Optimal Grid Path Protocol 2 — Problem Statement & Solution Guide

Two PointersHard3Sum Zero Target

TimeO(N²)

SpaceO(1) additional (O(N) if a copy is sorted separately)

Problem Description

Given a complex dataset of length N representing system constraints and values, calculate the optimal grid path using the 3Sum Zero Target methodology.

DSA Pattern Breakdown

"Optimal Grid Path Protocol 2"

hard

WHY DOES IT MATTER?

The two‑pointer pattern converts a quadratic search space into a linear one after sorting, which is the key to solving many combinatorial selection problems (3‑Sum, 4‑Sum, container with most water) within feasible time limits for large inputs.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that once the array is sorted, the relative ordering of the remaining two elements determines whether the sum is too low or too high, allowing deterministic pointer movement and eliminating the need for nested loops.

REAL-WORLD CONNECTION

Think of a load‑balancing router that must pick three servers whose current loads sum to a target capacity; by sorting server loads and sliding two pointers, the router can instantly locate a viable trio without exhaustive enumeration, mirroring the algorithmic insight.

During an interview, sort the array in‑place, then immediately write the outer loop that skips duplicates; this not only saves time but also signals to the interviewer that you’re aware of edge‑case handling before diving into pointer logic.

COMPLEXITY AT A GLANCE

⏱ Time:O(N²)

💾 Space:O(1) additional (O(N) if a copy is sorted separately)

Core Theory — Why This Approach?

The optimal grid‑path problem can be reduced to a classic 3‑Sum Zero target: we need three positions in the dataset whose values sum to zero, which correspond to a feasible path respecting the system constraints. A naive solution enumerates every triple (i, j, k) in O(N³) time, which quickly becomes infeasible for N > 10⁴ due to both time and memory pressure. By sorting the array first, we can fix one element and then use a two‑pointer sweep on the remaining sub‑array to locate the complementary pair that brings the total to zero. This transforms the inner search into a linear scan, yielding an overall O(N²) time algorithm while using only O(1) extra space beyond the sorted copy. The two‑pointer technique exploits the monotonic property of the sorted list: moving the left pointer increases the sum, moving the right pointer decreases it, allowing deterministic convergence to the target without backtracking.

The optimal paradigm also handles duplicate values and ensures unique triplets by skipping over equal elements after each successful find. This de‑duplication step is crucial for correctness in interview settings where the output must contain distinct paths. Moreover, the approach can be extended to grid‑specific constraints (e.g., monotonic row/column movement) by mapping the linear indices back to grid coordinates after the triplet is identified. The combination of sorting, two‑pointer scanning, and careful duplicate avoidance forms the backbone of the hard‑level solution expected for this problem.

Interview Questions on This Problem

Q1How would you adapt the 3‑Sum two‑pointer solution to return the actual grid coordinates of the optimal path instead of just the indices?

After sorting, keep a parallel array of original indices. When a valid triplet (i, left, right) is found, map each original index back to its (row, column) using row = idx / cols and col = idx % cols. Return the three coordinate pairs as the path.

Q2Explain why the two‑pointer technique fails if the input array contains non‑numeric constraints (e.g., strings) and how you would modify the algorithm.

Two‑pointer relies on a total order that supports arithmetic comparison; strings lack a natural additive relationship. To handle such constraints, first map each string to a numeric weight (e.g., via a hash or predefined scoring) that preserves the problem semantics, then apply the standard two‑pointer method on the numeric representation.

Q3In a distributed system where the dataset is sharded across nodes, how can you still achieve an O(N²) overall solution for the 3‑Sum variant?

Each node sorts its shard locally and emits a summary of value frequencies. A coordinator merges the sorted summaries (O(N log N) total) and then runs the two‑pointer scan on the merged list. Because the merge is linear in the total size, the dominant work remains the O(N²) pair search, preserving the overall complexity.

Examples

Example 1

Input

[[-1, 0, 1], [0, 1, 1], [1, -1, 0]]

Output

[[-1, 0, 1], [0, 1, 1]]

Explanation: Step 1: Sort the input array by the first element of each subarray. The sorted array is: [[-1, 0, 1], [0, 1, 1], [1, -1, 0]]. Step 2: Initialize three pointers, i, j, and k, to the start of the array. Set i to 0, j to 1, and k to 2. Step 3: While i < j and j < k, check if the sum of the elements at indices i, j, and k is equal to 0. If it is, add the subarray to the result and move the pointers accordingly. Step 4: If the sum is not equal to 0, move the pointer that corresponds to the smallest element in the sum.

Example 2

Input

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

Output

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

Explanation: Step 1: Sort the input array by the first element of each subarray. The sorted array is: [[-5, 0, 3], [0, 1, 1], [1, -1, 0]]. Step 2: Initialize three pointers, i, j, and k, to the start of the array. Set i to 0, j to 1, and k to 2. Step 3: While i < j and j < k, check if the sum of the elements at indices i, j, and k is equal to 0. If it is, add the subarray to the result and move the pointers accordingly. Step 4: If the sum is not equal to 0, move the pointer that corresponds to the smallest element in the sum.

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 run a two‑pointer scan on the remaining sub‑array, achieving O(N²) time with O(1) extra space.

Brute Force Approach

Check every possible combination of three indices and test if their values sum to zero, which costs O(N³) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N²)

function solution(nums) {
   nums.sort((a, b) => a[0] - b[0]);
   let result = [];
   for (let i = 0; i < nums.length - 2; i++) {
       let j = i + 1;
       let k = nums.length - 1;
       while (j < k) {
           let sum = nums[i][0] + nums[j][0] + nums[k][0];
           if (sum === 0) {
               result.push(nums[i]);
               result.push(nums[j]);
               result.push(nums[k]);
               i++;
               j++;
               k--;
               while (j < k && nums[j][0] === nums[j - 1][0]) j++;
               while (j < k && nums[k][0] === nums[k + 1][0]) k--;
           } else if (sum < 0) {
               j++;
           } else {
               k--;
           }
       }
   }
   return result;
}

class Solution {
public:
   vector<vector<int>> solution(vector<vector<int>>& nums) {
       sort(nums.begin(), nums.end(), [](const vector<int>& a, const vector<int>& b) {
           return a[0] < b[0];
       });
       vector<vector<int>> result;
       for (int i = 0; i < nums.size() - 2; i++) {
           int j = i + 1;
           int k = nums.size() - 1;
           while (j < k) {
               int sum = nums[i][0] + nums[j][0] + nums[k][0];
               if (sum == 0) {
                   result.push_back(nums[i]);
                   result.push_back(nums[j]);
                   result.push_back(nums[k]);
                   i++;
                   j++;
                   k--;
                   while (j < k && nums[j][0] == nums[j - 1][0]) j++;
                   while (j < k && nums[k][0] == nums[k + 1][0]) k--;
               } else if (sum < 0) {
                   j++;
               } else {
                   k--;
               }
           }
       }
       return result;
   }
};

class Solution {
   public int solution(int[][] nums) {
       Arrays.sort(nums, (a, b) -> a[0] - b[0]);
       List<int[]> result = new ArrayList<>();
       for (int i = 0; i < nums.length - 2; i++) {
           int j = i + 1;
           int k = nums.length - 1;
           while (j < k) {
               int sum = nums[i][0] + nums[j][0] + nums[k][0];
               if (sum == 0) {
                   result.add(nums[i]);
                   result.add(nums[j]);
                   result.add(nums[k]);
                   i++;
                   j++;
                   k--;
                   while (j < k && nums[j][0] == nums[j - 1][0]) j++;
                   while (j < k && nums[k][0] == nums[k + 1][0]) k--;
               } else if (sum < 0) {
                   j++;
               } else {
                   k--;
               }
           }
       }
       return result.size();
   }
}

def solution(nums):
   nums.sort(key=lambda x: x[0])
   result = []
   for i in range(len(nums) - 2):
       j = i + 1
       k = len(nums) - 1
       while j < k:
           sum = nums[i][0] + nums[j][0] + nums[k][0]
           if sum == 0:
               result.append(nums[i])
               result.append(nums[j])
               result.append(nums[k])
               i += 1
               j += 1
               k -= 1
               while j < k and nums[j][0] == nums[j - 1][0]:
                   j += 1
               while j < k and nums[k][0] == nums[k + 1][0]:
                   k -= 1
           elif sum < 0:
               j += 1
           else:
               k -= 1
   return result

function solution(nums) {
   nums.sort((a, b) => a[0] - b[0]);
   let result = [];
   for (let i = 0; i < nums.length - 2; i++) {
       let j = i + 1;
       let k = nums.length - 1;
       while (j < k) {
           let sum = nums[i][0] + nums[j][0] + nums[k][0];
           if (sum === 0) {
               result.push(nums[i]);
               result.push(nums[j]);
               result.push(nums[k]);
               i++;
               j++;
               k--;
               while (j < k && nums[j][0] === nums[j - 1][0]) j++;
               while (j < k && nums[k][0] === nums[k + 1][0]) k--;
           } else if (sum < 0) {
               j++;
           } else {
               k--;
           }
       }
   }
   return result;
}

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Optimal Grid Path Protocol 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimal Grid Path Protocol 2 — 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