Dynamic Path Weight — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the dynamic path weight according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Path Weight"
WHY DOES IT MATTER?
The "first true" binary‑search pattern appears in many real‑world scenarios: locating the earliest time a metric crosses a threshold, finding the minimal capacity to satisfy demand, or determining the earliest version that introduced a bug. Mastering this pattern equips engineers to turn O(N) scans into O(log N) lookups, a critical performance lever at scale.
OPTIMIZATION CHALLENGE
The key insight is to separate the problem into two stages: (1) build a monotonic representation (prefix sums) once, and (2) apply binary search on that representation for each query. Recognizing the monotonic predicate and avoiding recomputation of sums for every query cuts the time complexity from quadratic to near‑linear.
REAL-WORLD CONNECTION
Imagine a distributed monitoring system that records cumulative request latency over time. When an alert fires, you need the exact timestamp when latency first exceeded a SLA. The latency curve is monotonic (non‑decreasing), so a binary search on the stored cumulative values pinpoints the breach instantly, just like the dynamic path weight problem.
During an interview, write the prefix‑sum construction first, then clearly comment the binary‑search invariant (low, high, mid, condition). Use language‑provided binary‑search utilities when allowed, but be ready to implement the loop manually to demonstrate understanding of edge handling.
COMPLEXITY AT A GLANCE
O(N + Q·log N)O(N)Core Theory — Why This Approach?
Binary search thrives on monotonic predicates – functions that never decrease (or never increase) as the input index grows. When the problem asks for a "dynamic path weight" over a sequence, the natural monotonic property is the cumulative weight (prefix sum) which only grows as we move forward in the array. A naive scan computes the prefix sum for each query in O(N) time, which quickly becomes prohibitive for large N or many queries. By pre‑computing a single prefix‑sum array in O(N) time, each query reduces to finding the smallest index i such that prefix[i] ≥ target. This is a classic "first true" binary‑search problem on a sorted (non‑decreasing) array, delivering O(log N) query time. The optimal paradigm therefore combines linear preprocessing with logarithmic search, leveraging the fact that binary search can locate the boundary of a step‑function in logarithmic steps, dramatically shrinking the overall runtime for massive inputs.
The optimal solution also respects space constraints: the prefix‑sum array occupies O(N) additional memory, which is acceptable for most interview settings. Edge cases—such as targets larger than the total weight or negative numbers—must be handled by checking the bounds before invoking binary search. This approach illustrates a broader algorithmic pattern: transform a problem into a monotonic decision function, then apply binary search to locate the decision boundary efficiently.
Interview Questions on This Problem
Q1How would you find the first index in a sorted array where the cumulative sum exceeds a given target using binary search?
First compute the prefix‑sum array in O(N). Then perform a binary search on this array for the smallest index i where prefix[i] ≥ target. Return i or -1 if target exceeds total sum.
Q2Explain why a linear scan to compute the dynamic path weight for each query results in TLE for N = 10^5 and Q = 10^5.
A linear scan per query costs O(N) time, leading to O(N·Q) ≈ 10^10 operations, which exceeds typical time limits. Pre‑processing the prefix sums once (O(N)) and answering each query in O(log N) reduces total work to O(N + Q·log N), which comfortably fits within limits.
Q3Can you adapt the binary‑search solution to handle updates to individual elements (e.g., point updates) while still answering weight queries efficiently?
Yes. Replace the static prefix‑sum array with a Fenwick Tree (Binary Indexed Tree) or Segment Tree. Both support point updates in O(log N) and prefix‑sum queries in O(log N). To answer the first‑exceeding‑target query, perform a binary‑search on the tree using its built‑in "find‑by‑order" operation, achieving O(log^2 N) overall.
Examples
Input
[6, 7, 8, 9]
Output
30
Explanation: Step 1: Initialize the dynamic path weight to 0. Step 2: Iterate through the array from left to right. For each element, add it to the dynamic path weight. Step 3: Return the dynamic path weight. With input [6, 7, 8, 9], we do Step 1, then Step 2 (6 + 7 + 8 + 9), giving output 30.
Input
[2, 4]
Output
6
Explanation: Step 1: Initialize the dynamic path weight to 0. Step 2: Iterate through the array from left to right. For each element, add it to the dynamic path weight. Step 3: Return the dynamic path weight. With input [2, 4], we do Step 1, then Step 2 (2 + 4), giving output 6.
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
Pre‑compute a prefix‑sum array once, then binary‑search it for the smallest index where the sum ≥ target, achieving logarithmic query time.
Brute Force Approach
For each query, iterate from the start, accumulating the sum until it reaches or exceeds the target, returning the current index.
Verified Code Solutions
function solution(nums) {
let dynamicPathWeight = 0;
for (let num of nums) {
dynamicPathWeight += num;
}
return dynamicPathWeight;
}class Solution {
public:
int solution(vector<int>& nums) {
int dynamicPathWeight = 0;
for (int num : nums) {
dynamicPathWeight += num;
}
return dynamicPathWeight;
}
};class Solution {
public int solution(int[] nums) {
int dynamicPathWeight = 0;
for (int num : nums) {
dynamicPathWeight += num;
}
return dynamicPathWeight;
}
}def solution(nums):
dynamic_path_weight = 0
for num in nums:
dynamic_path_weight += num
return dynamic_path_weightfunction solution(nums) {
let dynamicPathWeight = 0;
for (let num of nums) {
dynamicPathWeight += num;
}
return dynamicPathWeight;
}Asked in Top Tech Interviews
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