#3028

Easy

LeetCode Problem Workspace

Ant on the Boundary

Solve the problem of counting how often an ant returns to a boundary based on the steps described in the input array.

Problem Statement

You are given an array of integers, nums, where each integer represents a movement of an ant. The ant starts at position 0 and moves according to the value of each element in the array. Positive numbers move the ant to the right and negative numbers move it to the left.

Your task is to return the number of times the ant's position reaches the boundary (position 0) during its movement across the array.

Examples

Example 1

Input: nums = [2,3,-5]

Output: 1

After the first step, the ant is 2 steps to the right of the boundary. After the second step, the ant is 5 steps to the right of the boundary. After the third step, the ant is on the boundary. So the answer is 1.

Example 2

Input: nums = [3,2,-3,-4]

Output: 0

After the first step, the ant is 3 steps to the right of the boundary. After the second step, the ant is 5 steps to the right of the boundary. After the third step, the ant is 2 steps to the right of the boundary. After the fourth step, the ant is 2 steps to the left of the boundary. The ant never returned to the boundary, so the answer is 0.

Constraints

1 <= nums.length <= 100
-10 <= nums[i] <= 10
nums[i] != 0

Solution Approach

Simulation with Position Tracking

To solve this problem, you can simulate the ant's movement step by step. Initialize a variable to track the current position of the ant, starting at 0. For each number in the array, update the position by adding the number's value. Keep a count of how many times the ant returns to position 0.

Efficient Counting with Prefix Sum

Another approach involves using a prefix sum. Instead of updating the position after each element, maintain a running total. For every element, check if the prefix sum equals 0, which indicates that the ant has returned to the boundary.

Handling Edge Cases

Ensure you handle arrays with both positive and negative values properly. Also, consider the case when the array has only one element. You should account for all possibilities, especially when the ant never reaches the boundary or does so multiple times.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the input array, as we must process each element once. The space complexity is O(1), as we only need a few variables to track the position and count of boundary returns.

What Interviewers Usually Probe

Evaluating the candidate's ability to simulate processes step-by-step and handle edge cases.
Looking for clarity in explaining the logic behind tracking the ant's position.
Assessing the candidate's ability to optimize the solution using prefix sums or similar techniques.

Common Pitfalls or Variants

Common pitfalls

Forgetting to reset the boundary count when the ant returns to position 0 multiple times.
Overcomplicating the problem by using unnecessary data structures or algorithms.
Not properly handling edge cases like arrays with only one element or extreme values.

Follow-up variants

Handling arrays where all elements are positive or negative.
Returning the final position of the ant instead of counting boundary returns.
Simulating multiple ants with different movement patterns on the same array.

FAQ

What is the key approach to solving the 'Ant on the Boundary' problem?

The key approach is to simulate the movement of the ant and track its position after each step. Count how many times the ant returns to the boundary (position 0).

How do I handle large input arrays in the 'Ant on the Boundary' problem?

The problem has a time complexity of O(n), where n is the length of the array, so large arrays can be processed efficiently within the constraints.

What is the significance of prefix sums in the 'Ant on the Boundary' problem?

Prefix sums allow you to maintain a running total of the ant's position, which can be used to check if the ant has returned to the boundary after each move.

How can I handle edge cases like a single-element array in this problem?

For a single-element array, the ant either returns to the boundary on the first step or doesn't, depending on the sign of the value. Ensure the solution correctly handles this case.

Why is the space complexity O(1) in the 'Ant on the Boundary' problem?

The space complexity is O(1) because we only need a few variables to track the current position and the boundary return count, regardless of the input array size.

terminal

Solution

Solution 1: Prefix Sum

Based on the problem description, we only need to calculate how many zeros are in all prefix sums of `nums`.

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class Solution:
    def returnToBoundaryCount(self, nums: List[int]) -> int:
        return sum(s == 0 for s in accumulate(nums))

Continue Practicing

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