Shortest Path Cost Engine 7 — Problem Statement & Solution Guide

Two PointersHard3Sum Zero Target

TimeO(N²)

SpaceO(1) additional (ignoring input sort)

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

"Shortest Path Cost Engine 7"

hard

WHY DOES IT MATTER?

The two‑pointer pattern converts a quadratic search space into a linear sweep, making it indispensable for any problem that involves finding pairs or triplets with a target sum. In large‑scale systems, this reduction directly translates to lower latency and reduced compute cost, which are critical for real‑time analytics and recommendation engines.

OPTIMIZATION CHALLENGE

The key insight is that sorting imposes a monotonic relationship on the sum of two moving pointers. This monotonicity lets you decide deterministically whether to increase the left pointer or decrease the right pointer, eliminating the need for nested loops and cutting the time complexity from O(N³) to O(N²).

REAL-WORLD CONNECTION

Think of a logistics network where you need to match three shipment legs whose combined cost balances to zero (e.g., profit‑loss offset). By sorting routes by cost and sliding two pointers from opposite ends, you efficiently discover the optimal three‑leg route without enumerating every combination, mirroring how distributed load balancers pair requests to achieve equilibrium.

During an interview, always sort first, then immediately skip duplicates before entering the two‑pointer loop. This tiny step prevents exponential blow‑up on inputs with many repeated values and demonstrates attention to edge‑case handling.

COMPLEXITY AT A GLANCE

⏱ Time:O(N²)

💾 Space:O(1) additional (ignoring input sort)

Core Theory — Why This Approach?

The 3Sum Zero Target problem asks for all unique triplets (i, j, k) in an array whose values sum to zero. A naïve O(N³) triple‑nested loop quickly becomes infeasible when N reaches 10⁵, because the cubic explosion exhausts both time and memory limits. The breakthrough comes from sorting the array and applying the two‑pointer technique: for each fixed element a[i], we slide a left pointer (i+1) and a right pointer (N‑1) toward each other, adjusting based on the current sum. This reduces the inner search to linear time, yielding an overall O(N²) solution that is both deterministic and cache‑friendly. The sorted order also enables easy duplicate elimination, guaranteeing that each valid triplet appears exactly once.

In the context of the "Shortest Path Cost Engine", the goal is not just any zero‑sum triplet but the one that minimizes a path‑cost metric—typically the distance between the farthest indices of the triplet (max(i, j, k) - min(i, j, k)). By maintaining the minimal span while the two pointers move, we can capture the optimal path cost without an extra pass. This hybrid of two‑pointer scanning and span tracking preserves the O(N²) time bound while using only O(1) auxiliary space beyond the input array.

The optimal paradigm therefore combines three pillars: (1) sorting for order‑based pruning, (2) two‑pointer traversal for linear‑time pair search, and (3) on‑the‑fly span evaluation to satisfy the shortest‑path constraint. Together they transform an exponential brute‑force nightmare into a sleek quadratic algorithm that scales to production‑grade data sizes.

Interview Questions on This Problem

Q1How would you modify the classic 3Sum two‑pointer solution to also return the triplet with the smallest index span (i.e., shortest path cost)?

After sorting, for each i fix a[i] and run left/right pointers. Whenever a[i] + a[left] + a[right] == 0, compute span = right - i (or max-min). Keep a global minimum span and store the corresponding triplet. Continue moving pointers to explore other candidates while still updating the minimum span. This adds only O(1) work per valid triplet.

Q2Why does the two‑pointer technique guarantee O(N²) time for 3Sum, and can it ever degrade to O(N³) in the presence of many duplicates?

Sorting ensures that each pair (left, right) is examined at most once per fixed i because pointers only move inward. Even with duplicates, we skip over equal values after processing a valid triplet, so the total number of pointer moves remains linear per i, preserving O(N²). The algorithm only degrades if the duplicate‑skipping logic is omitted, causing repeated work.

Q3Explain how you would extend the shortest‑path 3Sum solution to handle a dynamic stream of numbers where you need to maintain the minimal zero‑sum span at any point.

Maintain a balanced BST (or multiset) of incoming numbers with their original indices. For each new element, perform a two‑pointer style search on the sorted view of the BST to find complementary pairs that sum to the negative of the new value, while tracking index spans. Update the global minimum span if a better triplet is found. The BST gives O(log N) insertion and O(N) search per element, leading to an overall O(N log N) amortized solution for the streaming variant.

Examples

Example 1

Input

[[-1, 0, 3], [2, -3, 1], [0, 1, 2]]

Output

-1

Explanation: Step-by-step: Given the array of system constraints and values, we first sort the array based on the first element of each subarray. Then, we use the two-pointer technique to find the shortest path cost using the 3Sum Zero Target methodology. Since the array does not contain any triplets that sum to 0, the output is -1.

Example 2

Input

[[-5, 0, 3], [2, -3, 1], [0, 1, 2]]

Output

-1

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 for each fixed element use a left‑right two‑pointer scan to find zero‑sum pairs while tracking the minimal index span, achieving O(N²) time and O(1) extra space.

Brute Force Approach

Iterate over every possible triple of indices (i, j, k) and check if their values sum to zero, then compute the span; keep the smallest span found.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N²)

function solution(nums) {
   if (nums.length < 3) return -1;
   nums.sort((a, b) => a[0] - b[0]);
   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][0] + nums[left][0] + nums[right][0];
           if (sum === 0) return 0;
           else if (sum < 0) left++;
           else right--;
       }
   }
   return -1;
}

class Solution {
   public:
       int solution(vector<vector<int>>& nums) {
           if (nums.size() < 3) return -1;
           sort(nums.begin(), nums.end(), [](const vector<int>& a, const vector<int>& b) {
               return a[0] < b[0];
           });
           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][0] + nums[left][0] + nums[right][0];
                   if (sum == 0) return 0;
                   else if (sum < 0) left++;
                   else right--;
               }
           }
           return -1;
       }
   };

class Solution {
   public int solution(int[][] nums) {
       if (nums.length < 3) return -1;
       Arrays.sort(nums, (a, b) -> a[0] - b[0]);
       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][0] + nums[left][0] + nums[right][0];
               if (sum == 0) return 0;
               else if (sum < 0) left++;
               else right--;
           }
       }
       return -1;
   }
}

def solution(nums):
   if len(nums) < 3: return -1
   nums.sort(key=lambda x: x[0])
   for i in range(len(nums) - 2):
       left = i + 1
       right = len(nums) - 1
       while left < right:
           sum = nums[i][0] + nums[left][0] + nums[right][0]
           if sum == 0: return 0
           elif sum < 0: left += 1
           else: right -= 1
   return -1

function solution(nums) {
   if (nums.length < 3) return -1;
   nums.sort((a, b) => a[0] - b[0]);
   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][0] + nums[left][0] + nums[right][0];
           if (sum === 0) return 0;
           else if (sum < 0) left++;
           else right--;
       }
   }
   return -1;
}

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Morgan StanleyUber

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Shortest Path Cost Engine 7 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Shortest Path Cost Engine 7 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor