Shortest Path Cost Protocol 4 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the shortest path cost using the 3Sum Zero Target methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shortest Path Cost Protocol 4"
WHY DOES IT MATTER?
The two‑pointer pattern converts a potentially exponential search space into a deterministic linear scan by leveraging sorted order. This is essential for any problem that asks for combinations meeting a numeric constraint, especially when the input size scales to hundreds of thousands.
OPTIMIZATION CHALLENGE
The breakthrough insight is that once the array is sorted, the relative ordering of values dictates how the sum evolves as pointers move. This monotonic behavior allows you to discard large swaths of impossible pairs in O(1) per step, collapsing O(N³) to O(N²).
REAL-WORLD CONNECTION
Think of a logistics network where you need to find three warehouses whose combined inventory balances to zero net surplus. By sorting warehouses by inventory level and moving two pointers inward, you can instantly locate the optimal trio without checking every possible combination, mirroring how load‑balancing algorithms prune search space.
During an interview, sort the array first and immediately write the two‑pointer skeleton. Then, before diving into duplicate handling, add a quick check for early termination (e.g., if the smallest possible sum exceeds zero). This shows you understand both correctness and performance.
COMPLEXITY AT A GLANCE
O(N²)O(1) additional (ignoring input sort)Core Theory — Why This Approach?
The core of the Shortest Path Cost Protocol 4 problem lies in recognizing that the task is a constrained variant of the classic 3‑Sum problem. In a naïve setting, one would examine every possible triplet of indices (i, j, k) in O(N³) time, checking whether the sum of the three values equals the target (zero) and then computing the associated path cost. This brute‑force method quickly becomes infeasible for N in the order of 10⁵, which is typical for modern system‑level datasets, because the cubic explosion overwhelms both CPU and memory resources.
To achieve optimal performance, the problem is transformed by first sorting the dataset, which enables the two‑pointer technique. After fixing the first element of a potential triplet, two pointers are positioned at the remaining sub‑array's extremes. By moving the pointers inward based on the current sum relative to the zero target, we can enumerate all valid triples in linear time per fixed element. This reduces the overall time complexity to O(N²) while preserving the ability to compute the minimal path cost for each qualifying triplet. The sorted order also guarantees that duplicate triplets can be skipped efficiently, further tightening the runtime.
The two‑pointer paradigm is a manifestation of the broader “sliding window” and “two‑sum” patterns, which exploit monotonicity in sorted data to replace nested loops with pointer arithmetic. The optimal solution therefore hinges on three pillars: sorting (O(N log N)), a deterministic linear scan with two pointers (O(N) per outer iteration), and careful duplicate handling. Together they deliver a scalable algorithm suitable for production‑grade workloads.
Interview Questions on This Problem
Q1How would you adapt the two‑pointer 3‑Sum solution to also return the minimum "path cost" defined as the sum of absolute differences between the three indices?
After sorting the array, for each valid triplet (i, j, k) that sums to zero, compute the cost as |i‑j| + |j‑k| + |i‑k|. Keep a global minimum that updates whenever a lower cost is found. Because the indices are derived from the sorted order, you must map back to original positions if the cost depends on original indices.
Q2Why does the two‑pointer approach guarantee O(N²) time even when the input contains many duplicate values?
Sorting clusters duplicates together, and the inner while‑loops advance the left or right pointer on every iteration, never revisiting the same pair. Skipping over consecutive duplicates after a valid triplet is found prevents redundant work, ensuring each pair is examined at most once per outer index, preserving the quadratic bound.
Q3In a distributed system where each node holds a partition of the dataset, how could you parallelize the 3‑Sum search while still respecting the O(N²) overall complexity?
Each node can sort its local partition and then exchange boundary elements needed for cross‑partition triples. By assigning distinct ranges of the fixed first element to different nodes, the two‑pointer scans become embarrassingly parallel. A final reduction step merges the local minimum path costs, yielding the global optimum without increasing asymptotic complexity.
Examples
Input
[2, -5, 7, 12, 0]
Output
-14
Explanation: Step-by-step: Sort the array in ascending order. Initialize three pointers, one at the start, one at the end, and one at the middle. Move the pointers based on the sum of the elements at the pointers. If the sum is zero, return the sum. If the sum is less than zero, move the left pointer to the right. If the sum is greater than zero, move the right pointer to the left.
Input
[1, 2, 3]
Output
0
Explanation: Step-by-step: Sort the array in ascending order. Initialize three pointers, one at the start, one at the end, and one at the middle. Move the pointers based on the sum of the elements at the pointers. If the sum is zero, return the sum. If the sum is less than zero, move the left pointer to the right. If the sum is greater than zero, move the right pointer to the left.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Sort the array and use a fixed‑element loop combined with a two‑pointer scan to locate zero‑sum triples in linear time per iteration, updating the minimum path cost on the fly.
Brute Force Approach
Check every possible triplet of indices, verify if their values sum to zero, and compute the path cost for each valid triple.
Verified Code Solutions
function solution(nums) {
if (nums.length < 3) return 0;
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length - 2; i++) {
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let sum = nums[i] + nums[left] + nums[right];
if (sum === 0) return sum;
else if (sum < 0) left++;
else right--;
}
}
return 0;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() < 3) return 0;
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size() - 2; i++) {
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) return sum;
else if (sum < 0) left++;
else right--;
}
}
return 0;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length < 3) return 0;
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) return sum;
else if (sum < 0) left++;
else right--;
}
}
return 0;
}
}def solution(nums):
if len(nums) < 3: return 0
nums.sort()
for i in range(len(nums) - 2):
left = i + 1
right = len(nums) - 1
while left < right:
sum = nums[i] + nums[left] + nums[right]
if sum == 0: return sum
elif sum < 0: left += 1
else: right -= 1
return 0function solution(nums) {
if (nums.length < 3) return 0;
nums.sort((a, b) => a - b);
for (let i = 0; i < nums.length - 2; i++) {
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let sum = nums[i] + nums[left] + nums[right];
if (sum === 0) return sum;
else if (sum < 0) left++;
else right--;
}
}
return 0;
}Asked in Top Tech Interviews
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