Optimal Grid Path Engine 5 — Problem Statement & Solution Guide
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
DSA Pattern Breakdown
"Optimal Grid Path Engine 5"
WHY DOES IT MATTER?
Two Pointers is essential for solving problems involving sorted arrays where you need to find pairs or triplets with specific sum properties. It is the standard optimal solution for 3Sum and its variations, which are frequent in system design and algorithmic interviews.
OPTIMIZATION CHALLENGE
The key insight is that sorting the array allows the use of directional movement. Without sorting, you cannot predict whether moving a pointer will increase or decrease the sum, forcing a brute-force search. Sorting enables the 'pruning' of search paths.
REAL-WORLD CONNECTION
This pattern is analogous to balancing a scale in a logistics center. If the weight on the left is too heavy, you move the lighter item from the right to the left; if it's too light, you move the heavier item from the right to the left. This systematic adjustment ensures balance (target sum) is found efficiently.
In interviews, always mention the trade-off: sorting takes O(N log N) time, but the two-pointer traversal is O(N^2). This is significantly better than the O(N^3) brute force. Also, explicitly state how you handle duplicates to show attention to detail.
COMPLEXITY AT A GLANCE
O(N^2)O(1)Core Theory — Why This Approach?
The problem leverages the Two Pointers technique, specifically the variant used for the 3Sum problem, to efficiently find triplets in a sorted array that sum to a target value (zero in this case). The core theoretical foundation relies on the monotonic property of sorted arrays: as one pointer moves right, the sum increases, and as it moves left, the sum decreases. This allows the algorithm to prune large portions of the search space without exhaustive checking, transforming a cubic complexity problem into a quadratic one.
Interview Questions on This Problem
Q1At a fintech platform, how would you adapt the 3Sum logic to find three transactions that net out to zero for fraud detection, while handling duplicate transaction IDs?
First, sort the transactions by amount. Use a fixed pointer for the first element, then use two pointers for the remaining subarray. To handle duplicates, skip identical values for the first pointer and adjust the left/right pointers inward when a valid triplet is found to avoid counting the same combination multiple times.
Q2In a high-growth startup's recommendation engine, how does the Two Pointers approach optimize memory usage compared to using a HashSet for 3Sum?
The Two Pointers approach requires O(1) extra space (excluding the output array) because it manipulates indices directly on the sorted array. In contrast, a HashSet approach requires O(N) space to store intermediate sums or elements, which can become a bottleneck in memory-constrained distributed systems.
Q3For a global product company's data pipeline, how would you modify the 3Sum algorithm to find triplets that sum to a non-zero target K?
The logic remains identical, but the condition for moving pointers changes. Instead of checking if sum == 0, check if sum == K. If sum < K, move the right pointer right to increase the sum; if sum > K, move the left pointer left to decrease the sum.
Examples
Input
[0, -1, 2, 3]
Output
2
Explanation: Step-by-step: with input [0, -1, 2, 3], we sort the array to get [-1, 0, 2, 3]. Then, we use two pointers to find all unique triplets that sum to zero. The triplets are [-1, 0, 3] and [2, -3, 1]. Therefore, the output is 2.
Input
[0, 0, 0]
Output
1
Explanation: Step-by-step: with input [0, 0, 0], we sort the array to get [0, 0, 0]. Then, we use two pointers to find all unique triplets that sum to zero. The triplet is [0, 0, 0]. Therefore, the output is 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 use a fixed pointer for the first element, then apply the two-pointer technique on the remaining subarray to find pairs that sum to the target. This reduces the time complexity to O(N^2) with O(1) extra space.
Brute Force Approach
Use three nested loops to check every possible triplet in the array. This results in a time complexity of O(N^3), which is too slow for large datasets.
Verified Code Solutions
function solution(nums) {
nums.sort((a, b) => a - b);
let count = 0;
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let sum = nums[i] + nums[left] + nums[right];
if (sum < 0) left++;
else if (sum > 0) right--;
else {
count++;
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
left++;
right--;
}
}
}
return count;
}class Solution {
public:
int solution(vector<int>& nums) {
sort(nums.begin(), nums.end());
int count = 0;
for (int i = 0; i < nums.size() - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum < 0) left++;
else if (sum > 0) right--;
else {
count++;
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
}
}
}
return count;
}
};class Solution {
public int solution(int[] nums) {
Arrays.sort(nums);
int count = 0;
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum < 0) left++;
else if (sum > 0) right--;
else {
count++;
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
}
}
}
return count;
}
}def solution(nums):
nums.sort()
count = 0
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left = i + 1
right = len(nums) - 1
while left < right:
sum = nums[i] + nums[left] + nums[right]
if sum < 0:
left += 1
elif sum > 0:
right -= 1
else:
count += 1
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
return countfunction solution(nums) {
nums.sort((a, b) => a - b);
let count = 0;
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1;
let right = nums.length - 1;
while (left < right) {
let sum = nums[i] + nums[left] + nums[right];
if (sum < 0) left++;
else if (sum > 0) right--;
else {
count++;
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
left++;
right--;
}
}
}
return count;
}Asked in Top Tech Interviews
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