LeetCode 3028 - Ant on the Boundary

LeetCode Problem 3028

Difficulty: 🟢 Easy
Topics: Array, Simulation, Prefix Sum

Solution

Problem Understanding

The problem describes an ant moving along a one-dimensional infinite line, starting at a boundary point, which we can consider as position 0. The ant reads through an array of non-zero integers, nums. Each integer represents a movement: positive numbers move the ant to the right and negative numbers move it to the left. After every movement corresponding to an element in nums, we check whether the ant exactly lands on the boundary. If it does, we count it as a return. Crossing the boundary without landing exactly on it does not count.

The input is an array of integers, nums, with lengths between 1 and 100, and each value between -10 and 10 (excluding 0). These constraints tell us that a direct simulation of the ant's movements is feasible, as the array is short and the numbers are small.

Important edge cases include movements that immediately return the ant to the boundary, sequences that never bring the ant back, and alternating left and right moves that sum to zero over multiple steps but never land exactly on the boundary in between.

The output is a single integer representing the number of times the ant returns to the boundary after completing a movement in nums.

Approaches

The brute-force approach is straightforward: simulate the ant's movements step by step. Start at position 0, iterate through the nums array, and update the ant's position according to the value of the current element. After each movement, check if the position is exactly zero, and if so, increment a counter. This works because the array and movement values are small, and simulating every move ensures correctness.

The optimal approach relies on the observation that the ant's position is simply the cumulative sum of all moves. At each step, adding nums[i] to a running sum represents the ant's current position. Whenever the sum equals zero, the ant has returned to the boundary. This eliminates the need for any complicated logic or data structures, as simple iteration suffices.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(1)	Simulate each movement, check boundary after each move
Optimal	O(n)	O(1)	Keep cumulative sum of movements, increment counter when sum equals zero

Both approaches have the same time and space complexity for this problem due to the small input size, but the optimal approach is simpler and more readable.

Algorithm Walkthrough

1. Initialize a variable position to 0 to represent the ant's current location on the line.
2. Initialize a variable count to 0 to track how many times the ant returns to the boundary.
3. Iterate through each element move in nums.
4. Update position by adding move. Positive moves increase position (right), negative moves decrease it (left).
5. After each movement, check if position equals 0. If it does, increment count by 1.
6. Continue until all elements in nums are processed.
7. Return count.

Why it works: The cumulative sum of movements correctly tracks the ant's position along the line. The invariant is that position always represents the exact location of the ant after each step. Checking if position equals zero ensures we only count exact returns to the boundary, not crossings.

Python Solution

from typing import List

class Solution:
    def returnToBoundaryCount(self, nums: List[int]) -> int:
        position = 0
        count = 0
        for move in nums:
            position += move
            if position == 0:
                count += 1
        return count

This implementation initializes position and count, then iterates over each move. The position variable tracks the ant's location, and each time it equals zero, count is incremented. Finally, count is returned, representing the total number of returns to the boundary.

Go Solution

func returnToBoundaryCount(nums []int) int {
    position := 0
    count := 0
    for _, move := range nums {
        position += move
        if position == 0 {
            count++
        }
    }
    return count
}

The Go solution mirrors the Python implementation. We use a range loop to iterate over nums, updating position and incrementing count whenever position equals zero. Go handles integer operations safely for the constraints given, and slices are used directly without additional memory allocation.

Worked Examples

Example 1

Input: [2, 3, -5]

Step	Move	Position	Boundary Return?	Count
1	2	0 + 2 = 2	No	0
2	3	2 + 3 = 5	No	0
3	-5	5 - 5 = 0	Yes	1

Output: 1

Example 2

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

Step	Move	Position	Boundary Return?	Count
1	3	0 + 3 = 3	No	0
2	2	3 + 2 = 5	No	0
3	-3	5 - 3 = 2	No	0
4	-4	2 - 4 = -2	No	0

Output: 0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Iterate through the array once, performing constant-time operations per element
Space	O(1)	Only two variables are used to track position and count, independent of input size

Since n <= 100, this is highly efficient and well within acceptable performance limits.

Test Cases

# Provided examples
assert Solution().returnToBoundaryCount([2,3,-5]) == 1  # returns once
assert Solution().returnToBoundaryCount([3,2,-3,-4]) == 0  # never returns

# Edge cases
assert Solution().returnToBoundaryCount([1, -1, 1, -1]) == 2  # alternates and returns twice
assert Solution().returnToBoundaryCount([-1, 1, -1, 1]) == 2  # starts left, returns twice
assert Solution().returnToBoundaryCount([10, -10]) == 1  # large single moves
assert Solution().returnToBoundaryCount([1]*100) == 0  # never returns
assert Solution().returnToBoundaryCount([-10]*10) == 0  # moves only left, never returns

Test	Why
[2,3,-5]	Simple case with single return
[3,2,-3,-4]	No return, confirms correct non-counting
[1,-1,1,-1]	Alternating moves, multiple returns
[10,-10]	Large steps, single return
[1]*100	Stresses upper bound, no return
[-10]*10	Moves only in one direction, boundary never reached

Edge Cases

Alternating moves: If the ant alternates moves that sum to zero over multiple steps, the implementation correctly counts only those steps where the cumulative sum exactly reaches zero. This prevents overcounting returns.

Single-step returns: If the first movement immediately returns the ant to the boundary (for example [-0] is invalid but [-1] or [1] followed by a compensating move), the algorithm handles it because the cumulative sum is checked after each move.

All moves in one direction: If all moves are positive or negative, the ant will never return to the boundary. The solution handles this gracefully by simply never incrementing the counter, returning zero.

This approach is robust, simple, and directly mirrors the problem's rules.