Is Ant on the Boundary easy or hard?

Ant on the Boundary is classified as an Easy problem on LeetCode with a high acceptance rate around 74%. The main challenge is recognizing that the ant's position is simply the prefix sum of the movement array.

Ant on the Boundary Python/Java solution

In both Python and Java, the solution iterates through the array while maintaining a running position variable. Each time the cumulative position becomes zero, a counter increases. The implementation is short, runs in O(n) time, and uses O(1) additional space.

How to solve Ant on the Boundary in O(n)?

Maintain a running variable representing the ant's current position. Iterate through the array, add each move to the position, and check if it equals zero after the update. Each step involves constant work, so the entire algorithm runs in O(n) time.

What is the best approach for Ant on the Boundary?

The best approach is cumulative position tracking using a running prefix sum. Iterate through the moves, update the current position, and count how many times it becomes zero. This solution runs in O(n) time and O(1) space, making it the most efficient and interview-friendly method.

Is Ant on the Boundary asked at Google/Amazon/Meta?

Problems involving prefix sums and running cumulative calculations frequently appear in interviews at companies like Amazon, Google, and Meta. While this exact problem may not always appear, the underlying technique—tracking prefix sums and detecting specific states—is commonly tested.

What data structure is used in Ant on the Boundary?

The problem primarily uses arrays and prefix sum logic. Most solutions maintain a single integer representing the cumulative position, though some implementations store prefix positions in an additional array for analysis.

What is the time complexity of Ant on the Boundary?

The optimal solution runs in O(n) time because you process each move exactly once. Space complexity can be O(1) if you maintain only a running position, or O(n) if you store all prefix positions in an auxiliary array.

Description Hints 1 Solution Video 3:58

3028. Ant on the Boundary

Easy74.6% acceptance3 companies

An ant is on a boundary. It sometimes goes left and sometimes right.

You are given an array of non-zero integers nums. The ant starts reading nums from the first element of it to its end. At each step, it moves according to the value of the current element:

If nums[i] < 0, it moves left by -nums[i] units.
If nums[i] > 0, it moves right by nums[i] units.

Return the number of times the ant returns to the boundary.

Notes:

There is an infinite space on both sides of the boundary.
We check whether the ant is on the boundary only after it has moved |nums[i]| units. In other words, if the ant crosses the boundary during its movement, it does not count.

Example 1:

Input: nums = [2,3,-5]
Output: 1
Explanation: 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
Explanation: 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

Asked at 3 companies

Accenture

Google Amazon

Topics

Hints open one at a time

Hint 1

Define a variable and add nums[i] to it in each step.

3:58

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

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1class Solution:
2    def returnToBoundaryCount(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[2,3,-5]

Input

[3,2,-3,-4]

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