Shortest Path Cost Engine — Problem Statement & Solution Guide

Two PointersHard3Sum Zero Target

TimeO(N^2)

SpaceO(1) additional

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"

hard

WHY DOES IT MATTER?

The two‑pointer pattern transforms a quadratic search space into a linear sweep after sorting, turning an otherwise exponential‑time brute force into a tractable O(N²) solution. Mastery of this pattern is essential because many real‑world constraints (e.g., budget limits, latency windows) map naturally to sum‑based conditions.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that once the array is sorted, the relative ordering of elements gives a monotonic relationship to the target sum, allowing us to discard large swaths of the search space with a single pointer move instead of enumerating every pair.

REAL-WORLD CONNECTION

Think of a logistics network where you need to pair two shipments with a third to exactly fill a container's weight capacity. Sorting shipments by weight and then sliding two pointers from lightest and heaviest mirrors the two‑pointer technique, enabling rapid identification of perfect load combinations.

During an interview, always sort first and then treat the problem as "fix‑one, find‑two"; this mental model prevents you from falling back into triple loops and makes it easy to handle duplicates and edge cases on the fly.

COMPLEXITY AT A GLANCE

⏱ Time:O(N^2)

💾 Space:O(1) additional

Core Theory — Why This Approach?

The core of this problem lies in the classic 3‑Sum pattern, which asks for three elements whose sum equals a target (zero in this case). A naïve triple‑nested loop runs in O(N³) and quickly becomes infeasible for N ≥ 10⁴, the typical upper bound in modern coding interviews. By first sorting the array, we can fix one element and then use a two‑pointer sweep on the remaining sub‑array to locate the complementary pair in linear time. This reduces the overall complexity to O(N²) while preserving correctness because the sorted order guarantees that moving the left pointer increases the sum and moving the right pointer decreases it, allowing systematic elimination of impossible pairs. The optimal paradigm therefore combines sorting (O(N log N)) with the two‑pointer technique, yielding a deterministic, in‑place solution that also uses only O(1) extra space beyond the input.

Interview Questions on This Problem

Q1How would you adapt the 3‑Sum solution to return the triplet with the minimum absolute sum when an exact zero does not exist?

After sorting, keep track of the best triplet seen so far (minimum |sum|). For each fixed element, run the two‑pointer scan as usual, updating the best triplet whenever a smaller absolute sum is encountered. The algorithm remains O(N²) because the extra comparison does not affect the pointer movement.

Q2Explain how you would modify the algorithm to handle duplicate values and ensure each unique triplet is reported only once.

Skip over duplicate values for the fixed index i by checking if nums[i] == nums[i‑1] (i > 0). Inside the two‑pointer loop, after finding a valid triplet, increment left while nums[left] == nums[left‑1] and decrement right while nums[right] == nums[right+1] to bypass identical elements before moving the pointers.

Q3In a distributed system where the dataset is sharded across multiple nodes, how could you compute the global 3‑Sum zero triplet efficiently?

Each node locally sorts its shard and computes a local hash set of values. Nodes then exchange summary statistics (e.g., min, max, frequency maps) and collaboratively run a coordinated two‑pointer scan across the merged sorted view using a merge‑like iterator, reducing network traffic. The overall complexity stays near O(N²) but is distributed, and careful partitioning ensures load balancing.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Output

Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we need to find the shortest path cost using the 3Sum Zero Target methodology. We can achieve this by sorting the array and using two pointers to find the closest sum to zero. In this case, the closest sum to zero is 0, which is achieved by the elements at indices 1, 3, and 5 (2, 4, and 6). Therefore, the output is 0.

Example 2

Input

[-1, 0, 1, 2, -1, -4]

Output

Explanation: Step-by-step: Given the array [-1, 0, 1, 2, -1, -4], we need to find the shortest path cost using the 3Sum Zero Target methodology. We can achieve this by sorting the array and using two pointers to find the closest sum to zero. In this case, the closest sum to zero is 2, which is achieved by the elements at indices 1, 3, and 5 (0, 1, and 1). Therefore, the output is 2.

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, fix one element, and use a two‑pointer sweep on the remaining sub‑array to find complementary pairs in linear time.

Brute Force Approach

Check every possible triple with three nested loops and test if their sum equals zero.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N^2)

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

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

class Solution {
   public int solution(int[] nums) {
       Arrays.sort(nums);
       int left = 0;
       int right = nums.length - 1;
       int minCost = Integer.MAX_VALUE;
       while (left < right) {
           int sum = nums[left] + nums[right] + nums[right - 1];
           if (Math.abs(sum) < Math.abs(minCost)) {
               minCost = sum;
           }
           if (sum < 0) {
               left++;
           } else if (sum > 0) {
               right--;
           } else {
               return 0;
           }
       }
       return minCost;
   }
}

def solution(nums):
   nums.sort()
   left = 0
   right = len(nums) - 1
   minCost = float('inf')
   while left < right:
       sum = nums[left] + nums[right] + nums[right - 1]
       if abs(sum) < abs(minCost):
           minCost = sum
       if sum < 0:
           left += 1
       elif sum > 0:
           right -= 1
       else:
           return 0
   return minCost

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

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Shortest Path Cost Engine 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 — 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